Added(A)/Deleted(D) following books

A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/README.txt A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch2_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch3_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch4_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch5_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch6_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch7_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch8_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/ch9_1.ipynb A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch2_1.png A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/ch7_1.png A A_Course_In_Mechanical_Measurements_And_Instrumentation_by_A._K._Sawhney_And_P._Sawhney/screenshots/kVSv5.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER2_1.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER4_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER5_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER6_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/CHAPTER7_2.ipynb A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/README.txt A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP2.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP4.png A A_First_Course_on_Electrical_Drives_by_S._K._Pillai/screenshots/CHAP5.png D A_First_course_in_Programming_with_C/Chapter14.ipynb M A_First_course_in_Programming_with_C_by_T_Jeyapoovan/Chapter14_2.ipynb A A_First_course_in_Programming_with_C_by_T_Jeyapoovan/README.txt A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1.7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT1_7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.10.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.11.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.13.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.14.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.15.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.16.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.17.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.18.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.2.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.3.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.4.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.5.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.6.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.7.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.8.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2.9.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_10.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_11.ipynb A A_Textbook_on_Power_System_Engineering_by_A_Chakrabarti,_M_L_Soni,_P_V_Gupta,_U_S_Bhatnagar/CHAPT2_13.ipynb A 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Internal_Combustion_Engines_by_H._B._Keswani/screenshots/ch9.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter10_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter11_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter12_3.ipynb A 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Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter3_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter4_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter5_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter6_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter7_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter8_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_1.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_2.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_3.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/Chapter9_4.ipynb A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/README.txt A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter10_4.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter3_4.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_1.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_2.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_3.png A Introduction_To_Fluid_Mechanics_by_R._W._Fox_And_A._T._McDonald/screenshots/chapter4_4.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter1.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter2(PartB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter2.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter3(partB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter3.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter4(PartB).ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter4.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter5.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter6.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/Chapter7.ipynb A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter1.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter2.png A Introduction_To_Mechanical_Engineering_by_S._Chandra_And_O._Singh/screenshots/chapter3.png A Introduction_to_Electric_Drives_by_J._S._Katre/AppendixB.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/AppendixB_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter10_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter1_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter2_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter3_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter5_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter6_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter8_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9_1.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/chapter9_2.ipynb A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_VLdc_VLrms_2.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch6_variation_of_RF_FF_2.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V_1.png A Introduction_to_Electric_Drives_by_J._S._Katre/screenshots/ch_3_variation_avg_rms_load_V_2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex1.2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex3.13.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex6.7.png A Linear_Integrated_Circuits_by_J._B._Gupta/README.txt A Linear_Integrated_Circuits_by_J._B._Gupta/chapter01_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter01_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter02_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter02_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter03_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter03_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter04_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter04_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter05_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter05_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter06_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter06_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter07_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter07_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter08_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter08_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter09_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter09_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter10_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter10_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter11_1.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/chapter11_2.ipynb A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_14.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_14_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_15.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/5_15_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/per_error_1.png A Linear_Integrated_Circuits_by_J._B._Gupta/screenshots/per_error_2.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER10.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER13.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER14.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER15.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER16.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER18.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER19.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER2.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER21.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER23.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER24_.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER26.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER30.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER31.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER33.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER36.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/CHAPTER9.ipynb A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP10.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP16.png A Manufacturing_Engineering_&_Technology_by__S._Kalpakjian_and_S._R._Schmid/screenshots/CHAP19.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch2_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch3_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch4_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch5_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch6_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/ch7_1.ipynb A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/FricCoeff.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/FricCoeff_1.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/fillingtime.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/fillingtime_1.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/millPOwer.png A Manufacturing_Science_by_A._Ghosh_And_A._K._Mallik/screenshots/millPOwer_1.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter10.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter11.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter12.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter13.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter14.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter15.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter16.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter17.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter2.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter3.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter4.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter5.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter6.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter7.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter8.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/Chapter9.ipynb A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter10.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter11.png A Materials_Science_and_Engineering_-_A_First_Course_by_V._Raghavan/screenshots/Chapter12.png A Materials_Science_by_Dr._M._Arumugam/Chapter10_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter12_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter1_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter2_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter3_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter4_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter5_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter6_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter7_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter8_1.ipynb A Materials_Science_by_Dr._M._Arumugam/Chapter9_1.ipynb A Materials_Science_by_Dr._M._Arumugam/README.txt A Materials_Science_by_Dr._M._Arumugam/screenshots/11.png A Materials_Science_by_Dr._M._Arumugam/screenshots/22.png A Materials_Science_by_Dr._M._Arumugam/screenshots/33.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_2_Generalized_Configurations_and_Functional_Descriptions_of_Measuring_Instruments.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_3_Generalized_Performance_Characteristics_of_Instruments.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_4_Relative_Velocity_Translational_and_Rotational.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_5_Force_Torque_and_Shaft_power_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_6_Pressure_and_Sound_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_7_Flow_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/Chapter_8_Temperature_and_Heat-Flux_Measurement.ipynb A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha3.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha4.png A Measurement_Systems_by_E._O._Doebelin_And_D._N._Manik/screenshots/cha5.png A Mechanical_Metallurgy_by_George_E._Dieter/README.txt A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/AppendixA_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_10.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_11.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_12.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_13.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_14.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_4.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_5.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_6.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_7.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_8.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter01_9.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter02_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter03_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter04_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter05_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter06_3.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_1.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_2.ipynb A Mechanics_of_Materials_by_Pytel_and_Kiusalaas/Chapter07_3.ipynb A 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Special_Electrical_Machines_by_S.P._Burman/chapter03.ipynb A Special_Electrical_Machines_by_S.P._Burman/chapter04.ipynb A Special_Electrical_Machines_by_S.P._Burman/screenshots/ResolShaftSpeed3.png A Special_Electrical_Machines_by_S.P._Burman/screenshots/TorqLossEff1.png A Special_Electrical_Machines_by_S.P._Burman/screenshots/Torq_Speed1.png A Strength_Of_Materials_by_B_K_Sarkar/Chapter01.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter02.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter03.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter04.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter05.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter06.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter07.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter08.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter09.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter10.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter11.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter12.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter13.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter14.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter15.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter16.ipynb A Strength_Of_Materials_by_B_K_Sarkar/Chapter17.ipynb A Strength_Of_Materials_by_B_K_Sarkar/README.txt A Strength_Of_Materials_by_B_K_Sarkar/chapter_10_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_10_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_11_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_12_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_13_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_14_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_15_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_16_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_17_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_1_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_2_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_3_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_4_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_5_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_6_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_7_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_8_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som.ipynb A Strength_Of_Materials_by_B_K_Sarkar/chapter_9_som_1.ipynb A Strength_Of_Materials_by_B_K_Sarkar/screenshots/B.M.D_1.JPG A Strength_Of_Materials_by_B_K_Sarkar/screenshots/B.M.D_2.JPG A Strength_Of_Materials_by_B_K_Sarkar/screenshots/BMD2.png A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_1.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_1_1.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_2.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/S.F.D_4.jpg A Strength_Of_Materials_by_B_K_Sarkar/screenshots/SFD.png A Strength_Of_Materials_by_B_K_Sarkar/screenshots/SFD3.png M The_C_Book/Chapter2.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/README.txt R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch10_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch2_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch3_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch4_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch5_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch6_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch7_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch8_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9.ipynb -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9.ipynb A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/ch9_1.ipynb R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same3_7_1.png R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/same7_1.png R _Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7.png -> Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7.png A Vector_Mechanics_for_Engineers:_Stastics_And_Dynamics_by_F._P._Beer,_E._R._Johnston,_D._F._Mazurek,_P._J._Cornwell_And_E._R._Eisenberg/screenshots/shearAndBendingMoment7_1.png A Wireless_Communications_and_Networking_by_V._Garg/README.txt A Wireless_Communications_and_Networking_by_V._Garg/ch10_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch11_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch12_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch13_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch14_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch17_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch19_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch21_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch2_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch3_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch4_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch5_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch6_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch8_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/ch9_1.ipynb A Wireless_Communications_and_Networking_by_V._Garg/screenshots/EbbyNo_1.png A Wireless_Communications_and_Networking_by_V._Garg/screenshots/comparision_of_models_1.png A Wireless_Communications_and_Networking_by_V._Garg/screenshots/multiplexing_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/README.txt A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch10_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch11_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch12_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch1_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch2_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch3_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch4_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch5_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch6_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch7_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch8_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/ch9_1.ipynb A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/energy_stored3_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/magnetic_flux12_1.png A _Electric_Machinery_And_Transformers_by_B._S._Guru_And_H._R._Hiziroglu/screenshots/pitch_factor7_1.png A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_10_Solid_Solutions_and_Phase_Equilibrium_3.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_11_Dispertion_Strengthening_and_Eutectic_Phase_Diagrams_3.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_2.ipynb A _Essentials_of_Materials_Science_and_Engineering_by__D._R._Askeland_and_P._P._Phule/Chapter_12_Dispersion_Strengthening__by_Phase_Transmission_and_Heat_Treatment_3.ipynb A 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author: FOSSEE SysAds 2015-12-08 15:04:13 +0600
committer: FOSSEE SysAds 2015-12-08 15:04:13 +0600
commit: af2e802cfaf390cbc43b476ec9f0c73c7e651d9f (patch)
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Operational Amplifier Fundamentals"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.1, Page 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "%matplotlib inline"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "The overall gain is  60.0 V/V\n",
+      "The input load is  80.0 % of it's unloaded value\n",
+      "The output load is  75.0 % of it's unloaded value\n",
+      "b)\n",
+      "The overall gain is  53.3 V/V\n",
+      "The input load is  66.7 % of it's unloaded value\n",
+      "The output load is  80.0 % of it's unloaded value\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable declaration\n",
+    "R0 = 1.0       #ohm\n",
+    "Ri = 100.0     #kilo ohm\n",
+    "Aoc = 100.0    #volts per volts\n",
+    "Rs=0.0             #kilo ohm\n",
+    "Rl=0.0             #ohm\n",
+    "gain=0.0\n",
+    "input_load=0.0\n",
+    "output_load=0.0\n",
+    "\n",
+    "def calculate():                       #returns gain\n",
+    "    global input_load, output_load\n",
+    "    input_load = (Ri/(Rs+Ri))\n",
+    "    output_load = (Rl/(R0+Rl))\n",
+    "    ans=input_load*Aoc*output_load     # in V/V\n",
+    "    return ans\n",
+    "#answer part (a)\n",
+    "Rs=25.0\n",
+    "Rl=3.0\n",
+    "gain=calculate()\n",
+    "print \"a)\"\n",
+    "print \"The overall gain is \",round(gain,1),\"V/V\"\n",
+    "print \"The input load is \",input_load*100,\"% of it's unloaded value\"\n",
+    "print \"The output load is \",output_load*100,\"% of it's unloaded value\"\n",
+    "\n",
+    "#answer part (b)\n",
+    "Rs=50.0\n",
+    "Rl=4.0\n",
+    "gain=calculate()\n",
+    "print \"b)\"\n",
+    "print \"The overall gain is \",round(gain,1),\"V/V\"\n",
+    "print \"The input load is \",round(input_load*100,1),\"% of it's unloaded value\"\n",
+    "print \"The output load is \",round(output_load*100,1),\"% of it's unloaded value\"\n",
+    "             \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.2, Page 9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)Vo =  9.17431 V\n",
+      "b)Vo =  9.99101 V\n",
+      "c)Vo =  9.99991 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable declaration\n",
+    "\n",
+    "vt = 1.0     # in volt\n",
+    "R1 = 2.0     # in kilo ohm\n",
+    "R2 = 18.0    #in kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate(a):               #returns Vo\n",
+    "    global vt,R1,R2\n",
+    "    ans=vt*(1+(R2/R1))/(1+((R2/R1)/a))         #equation 1.11\n",
+    "    return ans\n",
+    "\n",
+    "#answer\n",
+    "print \"a)Vo = \",round(calculate(10**2),5),\"V\"\n",
+    "print \"b)Vo = \",round(calculate(10**4),5),\"V\"\n",
+    "print \"c)Vo = \",round(calculate(10**6),5),\"V\"\n",
+    "\n",
+    "#textbook contains precision error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.4, Page 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  20 kilo ohm\n",
+      "R2 =  15 kilo ohm\n",
+      "R3 =  30 kilo ohm\n",
+      "Rf =  120 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "rf1 = 3     # coefficient of V1\n",
+    "rf2 = 4     # coefficient of V2\n",
+    "rf3 = 2     # coefficient of V3\n",
+    "\n",
+    "#Calculations\n",
+    "\n",
+    "rf1*=2      # Common factor 2\n",
+    "rf2*=2      # Common factor 2\n",
+    "rf3*=2      # Common factor 2\n",
+    "r1=20       # assumption\n",
+    "rf=r1*rf1\n",
+    "r2=rf/rf2\n",
+    "r3=rf/rf3\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"R3 = \",r3,\"kilo ohm\"\n",
+    "print \"Rf = \",rf,\"kilo ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.5, Page 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  10 kilo ohm\n",
+      "R2 =  300 kilo ohm\n",
+      "Rf =  100 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "r1,r2,rf          #vo=10*v1+5=-(rf/r1*v1)-rf/r2*(-15)\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "r1=10\n",
+    "rf=10*r1;       #-rf/r1*v1=10*v1\n",
+    "r2=rf*15/5      #-rf/r2*(-15)=5\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"Rf = \",rf,\"kilo ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.6, Page 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  100 kilo ohm\n",
+      "R2 =  300 kilo ohm\n",
+      "R3 =  25 kilo ohm\n",
+      "R4 =  75 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "ri1=100       # in kilo ohm\n",
+    "ri2=100      # in kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "r1=ri1;\n",
+    "r2=3*r1;    #r2/r1=3\n",
+    "# r3 + r4 = ri2 and (1+r1/r2)/(1+r3/r4)=1\n",
+    "#Solving the above two\n",
+    "r3=ri2/4;\n",
+    "r4=ri2-r3\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"R3 = \",r3,\"kilo ohm\"\n",
+    "print \"R4 = \",r4,\"kilo ohm\"\n",
+    "\n",
+    "#in textbook r3 and r4 values are reversed which doesn't satisfy the equations"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.7, Page 25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)T >=  1000\n",
+      "b)a >=  100000\n",
+      "a)Beta =  0.00999\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable Declaration\n",
+    "\n",
+    "A=100 \n",
+    "accuracy=0.1\n",
+    "\n",
+    "#Calcualtion\n",
+    "\n",
+    "T=100/accuracy\n",
+    "beta=1.0/100.0   # A_ideal=i/beta=100\n",
+    "a=(10**3)/beta\n",
+    "beta=(a/100-1)/a    # A=a/(1+(a*beta))\n",
+    "\n",
+    "#answer\n",
+    "print \"a)T >= \",int(T)\n",
+    "print \"b)a >= \",int(a)\n",
+    "print \"a)Beta = \",beta"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.8, Page 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a) A changes by (+-) 0.09901 %\n",
+      "b) A changes by (+-) 0.0001 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "a = 10**5       \n",
+    "beta\n",
+    "T\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate():\n",
+    "    global a,beta,T\n",
+    "    T=a*beta\n",
+    "    ans=10.0/(1+T)    # for a +- 10% change in a\n",
+    "    return ans\n",
+    "\n",
+    "#answer\n",
+    "beta=10**(-3)                                                    #given\n",
+    "desensitivity_factor=calculate();                                # stores the answer\n",
+    "print \"a) A changes by (+-)\",round(desensitivity_factor,6),\"%\"   #part a\n",
+    "\n",
+    "beta=1                                                           #given\n",
+    "desensitivity_factor=calculate();\n",
+    "print \"b) A changes by (+-)\",round(desensitivity_factor,6),\"%\"   #part b"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.9, Page 33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  A =  995.024876 V/V\n",
+      "  Ro =  373.134 mili ohm\n",
+      "  Ri =  402.0 Mega ohm\n",
+      "b)\n",
+      "  A =  0.999995 V/V\n",
+      "  Ro =  0.375 mili ohm\n",
+      "  Ri =  400002.0 Mega ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 2.0    # Mega ohm\n",
+    "ro = 75.0   # ohm\n",
+    "a = 200000.0   # V/V\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate(R1,R2):\n",
+    "    global a,ro,rd\n",
+    "    beta=R1/(R1+R2)\n",
+    "    if(R1==float(\"inf\")):      # for infinty\n",
+    "        beta=1\n",
+    "    T=a*beta\n",
+    "    A=(1+(R2/R1))/(1+(1/T))    # equation 1.55\n",
+    "    if(R1==float(\"inf\")):      # for infinity\n",
+    "        A=1/(1+(1/T))\n",
+    "    Ro=ro/(1+T)                # equation 1.61\n",
+    "    Ri=rd*(1+T)                # equation 1.59\n",
+    "    print \"  A = \",round(A,6),\"V/V\"\n",
+    "    print \"  Ro = \",round(Ro*(10**3),3),\"mili ohm\"\n",
+    "    print \"  Ri = \", round(Ri,3),\"Mega ohm\"\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "calculate(1.0,999)\n",
+    "print \"b)\"\n",
+    "calculate(float(\"inf\"),1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.10, Page 35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      " A =  -0.99999 V/V\n",
+      " Rn =  0.5 ohm\n",
+      " Ri =  100000.0 ohm\n",
+      " Ro =  0.00075 ohm\n",
+      "b)\n",
+      " A =  -995.01993 V/V\n",
+      " Rn =  4.99998 ohm\n",
+      " Ri =  1000.0 ohm\n",
+      " Ro =  0.37351 ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "a = 200000.0    # V/V\n",
+    "ro = 75         # ohm\n",
+    "\n",
+    "#Calculating function\n",
+    "\n",
+    "def calculate(R1,R2):\n",
+    "    global a,ro\n",
+    "    T=a*(R1/(R1+R2)) \n",
+    "    A=(-1)*(R2/R1)/(1+(1/T))   # equation 1.63\n",
+    "    Rn=R2/(1+a)               # equation 1.67b\n",
+    "    Ri=R1                     # equation 1.68\n",
+    "    Ro=ro/(1+T)\n",
+    "    print \" A = \",round(A,5),\"V/V\"\n",
+    "    print \" Rn = \",round(Rn,5),\"ohm\"\n",
+    "    print \" Ri = \",round(Ri,5),\"ohm\"\n",
+    "    print \" Ro = \",round(Ro,5),\"ohm\"\n",
+    "                    \n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "calculate(100000.0,100000.0)\n",
+    "print \"b)\"\n",
+    "calculate(1000.0,1000000.0)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.11, Page 38"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "317.440162529\n",
+      "a) A_ideal = -101.1 V/V\n",
+      "b) A = -100.78 V/V\n",
+      "Deviation from ideal = 0.31 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "R1 = 1000000.0     # ohm\n",
+    "R2 = 1000000.0     # ohm\n",
+    "R3 = 100000.0      # ohm\n",
+    "R4 = 1000.0        # ohm\n",
+    "RL = 2000.0        # ohm\n",
+    "rd = 1000000.0     #ohm\n",
+    "a = 10**5        # V/V\n",
+    "ro = 100.0         # ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "A_ideal = (-1)*(R2/R1)*(1+(R3/R2)+(R3/R4))       # ideal op-amp and summing  currents at node  v1\n",
+    "T = a/(1+(R2/R1)+(R2/rd))/(1+(ro/(R2+(R1*rd/(R1+rd))))+(ro/RL))/100  #equation 1.73\n",
+    "A = A_ideal/(1+(1/T))                   \n",
+    "dev=(A_ideal-A)/A_ideal*100\n",
+    "\n",
+    "#answer\n",
+    "print T\n",
+    "print \"a) A_ideal =\",A_ideal,\"V/V\"\n",
+    "print \"b) A =\",round(A,2),\"V/V\"\n",
+    "print \"Deviation from ideal =\",round(dev,2),\"%\"\n",
+    "\n",
+    "#book example has precision error so answer is 0.32%"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.12, Page 40"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  Beta = 0.016911 V/V\n",
+      "  T = 169.1\n",
+      "b)\n",
+      "  Vo= -( 29.82 V1 + 14.91 V2 + 9.94 V3 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 1000.0          # kilo ohm\n",
+    "a = 10**4            # V/V\n",
+    "ro = 100.0           #ohm\n",
+    "R1 = 10.0            # kilo ohm\n",
+    "R2 = 20.0            # kilo ohm\n",
+    "R3 = 30.0            # kilo ohm\n",
+    "R4 = 300.0           # kilo ohm\n",
+    "RL = 2.0             # kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def parallel(a,b):\n",
+    "    ans=a*b/(a+b)\n",
+    "    return ans\n",
+    "\n",
+    "Ra = parallel(R1,parallel(R2,parallel(R3,rd)))\n",
+    "Rb=Ra+R4\n",
+    "Rc=parallel(Rb,RL)                      #After suppressing all input sources\n",
+    "Rd=Rc+ro/1000                           #replacing the op-amp with it's terminal resistances\n",
+    "Vn=Rb/Ra                                #and applying a test voltage and analysing the circuit\n",
+    "Vt=Rd/Rc\n",
+    "beta=1/Vn/Vt\n",
+    "T=a*beta\n",
+    "v1=R4/R1\n",
+    "v2=R4/R2\n",
+    "v3=R4/R3\n",
+    "A=1/(1+1/T)\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "print \"  Beta =\",round(beta,6),\"V/V\"\n",
+    "print \"  T =\",round(T,1)\n",
+    "print \"b)\"\n",
+    "print \"  Vo= -(\",round(A*v1,2),\"V1 +\",round(A*v2,2),\"V2 +\",round(A*v3,2),\"V3 )\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.13, Page 41"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Beta = 0.8101 V/V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 100.0        # kilo ohm\n",
+    "ro = 100.0        # ohm\n",
+    "R1 = 30.0         # kilo ohm\n",
+    "R2 = 20.0         # kilo ohm\n",
+    "R3 = 10.0         # kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def parallel(a,b):\n",
+    "    ans=a*b/(a+b)\n",
+    "    return ans\n",
+    "\n",
+    "beta_n = (parallel(R1,rd)+R1)/((ro/1000)+R2+parallel(R1,rd)+R3)    # from circuit 1.35 after appyling\n",
+    "beta_p = R3/((ro/1000)+R2+parallel(R1,rd)+R3)                      # voltage divide formula twice\n",
+    "beta=beta_n-beta_p                  #equation 1.76\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"Beta =\",round(beta,4),\"V/V\"\n",
+    "\n",
+    "# beta_n calculation in book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.14, Page 43"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  Icc = 0.5 mA\n",
+      "  Iee = 3.5 mA\n",
+      "  I0 = 3 mA\n",
+      "b)\n",
+      "  Power Poa = 42.0 mW\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration \n",
+    "\n",
+    "R1 = 10     #kilo ohm\n",
+    "R2 = 20     #kilo ohm\n",
+    "V1 = 3      # V\n",
+    "Iq = 0.5    # mA\n",
+    "RL = 2      #kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "V0 = (-1)*R2/R1*V1\n",
+    "It = abs(V0)/RL                     # Currents through R1,R2,Rt are i1,i2,It respectively\n",
+    "i1 = It/R1\n",
+    "i2 = i1                             # applying voltage divider rule\n",
+    "i0 = i2+It\n",
+    "icc = Iq\n",
+    "iee = icc+ i0\n",
+    "Poa = 30*Iq+((V0+15)*i0)            #Whenever current passes through voltage drop, power = vi\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "print \"  Icc =\",icc,\"mA\"\n",
+    "print \"  Iee =\",iee,\"mA\"\n",
+    "print \"  I0 =\",i0,\"mA\"\n",
+    "print \"b)\"\n",
+    "print \"  Power Poa =\",Poa,\"mW\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.15, Page 43"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "b)\n",
+      "  Change in v = 3.75 micro Volt -> quite a small change\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable Declaration\n",
+    "\n",
+    "ro = 75.0      #kilo ohm\n",
+    "T = 200000.0\n",
+    "Vs = 10.0      # V\n",
+    "Rl = 1.0       #kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "iL = Vs/Rl\n",
+    "Ro = ro/(1+T)\n",
+    "del_v = Ro*10*(10**(-3))\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"b)\"\n",
+    "print \"  Change in v =\",round(del_v*(10**6),2),\"micro Volt -> quite a small change\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.16, Page 46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The op-amp saturates at Vo=+-13 V\n",
+      "With Vn= 20/3-13/3 = 2.3333 V\n"
+     ]
+    },
+    {
+     "data": {
+      "text/plain": [
+       "[<matplotlib.lines.Line2D at 0xa087c88>]"
+      ]
+     },
+     "execution_count": 16,
+     "metadata": {},
+     "output_type": "execute_result"
+    },
+    {
+     "data": {
+      "image/png": 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AxIlhz4bvfleVRGt07Ag33BCmDP/kJ2Hgf9mypKMSqR65ueQaGzcsl1wqKwqgMzDRzKYD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\niEhBqihERKQgVRQiIlLQ/wfqhyu+stXumgAAAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x9e977f0>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "import matplotlib.pyplot as plt\n",
+    "import scipy as np\n",
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "A = -2               #V/V\n",
+    "peak = 10           # V\n",
+    "\n",
+    "#Calculation \n",
+    "\n",
+    "output = np.absolute(A) * peak\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"The op-amp saturates at Vo=+-13 V\"\n",
+    "print \"With Vn= 20/3-13/3 =\",round((20.0/3)-(13.0/3),4),\"V\"\n",
+    "\n",
+    "#Graphs\n",
+    "\n",
+    "t1 = np.arange(0,1,.0005)               #  Triangular waveform\n",
+    "t2 = np.arange(1,3,.0005)\n",
+    "t3 = np.arange(3,5,.0005)\n",
+    "\n",
+    "m1 = np.arange(0,0.65,.0005)\n",
+    "m2 = np.arange(.65,1.35,.0005)\n",
+    "m3 = np.arange(1.35,2.65,.0005)            # Output Vo wave\n",
+    "m4 = np.arange(2.65,3.35,.0005)\n",
+    "m5 = np.arange(3.35,4.65,.0005)\n",
+    "m6 = np.arange(4.65,5,.0005)                # Output Vn wave\n",
+    "m7 = np.arange(0.65,1,.0005)\n",
+    "m8 = np.arange(1,1.35,.0005)\n",
+    "m9 = np.arange(2.65,3,.0005)\n",
+    "m10 = np.arange(3, 3.35, .0005)\n",
+    "\n",
+    "plt.subplot(2,1,1)\n",
+    "\n",
+    "plt.suptitle(\"Vt (Blue), Vo (Red) and Vn (Green) Graphs\")\n",
+    "plt.xlim(0,4.5)\n",
+    "plt.xlabel(\"time->\")\n",
+    "plt.ylabel(\"V->\")\n",
+    "plt.plot(t1,peak*t1,\"b\",)\n",
+    "plt.plot(t2,(-1)*peak*t2+2*(peak),\"b\",)\n",
+    "plt.plot(t3,peak*t3-4*(peak),\"b\",)\n",
+    "\n",
+    "plt.subplot(2,1,2)\n",
+    "\n",
+    "plt.xlim(0,4.5)\n",
+    "plt.xlabel(\"time->\")\n",
+    "plt.ylabel(\"V->\")\n",
+    "plt.plot(m1,-20*m1,\"r\")\n",
+    "plt.plot(m2,np.full(len(m2),-13),\"r\")\n",
+    "plt.plot(m3,20*m3-40,\"r\")\n",
+    "plt.plot(m4,np.full(len(m4),13),\"r\")\n",
+    "plt.plot(m5,-20*m5+80,\"r\")\n",
+    "plt.plot(m6,np.full(len(m6),-13),\"r\")\n",
+    "\n",
+    "plt.plot(m1,np.full(len(m1),0),\"g\",)\n",
+    "plt.plot(m7,6.665*m7-4.4,\"g\")\n",
+    "plt.plot(m8,-6.665*m8+8.8,\"g\")\n",
+    "plt.plot(m3,np.full(len(m3),0),\"g\")\n",
+    "plt.plot(m9,6.665*m9-17.6,\"g\")\n",
+    "plt.plot(m10,-6.665*m10+22.4,\"g\")\n",
+    "plt.plot(m5,np.full(len(m5),0),\"g\")\n",
+    "plt.plot(m6,np.full(len(m6),0),\"g\")"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/ARIJITCHATTERJEE/chapter1_1.ipynb b/sample_notebooks/ARIJITCHATTERJEE/chapter1_1.ipynb
new file mode 100755
index 00000000..e66c292c
--- /dev/null
+++ b/sample_notebooks/ARIJITCHATTERJEE/chapter1_1.ipynb
@@ -0,0 +1,895 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Operational Amplifier Fundamentals"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.1, Page 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "The overall gain is  60.0 V/V\n",
+      "The input load is  80.0 % of it's unloaded value\n",
+      "The output load is  75.0 % of it's unloaded value\n",
+      "b)\n",
+      "The overall gain is  53.3 V/V\n",
+      "The input load is  66.7 % of it's unloaded value\n",
+      "The output load is  80.0 % of it's unloaded value\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable declaration\n",
+    "R0 = 1.0       #ohm\n",
+    "Ri = 100.0     #kilo ohm\n",
+    "Aoc = 100.0    #volts per volts\n",
+    "Rs=0.0             #kilo ohm\n",
+    "Rl=0.0             #ohm\n",
+    "gain=0.0\n",
+    "input_load=0.0\n",
+    "output_load=0.0\n",
+    "\n",
+    "def calculate():                       #returns gain\n",
+    "    global input_load, output_load\n",
+    "    input_load = (Ri/(Rs+Ri))\n",
+    "    output_load = (Rl/(R0+Rl))\n",
+    "    ans=input_load*Aoc*output_load     # in V/V\n",
+    "    return ans\n",
+    "#answer part (a)\n",
+    "Rs=25.0\n",
+    "Rl=3.0\n",
+    "gain=calculate()\n",
+    "print \"a)\"\n",
+    "print \"The overall gain is \",round(gain,1),\"V/V\"\n",
+    "print \"The input load is \",input_load*100,\"% of it's unloaded value\"\n",
+    "print \"The output load is \",output_load*100,\"% of it's unloaded value\"\n",
+    "\n",
+    "#answer part (b)\n",
+    "Rs=50.0\n",
+    "Rl=4.0\n",
+    "gain=calculate()\n",
+    "print \"b)\"\n",
+    "print \"The overall gain is \",round(gain,1),\"V/V\"\n",
+    "print \"The input load is \",round(input_load*100,1),\"% of it's unloaded value\"\n",
+    "print \"The output load is \",round(output_load*100,1),\"% of it's unloaded value\"\n",
+    "             \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.2, Page 9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)Vo =  9.17431 V\n",
+      "b)Vo =  9.99101 V\n",
+      "c)Vo =  9.99991 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable declaration\n",
+    "\n",
+    "vt = 1.0     # in volt\n",
+    "R1 = 2.0     # in kilo ohm\n",
+    "R2 = 18.0    #in kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate(a):               #returns Vo\n",
+    "    global vt,R1,R2\n",
+    "    ans=vt*(1+(R2/R1))/(1+((R2/R1)/a))         #equation 1.11\n",
+    "    return ans\n",
+    "\n",
+    "#answer\n",
+    "print \"a)Vo = \",round(calculate(10**2),5),\"V\"\n",
+    "print \"b)Vo = \",round(calculate(10**4),5),\"V\"\n",
+    "print \"c)Vo = \",round(calculate(10**6),5),\"V\"\n",
+    "\n",
+    "#textbook contains precision error"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.4, Page 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  20 kilo ohm\n",
+      "R2 =  15 kilo ohm\n",
+      "R3 =  30 kilo ohm\n",
+      "Rf =  120 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "rf1 = 3     # coefficient of V1\n",
+    "rf2 = 4     # coefficient of V2\n",
+    "rf3 = 2     # coefficient of V3\n",
+    "\n",
+    "#Calculations\n",
+    "\n",
+    "rf1*=2      # Common factor 2\n",
+    "rf2*=2      # Common factor 2\n",
+    "rf3*=2      # Common factor 2\n",
+    "r1=20       # assumption\n",
+    "rf=r1*rf1\n",
+    "r2=rf/rf2\n",
+    "r3=rf/rf3\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"R3 = \",r3,\"kilo ohm\"\n",
+    "print \"Rf = \",rf,\"kilo ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.5, Page 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  10 kilo ohm\n",
+      "R2 =  300 kilo ohm\n",
+      "Rf =  100 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "r1,r2,rf          #vo=10*v1+5=-(rf/r1*v1)-rf/r2*(-15)\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "r1=10\n",
+    "rf=10*r1;       #-rf/r1*v1=10*v1\n",
+    "r2=rf*15/5      #-rf/r2*(-15)=5\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"Rf = \",rf,\"kilo ohm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.6, Page 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1 =  100 kilo ohm\n",
+      "R2 =  300 kilo ohm\n",
+      "R3 =  25 kilo ohm\n",
+      "R4 =  75 kilo ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable declaration\n",
+    "\n",
+    "ri1=100       # in kilo ohm\n",
+    "ri2=100      # in kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "r1=ri1;\n",
+    "r2=3*r1;    #r2/r1=3\n",
+    "# r3 + r4 = ri2 and (1+r1/r2)/(1+r3/r4)=1\n",
+    "#Solving the above two\n",
+    "r3=ri2/4;\n",
+    "r4=ri2-r3\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"R1 = \",r1,\"kilo ohm\"\n",
+    "print \"R2 = \",r2,\"kilo ohm\"\n",
+    "print \"R3 = \",r3,\"kilo ohm\"\n",
+    "print \"R4 = \",r4,\"kilo ohm\"\n",
+    "\n",
+    "#in textbook r3 and r4 values are reversed which doesn't satisfy the equations"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.7, Page 25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)T >=  1000\n",
+      "b)a >=  100000\n",
+      "a)Beta =  0.00999\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable Declaration\n",
+    "\n",
+    "A=100 \n",
+    "accuracy=0.1\n",
+    "\n",
+    "#Calcualtion\n",
+    "\n",
+    "T=100/accuracy\n",
+    "beta=1.0/100.0   # A_ideal=i/beta=100\n",
+    "a=(10**3)/beta\n",
+    "beta=(a/100-1)/a    # A=a/(1+(a*beta))\n",
+    "\n",
+    "#answer\n",
+    "print \"a)T >= \",int(T)\n",
+    "print \"b)a >= \",int(a)\n",
+    "print \"a)Beta = \",beta"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.8, Page 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a) A changes by (+-) 0.09901 %\n",
+      "b) A changes by (+-) 0.0001 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "a = 10**5       \n",
+    "beta\n",
+    "T\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate():\n",
+    "    global a,beta,T\n",
+    "    T=a*beta\n",
+    "    ans=10.0/(1+T)    # for a +- 10% change in a\n",
+    "    return ans\n",
+    "\n",
+    "#answer\n",
+    "beta=10**(-3)                                                    #given\n",
+    "desensitivity_factor=calculate();                                # stores the answer\n",
+    "print \"a) A changes by (+-)\",round(desensitivity_factor,6),\"%\"   #part a\n",
+    "\n",
+    "beta=1                                                           #given\n",
+    "desensitivity_factor=calculate();\n",
+    "print \"b) A changes by (+-)\",round(desensitivity_factor,6),\"%\"   #part b"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.9, Page 33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  A =  995.024876 V/V\n",
+      "  Ro =  373.134 mili ohm\n",
+      "  Ri =  402.0 Mega ohm\n",
+      "b)\n",
+      "  A =  0.999995 V/V\n",
+      "  Ro =  0.375 mili ohm\n",
+      "  Ri =  400002.0 Mega ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 2.0    # Mega ohm\n",
+    "ro = 75.0   # ohm\n",
+    "a = 200000.0   # V/V\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def calculate(R1,R2):\n",
+    "    global a,ro,rd\n",
+    "    beta=R1/(R1+R2)\n",
+    "    if(R1==float(\"inf\")):      # for infinty\n",
+    "        beta=1\n",
+    "    T=a*beta\n",
+    "    A=(1+(R2/R1))/(1+(1/T))    # equation 1.55\n",
+    "    if(R1==float(\"inf\")):      # for infinity\n",
+    "        A=1/(1+(1/T))\n",
+    "    Ro=ro/(1+T)                # equation 1.61\n",
+    "    Ri=rd*(1+T)                # equation 1.59\n",
+    "    print \"  A = \",round(A,6),\"V/V\"\n",
+    "    print \"  Ro = \",round(Ro*(10**3),3),\"mili ohm\"\n",
+    "    print \"  Ri = \", round(Ri,3),\"Mega ohm\"\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "calculate(1.0,999)\n",
+    "print \"b)\"\n",
+    "calculate(float(\"inf\"),1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.10, Page 35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      " A =  -0.99999 V/V\n",
+      " Rn =  0.5 ohm\n",
+      " Ri =  100000.0 ohm\n",
+      " Ro =  0.00075 ohm\n",
+      "b)\n",
+      " A =  -995.01993 V/V\n",
+      " Rn =  4.99998 ohm\n",
+      " Ri =  1000.0 ohm\n",
+      " Ro =  0.37351 ohm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "a = 200000.0    # V/V\n",
+    "ro = 75         # ohm\n",
+    "\n",
+    "#Calculating function\n",
+    "\n",
+    "def calculate(R1,R2):\n",
+    "    global a,ro\n",
+    "    T=a*(R1/(R1+R2)) \n",
+    "    A=(-1)*(R2/R1)/(1+(1/T))   # equation 1.63\n",
+    "    Rn=R2/(1+a)               # equation 1.67b\n",
+    "    Ri=R1                     # equation 1.68\n",
+    "    Ro=ro/(1+T)\n",
+    "    print \" A = \",round(A,5),\"V/V\"\n",
+    "    print \" Rn = \",round(Rn,5),\"ohm\"\n",
+    "    print \" Ri = \",round(Ri,5),\"ohm\"\n",
+    "    print \" Ro = \",round(Ro,5),\"ohm\"\n",
+    "                    \n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "calculate(100000.0,100000.0)\n",
+    "print \"b)\"\n",
+    "calculate(1000.0,1000000.0)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.11, Page 38"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "317.440162529\n",
+      "a) A_ideal = -101.1 V/V\n",
+      "b) A = -100.78 V/V\n",
+      "Deviation from ideal = 0.31 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "R1 = 1000000.0     # ohm\n",
+    "R2 = 1000000.0     # ohm\n",
+    "R3 = 100000.0      # ohm\n",
+    "R4 = 1000.0        # ohm\n",
+    "RL = 2000.0        # ohm\n",
+    "rd = 1000000.0     #ohm\n",
+    "a = 10**5        # V/V\n",
+    "ro = 100.0         # ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "A_ideal = (-1)*(R2/R1)*(1+(R3/R2)+(R3/R4))       # ideal op-amp and summing  currents at node  v1\n",
+    "T = a/(1+(R2/R1)+(R2/rd))/(1+(ro/(R2+(R1*rd/(R1+rd))))+(ro/RL))/100  #equation 1.73\n",
+    "A = A_ideal/(1+(1/T))                   \n",
+    "dev=(A_ideal-A)/A_ideal*100\n",
+    "\n",
+    "#answer\n",
+    "print T\n",
+    "print \"a) A_ideal =\",A_ideal,\"V/V\"\n",
+    "print \"b) A =\",round(A,2),\"V/V\"\n",
+    "print \"Deviation from ideal =\",round(dev,2),\"%\"\n",
+    "\n",
+    "#book example has precision error so answer is 0.32%"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.12, Page 40"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  Beta = 0.016911 V/V\n",
+      "  T = 169.1\n",
+      "b)\n",
+      "  Vo= -( 29.82 V1 + 14.91 V2 + 9.94 V3 )\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 1000.0          # kilo ohm\n",
+    "a = 10**4            # V/V\n",
+    "ro = 100.0           #ohm\n",
+    "R1 = 10.0            # kilo ohm\n",
+    "R2 = 20.0            # kilo ohm\n",
+    "R3 = 30.0            # kilo ohm\n",
+    "R4 = 300.0           # kilo ohm\n",
+    "RL = 2.0             # kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def parallel(a,b):\n",
+    "    ans=a*b/(a+b)\n",
+    "    return ans\n",
+    "\n",
+    "Ra = parallel(R1,parallel(R2,parallel(R3,rd)))\n",
+    "Rb=Ra+R4\n",
+    "Rc=parallel(Rb,RL)                      #After suppressing all input sources\n",
+    "Rd=Rc+ro/1000                           #replacing the op-amp with it's terminal resistances\n",
+    "Vn=Rb/Ra                                #and applying a test voltage and analysing the circuit\n",
+    "Vt=Rd/Rc\n",
+    "beta=1/Vn/Vt\n",
+    "T=a*beta\n",
+    "v1=R4/R1\n",
+    "v2=R4/R2\n",
+    "v3=R4/R3\n",
+    "A=1/(1+1/T)\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "print \"  Beta =\",round(beta,6),\"V/V\"\n",
+    "print \"  T =\",round(T,1)\n",
+    "print \"b)\"\n",
+    "print \"  Vo= -(\",round(A*v1,2),\"V1 +\",round(A*v2,2),\"V2 +\",round(A*v3,2),\"V3 )\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.13, Page 41"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 27,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Beta = 0.8101 V/V\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "rd = 100.0        # kilo ohm\n",
+    "ro = 100.0        # ohm\n",
+    "R1 = 30.0         # kilo ohm\n",
+    "R2 = 20.0         # kilo ohm\n",
+    "R3 = 10.0         # kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "def parallel(a,b):\n",
+    "    ans=a*b/(a+b)\n",
+    "    return ans\n",
+    "\n",
+    "beta_n = (parallel(R1,rd)+R1)/((ro/1000)+R2+parallel(R1,rd)+R3)    # from circuit 1.35 after appyling\n",
+    "beta_p = R3/((ro/1000)+R2+parallel(R1,rd)+R3)                      # voltage divide formula twice\n",
+    "beta=beta_n-beta_p                  #equation 1.76\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"Beta =\",round(beta,4),\"V/V\"\n",
+    "\n",
+    "# beta_n calculation in book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.14, Page 43"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a)\n",
+      "  Icc = 0.5 mA\n",
+      "  Iee = 3.5 mA\n",
+      "  I0 = 3 mA\n",
+      "b)\n",
+      "  Power Poa = 42.0 mW\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#Variable Declaration \n",
+    "\n",
+    "R1 = 10     #kilo ohm\n",
+    "R2 = 20     #kilo ohm\n",
+    "V1 = 3      # V\n",
+    "Iq = 0.5    # mA\n",
+    "RL = 2      #kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "V0 = (-1)*R2/R1*V1\n",
+    "It = abs(V0)/RL                     # Currents through R1,R2,Rt are i1,i2,It respectively\n",
+    "i1 = It/R1\n",
+    "i2 = i1                             # applying voltage divider rule\n",
+    "i0 = i2+It\n",
+    "icc = Iq\n",
+    "iee = icc+ i0\n",
+    "Poa = 30*Iq+((V0+15)*i0)            #Whenever current passes through voltage drop, power = vi\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"a)\"\n",
+    "print \"  Icc =\",icc,\"mA\"\n",
+    "print \"  Iee =\",iee,\"mA\"\n",
+    "print \"  I0 =\",i0,\"mA\"\n",
+    "print \"b)\"\n",
+    "print \"  Power Poa =\",Poa,\"mW\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.15, Page 43"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "b)\n",
+      "  Change in v = 3.75 micro Volt -> quite a small change\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Variable Declaration\n",
+    "\n",
+    "ro = 75.0      #kilo ohm\n",
+    "T = 200000.0\n",
+    "Vs = 10.0      # V\n",
+    "Rl = 1.0       #kilo ohm\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "iL = Vs/Rl\n",
+    "Ro = ro/(1+T)\n",
+    "del_v = Ro*10*(10**(-3))\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"b)\"\n",
+    "print \"  Change in v =\",round(del_v*(10**6),2),\"micro Volt -> quite a small change\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.16, Page 46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The op-amp saturates at Vo=+-13 V\n",
+      "With Vn= 20/3-13/3 = 2.3333 V\n"
+     ]
+    },
+    {
+     "data": {
+      "text/plain": [
+       "[<matplotlib.lines.Line2D at 0xb4a42b0>]"
+      ]
+     },
+     "execution_count": 30,
+     "metadata": {},
+     "output_type": "execute_result"
+    },
+    {
+     "data": {
+      "image/png": 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AxIlhz4bvfleVRGt07Ag33BCmDP/kJ2Hgf9mypKMSqR65ueQaGzcsl1wqKwqgMzDRzKYD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\niEhBqihERKQgVRQiIlLQ/wfqhyu+stXumgAAAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x3ba3a90>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "%matplotlib inline\n",
+    "\n",
+    "import matplotlib.pyplot as plt\n",
+    "import scipy as np\n",
+    "import math\n",
+    "\n",
+    "#Variable Declaration\n",
+    "\n",
+    "A = -2               #V/V\n",
+    "peak = 10           # V\n",
+    "\n",
+    "#Calculation \n",
+    "\n",
+    "output = np.absolute(A) * peak\n",
+    "\n",
+    "#answer\n",
+    "\n",
+    "print \"The op-amp saturates at Vo=+-13 V\"\n",
+    "print \"With Vn= 20/3-13/3 =\",round((20.0/3)-(13.0/3),4),\"V\"\n",
+    "\n",
+    "#Graphs\n",
+    "\n",
+    "t1 = np.arange(0,1,.0005)               #  Triangular waveform\n",
+    "t2 = np.arange(1,3,.0005)\n",
+    "t3 = np.arange(3,5,.0005)\n",
+    "\n",
+    "m1 = np.arange(0,0.65,.0005)\n",
+    "m2 = np.arange(.65,1.35,.0005)\n",
+    "m3 = np.arange(1.35,2.65,.0005)            # Output Vo wave\n",
+    "m4 = np.arange(2.65,3.35,.0005)\n",
+    "m5 = np.arange(3.35,4.65,.0005)\n",
+    "m6 = np.arange(4.65,5,.0005)                # Output Vn wave\n",
+    "m7 = np.arange(0.65,1,.0005)\n",
+    "m8 = np.arange(1,1.35,.0005)\n",
+    "m9 = np.arange(2.65,3,.0005)\n",
+    "m10 = np.arange(3, 3.35, .0005)\n",
+    "\n",
+    "plt.subplot(2,1,1)\n",
+    "\n",
+    "plt.suptitle(\"Vt (Blue), Vo (Red) and Vn (Green) Graphs\")\n",
+    "plt.xlim(0,4.5)\n",
+    "plt.xlabel(\"time->\")\n",
+    "plt.ylabel(\"V->\")\n",
+    "plt.plot(t1,peak*t1,\"b\",)\n",
+    "plt.plot(t2,(-1)*peak*t2+2*(peak),\"b\",)\n",
+    "plt.plot(t3,peak*t3-4*(peak),\"b\",)\n",
+    "\n",
+    "plt.subplot(2,1,2)\n",
+    "\n",
+    "plt.xlim(0,4.5)\n",
+    "plt.xlabel(\"time->\")\n",
+    "plt.ylabel(\"V->\")\n",
+    "plt.plot(m1,-20*m1,\"r\")\n",
+    "plt.plot(m2,np.full(len(m2),-13),\"r\")\n",
+    "plt.plot(m3,20*m3-40,\"r\")\n",
+    "plt.plot(m4,np.full(len(m4),13),\"r\")\n",
+    "plt.plot(m5,-20*m5+80,\"r\")\n",
+    "plt.plot(m6,np.full(len(m6),-13),\"r\")\n",
+    "\n",
+    "plt.plot(m1,np.full(len(m1),0),\"g\",)\n",
+    "plt.plot(m7,6.665*m7-4.4,\"g\")\n",
+    "plt.plot(m8,-6.665*m8+8.8,\"g\")\n",
+    "plt.plot(m3,np.full(len(m3),0),\"g\")\n",
+    "plt.plot(m9,6.665*m9-17.6,\"g\")\n",
+    "plt.plot(m10,-6.665*m10+22.4,\"g\")\n",
+    "plt.plot(m5,np.full(len(m5),0),\"g\")\n",
+    "plt.plot(m6,np.full(len(m6),0),\"g\")"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Ashish  KumarSingh/Chapter_First.ipynb b/sample_notebooks/Ashish  KumarSingh/Chapter_First.ipynb
new file mode 100755
index 00000000..26e2a823
--- /dev/null
+++ b/sample_notebooks/Ashish  KumarSingh/Chapter_First.ipynb
@@ -0,0 +1,243 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Light"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.1, Page Number 10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Brewsters Angle of the Material is 56.31 Degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "n2=1.5 #Given Refractive Index of Glass in Air\n",
+    "n1=1   #Given Refractive Index of Air\n",
+    "\n",
+    "theta=0 #Brewster's Angle\n",
+    "#From Equation 1.13 (Brewsters angle= Tan Inverse (n2/n1))\n",
+    "theta=math.degrees(math.atan(1.5))\n",
+    "print \"The Brewsters Angle of the Material is \"+str(round(theta,2))+\" Degrees\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.2, Page Number 13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "In Coherant Sources The Maximum Irradiance is 16I\n",
+      "In Incoherant Sources The Maximum Irradiance is 4I\n"
+     ]
+    }
+   ],
+   "source": [
+    "n=4 #Total Number of Sources\n",
+    "\n",
+    "#For Coherant Sources\n",
+    "print \"In Coherant Sources The Maximum Irradiance is \"+str(n*n)+\"I\"  #Where I is the Irradiance at any point\n",
+    "#For Incoherant Sources\n",
+    "print \"In Incoherant Sources The Maximum Irradiance is \"+str(n)+\"I\"   "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.3, Page Number 23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(A)The Minimum Seperation Between the Sources is 0.0025 m\n",
+      "(B)The Minimum Wavelength Difference which may be resolved is 2.08333333333e-11 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "D=0.1 #Diameter of the Objective Lens\n",
+    "d1=500 #Distance from the source\n",
+    "l =0.000000500 #Wavelength Provided\n",
+    "p=1 #First Order\n",
+    "N=40*600 #The diffraction grating is 40 mm wide and has 600 lines/mm\n",
+    "\n",
+    "#From Equation 1.29 we Have\n",
+    "Smin=(d1*l)/D  #Where Smin is the minimum Seperation of the Sources\n",
+    "print \"(A)The Minimum Seperation Between the Sources is \"+str(Smin)+\" m\"\n",
+    "\n",
+    "#We know that Chromatic resolving power is given by l/dl where dl is the Minimum Wavelength Difference\n",
+    "#From Equation l/dl=p*N\n",
+    "dl=l/(N*p)\n",
+    "\n",
+    "print \"(B)The Minimum Wavelength Difference which may be resolved is \"+str(dl)+\" m\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 1.4, Page Number 29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Total Power Radiated from the Source is 6.3504 W\n"
+     ]
+    }
+   ],
+   "source": [
+    "em=0.7 #Emissivity Of the Surface\n",
+    "T=2000 #Temperature in Kelvin\n",
+    "A=0.00001 #Area in Meter Square\n",
+    "S=5.67*(10**-8) #Stefan-Boltzmann Constant\n",
+    "\n",
+    "W=S*A*em*(T**4) #Where W is the total power radiated\n",
+    "\n",
+    "print \"The Total Power Radiated from the Source is \"+str(W)+\" W\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 1.5, Page Number 31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Ionization Energy required to excite the electron from ground to Infinity 13.66 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "Z=1 #Atomic Number of Hydrogen\n",
+    "m=9.1*(10**-31) #Mass of a Electron\n",
+    "e=1.6*(10**-19) #Charge Of a Electron\n",
+    "p=6.6*(10**-34) #Plancks Constant\n",
+    "e1=8.85*(10**-12)#Permittivity of Free Space\n",
+    "#From Equation 1.43\n",
+    "E=(m*(Z**2)*(e**4))/(8*(p**2)*(e1**2)) #Where E is the Ionization Energy\n",
+    "E2=E/e #Converting in Electron Volts\n",
+    "E2=round(E2,2)\n",
+    "\n",
+    "print \"The Ionization Energy required to excite the electron from ground to Infinity \"+str(E2)+\" eV\"\n",
+    "       "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 1.6, Page Number 32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Required Work function is 4.5375 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "e=1.6*(10**-19) #Charge Of a Electron\n",
+    "h=6.6*(10**-34) #Plancks Constant\n",
+    "vo=1.1*(10**15) #Threshold Frequency in Hertz\n",
+    "\n",
+    "# We Know h*vo=phi*e where phi is the required Work Function\n",
+    "# We assume that the ejected electron has zero kinetic energy\n",
+    "\n",
+    "phi=h*vo/e\n",
+    "print \"The Required Work function is \"+str(phi)+\" eV\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/AshvaniKumar/Chapter2.ipynb b/sample_notebooks/AshvaniKumar/Chapter2.ipynb
new file mode 100755
index 00000000..ba79768d
--- /dev/null
+++ b/sample_notebooks/AshvaniKumar/Chapter2.ipynb
@@ -0,0 +1,356 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ca4b0079bde9d44d795cfee22e676408883fa63b320ce21fd0ba8c51e96fec7f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2 - Press Tool Design"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.1 - page 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from numpy import pi\n",
+      "D = 50 # Diameter of washer in mm\n",
+      "t = 4 # thickness of material in mm\n",
+      "d = 24 # diameter of hole in mm\n",
+      "p = 360 # shear strength of material in N/mm**2\n",
+      "F1 = pi*D*t*p # blanking pressure in N\n",
+      "F2 = pi*d*t*p # piercing pressure in N\n",
+      "F = F1 + F2 # total pressure in N\n",
+      "d1 = d + 0.4 # piercing die diameter in mm\n",
+      "d2 = D - 0.4 # blank punch diameter in mm\n",
+      "c = 0.8*F # press capacity in N\n",
+      "print \"Blanking pressure = %d kN\\nPiercing pressure = %0.3f KN\\nTotal pressure required = %0.1f KN\\n\" %(F1/1000,F2/1000,F/1000)\n",
+      "print \"Piercing punch diameter = %0.2f cm\\nBlanking punch diametre = %0.2f cm \\npress capacity = %0.2f kN\\n\"%(d1/10 , d2/10 , c/1000)\n",
+      "# Answers vary due to round off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Blanking pressure = 226 kN\n",
+        "Piercing pressure = 108.573 KN\n",
+        "Total pressure required = 334.8 KN\n",
+        "\n",
+        "Piercing punch diameter = 2.44 cm\n",
+        "Blanking punch diametre = 4.96 cm \n",
+        "press capacity = 267.81 kN\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.2 - page 101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "l1 = 76 - ( 2.3 + 0.90) # length1 in mm\n",
+      "l2 = 115 - (2.3 + 0.90) # length2 in mm\n",
+      "t = 2.3 # mm\n",
+      "r = 0.90 # inner radius in mm\n",
+      "k = t/3 # mm\n",
+      "B = 0.5*pi*(r + k) # bending allowance in mm\n",
+      "d = l1 + l2 + B # developed length in mm\n",
+      "print \"Developed length = %0.2f mm\" %(d)\n",
+      "# Answers vary due to round off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Developed length = 187.22 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.3 - page 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import floor\n",
+      "t = 3.2 # thickness of blank in mm\n",
+      "l = 900 # bending length in mm\n",
+      "sigma = 400 # ultimate tensile strength in N/mm**2\n",
+      "k = 0.67 # die opening factor\n",
+      "r1 = 9.5 # punch radius in mm\n",
+      "r2 = 9.5 # die edge radius in mm\n",
+      "c = 3.2 # clearance in mm\n",
+      "w = r1 + r2 + c # width between contact points batween die and punch in mm\n",
+      "F= (k*l*sigma*t**2)/w # bending force in N\n",
+      "F=floor(F/10)*10 # N\n",
+      "print \"bending force = %0.2f kN\" %(F/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "bending force = 111.25 kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.4 - page 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "k = 1.33 # die opening factor\n",
+      "l = 1200 # bend length in mm\n",
+      "sigma = 455 # ultimate tensile strength in N/mm**2\n",
+      "t = 1.6 # blank thickness in mm\n",
+      "w = 8*t # width of die opening in mm\n",
+      "F = k*l*sigma*t**2/w # bending force in N \n",
+      "print \"bending force  = %0.2f kN\"%(F/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "bending force  = 145.24 kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.5 - page 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "c = 1.25 # clearance in mm\n",
+      "r1 = 3 # die radius in mm\n",
+      "r2 = 1.5 # punch radius in mm\n",
+      "sigma = 315 # ultimate tensile strength in MPa\n",
+      "t = 1 # thickness in mm\n",
+      "l = 50 # width at bend in mm\n",
+      "w = r1 + r2 +c # width between contact points on die and punch in mm\n",
+      "F = 0.67*l*sigma*t**2/w # bending force in N\n",
+      "F_p = 0.67*sigma*l*t # pad force in N\n",
+      "sigma_c = 560 # setting pressure in MPa\n",
+      "b1 = 2 # beads on punch\n",
+      "b = b1*r1 # mm\n",
+      "F_b = sigma_c*l*b # bottoming force in N\n",
+      "F_o = F_p + F_b # Force required when bottoming is used in N\n",
+      "F_n = F +F_p # Force required when bottoming is not used in N\n",
+      "print \"  Force required when bottoming is used  = %0.1f tonnes\"%(F_o/(9.81*1000))\n",
+      "print \"  Force required when bottoming is not used  = %0.3f tonnes\" %(F_n/(9.81*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  Force required when bottoming is used  = 18.2 tonnes\n",
+        "  Force required when bottoming is not used  = 1.263 tonnes\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.6 - page 103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sin\n",
+      "w = 2 # width in mm\n",
+      "t = 5 # thickness in mm\n",
+      "theta=6 # shear in degrees\n",
+      "tau = 382.5 # ultimate shear stress in MPa\n",
+      "F = w*t*tau*1000 # cutting force in N\n",
+      "l = t/sin(theta*pi/180) # length to be cut in mm\n",
+      "F_i = l*t*tau # cutting force in N\n",
+      "print \" cutting force with parallel cutting edges = %0.3f MN\\n cutting force when shear is considered = %0.2f kN \"%(F/10**6 ,F_i/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " cutting force with parallel cutting edges = 3.825 MN\n",
+        " cutting force when shear is considered = 91.48 kN \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.7 - page 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "d1 = 105 # inside diameter in mm\n",
+      "h = 90 # depth in mm\n",
+      "t = 1 # thickness in mm\n",
+      "D = sqrt(d1**2+4*d1*h) # blank diameter in mm\n",
+      "tr = t*100/D # thickness ratio\n",
+      "# from table safe drawing ratio is 1.82\n",
+      "r = 1.82 # draw ratio\n",
+      "d2 = D/r # diameter for first draw in mm\n",
+      "d = 130 # Let diameter for first draw in mm\n",
+      "ratio1 = D/d # D/d for first draw \n",
+      "ratio2 = d/d1 # D/d for second draw\n",
+      "h1 =((D)**2-(d)**2)/(4*d) # Depth of first draw in mm\n",
+      "sigma = 415 # N/mm**2\n",
+      "c = 0.65 # constant\n",
+      "pc = pi*d*t*sigma*(D/d - c) # press capacity in kN\n",
+      "print \" Blank diameter = %d mm \\n Thickness ratio = %0.3f \\n Diameter for first draw = %d mm \\n Depth of first draw = %0.2f mm \\n Press capacity = %0.2f kN\"%(D,tr,d2,h1,pc/1000)\n",
+      "# Answers vary due to round off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Blank diameter = 220 mm \n",
+        " Thickness ratio = 0.453 \n",
+        " Diameter for first draw = 121 mm \n",
+        " Depth of first draw = 61.39 mm \n",
+        " Press capacity = 177.92 kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2.8 - page 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "d = 80 # diameter in mm\n",
+      "h = 250 # height in mm\n",
+      "D = sqrt((d**2+4*d*h))/10 # blank diameter in cm\n",
+      "D1 = 0.5*D # diameter after first draw in cm\n",
+      "# let reduction be 40% in second draw\n",
+      "D2 = D1-0.4*D1 # diameter after scond draw in cm\n",
+      "R = (1 - (d/(10*D2)))*100 # percentage reduction for third draw\n",
+      "l1 = ((D)**2-(D1)**2)/(4*D1) # height of cup after first draw in cm\n",
+      "l2 = ((D)**2-(D2)**2)/(4*D2) # height of cup after first draw in cm\n",
+      "l3 = ((D)**2-(d/10)**2)/(4*d/10) # height of cup after first draw in cm\n",
+      "t = 3 # mm\n",
+      "sigma = 250 # N/mm**2\n",
+      "C = 0.66\n",
+      "F = pi*d/10*t*sigma*((D*10/d)-C) # drawing force in kN\n",
+      "print \" Diameter after first draw = %0.1f \\n Diameter after second draw = %0.2f \\n Percentage reduction after third draw = %d percent\"%(D1,D2,R)\n",
+      "print \" Height of cup after first draw = %0.2f cm\\n Height of cup after second draw = %0.2f cm\\n Height of cup after third draw = %0.2f cm\"%(l1,l2,l3)\n",
+      "print \" Drawing force = %0.3f kN\"%(F/1000)\n",
+      "# Answers vary due to round off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Diameter after first draw = 14.7 \n",
+        " Diameter after second draw = 8.82 \n",
+        " Percentage reduction after third draw = 9 percent\n",
+        " Height of cup after first draw = 11.02 cm\n",
+        " Height of cup after second draw = 22.29 cm\n",
+        " Height of cup after third draw = 25.00 cm\n",
+        " Drawing force = 56.817 kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/AviralYadav/Chapter9.ipynb b/sample_notebooks/AviralYadav/Chapter9.ipynb
new file mode 100755
index 00000000..0655eef8
--- /dev/null
+++ b/sample_notebooks/AviralYadav/Chapter9.ipynb
@@ -0,0 +1,144 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:349ae7afdee1d1b3c3dc4037b8dc3bb200738707d16369e5edfee0d065859f9b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Signal Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.1:pg-277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# To find dynamic range of spectrum analyser\n",
+      "\n",
+      "# Given data\n",
+      "I_p = +25.0; #Third order intercept point in dBm\n",
+      "MDS = -85.0; #noise level in dBm\n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "dynamic_range = 2/3.0*(I_p -MDS);\n",
+      "print \"The dynamic range of the spectrum analyser =\",int(dynamic_range),\" dB\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dynamic range of the spectrum analyser = 73  dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.2:pg-277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# To find minimum detectable signal\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "# Given data\n",
+      "NF = 20.0; #Noise figure in dB\n",
+      "BW = 1*10.0**3; #Bandwidth in Hz\n",
+      "\n",
+      "#Calculations\n",
+      "MDS=-114+10*math.log10((BW/(1*10.0**6)))+NF\n",
+      "print \"The minimum detectable signal of the spectrum analyser = \",int(MDS),\" dBm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum detectable signal of the spectrum analyser =  -124  dBm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9.3:pg-285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# To find dynamic range and total frequency display\n",
+      "\n",
+      "import math\n",
+      "# Given data\n",
+      "T = 4.0; #Sample window in s\n",
+      "f_s = 20*10.0**3; # sample frequency in Hz\n",
+      "N = 10.0; #no of bits\n",
+      "\n",
+      "#Calculations\n",
+      "f_r = 1/T;\n",
+      "f_h = f_s/2.0; \n",
+      "R_d = 20*math.log10(2.0**N);\n",
+      "\n",
+      "print \"The ratio of the spectral calculation = \",round(f_r,2),\" Hz\\n\"\n",
+      "print \"The maximum calculated spectral frequency = \",int(f_h),\" Hz\\n\"\n",
+      "print \"The dynamic range = \",int(R_d),\" dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ratio of the spectral calculation =  0.25  Hz\n",
+        "\n",
+        "The maximum calculated spectral frequency =  10000  Hz\n",
+        "\n",
+        "The dynamic range =  60  dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/BhavithaInnamuri/Chapter_1_CRYSTAL_STRUCTURES.ipynb b/sample_notebooks/BhavithaInnamuri/Chapter_1_CRYSTAL_STRUCTURES.ipynb
new file mode 100755
index 00000000..cd376de8
--- /dev/null
+++ b/sample_notebooks/BhavithaInnamuri/Chapter_1_CRYSTAL_STRUCTURES.ipynb
@@ -0,0 +1,677 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 CRYSTAL STRUCTURES"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_4 pgno:10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1.0\n",
+      "r=a/2 =  0.5\n",
+      "Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)=  0.523598775598\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.0\n",
+      "Fp(S.C)=(v∗100/V)=  52.3598775598\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.4\n",
+    "from math import pi\n",
+    "a=1.\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=a/2.\n",
+    "print \"r=a/2 = \",r # initializing value of radius of atom for simple cubic .\n",
+    "v=((4*pi*(r**3))/3)\n",
+    "print \"Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)= \",v # calcuation . \n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(S.C)=(v∗100/V)= \",Fp,# calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_5 pgno:11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1.0\n",
+      "Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) =  0.433012701892\n",
+      "Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 =  0.680174761588\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.0\n",
+      "Fp(B.C.C)=(v∗100/V)=  68.0174761588 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.5\n",
+    "from math import sqrt\n",
+    "a=1.\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=(sqrt(3)*(a**2/4))\n",
+    "print \"Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) = \",r # initializing value of radius of atom for BCC.\n",
+    "v=((4*pi*(r**3))/3)*2\n",
+    "print \"Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 = \",v # calcuation \n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(B.C.C)=(v∗100/V)= \",Fp,\"%\" # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_6 pgno:12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1\n",
+      "Radius of the atom,r=(a/(2∗sqrt(2)))=  0.353553390593\n",
+      "Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)=  0.740480489693\n",
+      "Total volume of the cube ,V=aˆ3=  2\n",
+      "Fp(F.C.C)=(v∗100/V)=  37.0240244847 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.6\n",
+    "a=1\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=(a/(2*sqrt(2)))\n",
+    "print \"Radius of the atom,r=(a/(2∗sqrt(2)))= \",r  # initializing value of radius of atom for FCC .\n",
+    "v=(((4*pi*(r**3))/3)*4)\n",
+    "print \"Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)= \",v # calcuation \n",
+    "V=a^3\n",
+    "print \"Total volume of the cube ,V=aˆ3= \",V  #  calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(F.C.C)=(v∗100/V)= \",Fp,\"%\"  # calculation\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_8 pgno:14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1\n",
+      "Radius of the atom , r=(sqrt (3)∗a/8))=  0.216506350946\n",
+      "v=(((4∗pi∗(rˆ3))/3)∗8) =  0.340087380794\n",
+      "V=aˆ3=  2\n",
+      "Fp(Diamond)=(v∗100/V) =  17.0043690397 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 1.8 \n",
+    "a=1\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=((sqrt(3)*a/8))\n",
+    "print \"Radius of the atom , r=(sqrt (3)∗a/8))= \",r # initializing value of radius of atom for diamond .\n",
+    "v=(((4*pi*(r**3))/3)*8)\n",
+    "print \"v=(((4∗pi∗(rˆ3))/3)∗8) = \",v # calcuation .\n",
+    "V=a^3\n",
+    "print \"V=aˆ3= \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(Diamond)=(v∗100/V) = \",Fp,\"%\" # calculation\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_9 pgno:14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  5e-08  cm\n",
+      "Radius of the atom,r=(sqrt(3)∗(a/4))=  2.16506350946e-08\n",
+      "Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2=  8.50218451985e-23\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.25e-22\n",
+      "Fp(B.C.C)=(v∗100/V) =  68.0174761588 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.9\n",
+    "a=5*10**-8\n",
+    "print \"a = \",a,\" cm\" # initializing value of lattice constant .\n",
+    "r=(sqrt(3)*(a/4))\n",
+    "print \"Radius of the atom,r=(sqrt(3)∗(a/4))= \",r  # initializing value of radius of atom for BCC.\n",
+    "v=((4*pi*(r**3))/3)*2\n",
+    "print \"Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2= \",v # calcuation .\n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(B.C.C)=(v∗100/V) = \",Fp,\"%\"  #  calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_10 pgno:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  1\n",
+      "y intercept =  inf\n",
+      "z intercept =  inf\n",
+      "miller indices ,h=(1/x )=  [1]\n",
+      "k=(1/y)=  [0.0]\n",
+      "l=(1/z) =  [0.0]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.10\n",
+    "x=1\n",
+    "print \"x intercept = \",x # initializing value of x intercept .\n",
+    "y=float('inf')\n",
+    "print \"y intercept = \",y # initializing value of y intercept .\n",
+    "z=float('inf')\n",
+    "print \"z intercept = \",z # initializing value of z intercept .\n",
+    "h=[1/x]\n",
+    "print \"miller indices ,h=(1/x )= \",h # calculation\n",
+    "k=[1/y]\n",
+    "print \"k=(1/y)= \",k # calculation\n",
+    "l=[1/z]\n",
+    "print \"l=(1/z) = \",l # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_11 pgno:15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  inf\n",
+      "y intercept =  inf\n",
+      "z intercept =  1\n",
+      "miller indices ,h=[1/x] =  [0.0]\n",
+      "k=[1/y] =  [0.0]\n",
+      "l=[1/z] =  [1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.11\n",
+    "x=float('inf')\n",
+    "print \"x intercept = \",x # initializing of x intercept .\n",
+    "y=float('inf') \n",
+    "print\"y intercept = \",y # initializing of Y intercept .\n",
+    "z=1\n",
+    "print \"z intercept = \",z # initializing of Z intercept .\n",
+    "h=[1/x]\n",
+    "print \"miller indices ,h=[1/x] = \",h # calculation\n",
+    "k=[1/y]\n",
+    "print \"k=[1/y] = \",k # calculation \n",
+    "l=[1/z]\n",
+    "print \"l=[1/z] = \",l # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_12 pgno: 16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  inf\n",
+      "y intercept =  1\n",
+      "z intercept =  inf\n",
+      "miller indices ,h=[1/x] =  [0.0]\n",
+      "k=[1/y] =  [1]\n",
+      "l=[1/z] =  [0.0]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.12\n",
+    "x=float('inf') \n",
+    "print \"x intercept = \",x  # initializing of X intercept .\n",
+    "y=1\n",
+    "print \"y intercept = \",y  # initializing of X intercept .\n",
+    "z=float('inf') \n",
+    "print \"z intercept = \",z  # initializing of X intercept .\n",
+    "h=[1/x]\n",
+    "print \"miller indices ,h=[1/x] = \",h # calculation\n",
+    "k=[1/y]\n",
+    "print \"k=[1/y] = \",k  # calculation \n",
+    "l=[1/z]\n",
+    "print \"l=[1/z] = \",l #calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_13 pgno:16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  1\n",
+      "y intercept =  1\n",
+      "z intercept =  inf\n",
+      "miller indices ,h=[1/x] =  [1]\n",
+      "k=[1/y] =  [1]\n",
+      "l=[1/z] =  [0.0]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.13\n",
+    "x=1\n",
+    "print \"x intercept = \",x  # initializing of X intercept .\n",
+    "y=1\n",
+    "print \"y intercept = \",y  # initializing of X intercept .\n",
+    "z=float('inf') \n",
+    "print \"z intercept = \",z  # initializing of X intercept .\n",
+    "h=[1/x]\n",
+    "print \"miller indices ,h=[1/x] = \",h # calculation\n",
+    "k=[1/y]\n",
+    "print \"k=[1/y] = \",k  # calculation \n",
+    "l=[1/z]\n",
+    "print \"l=[1/z] = \",l #calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_14 pgno:17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  inf\n",
+      "y intercept =  1\n",
+      "z intercept =  1\n",
+      "miller indices ,h=[1/x] =  [0.0]\n",
+      "k=[1/y] =  [1]\n",
+      "l=[1/z] =  [1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.14\n",
+    "x=float('inf') \n",
+    "print \"x intercept = \",x  # initializing of X intercept .\n",
+    "y=1\n",
+    "print \"y intercept = \",y  # initializing of X intercept .\n",
+    "z=1\n",
+    "print \"z intercept = \",z  # initializing of X intercept .\n",
+    "h=[1/x]\n",
+    "print \"miller indices ,h=[1/x] = \",h # calculation\n",
+    "k=[1/y]\n",
+    "print \"k=[1/y] = \",k  # calculation \n",
+    "l=[1/z]\n",
+    "print \"l=[1/z] = \",l #calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_15 pgno:18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x intercept =  2\n",
+      "y intercept =  2\n",
+      "z intercept =  2\n",
+      "common factor of all the intercept=  2\n",
+      "miller indices ,h=[c/x] =  [1]\n",
+      "k=[c/y] =  [1]\n",
+      "l=[c/z] =  [1]\n"
+     ]
+    }
+   ],
+   "source": [
+    "x=2\n",
+    "print \"x intercept = \",x  # initializing of X intercept .\n",
+    "y=2\n",
+    "print \"y intercept = \",y  # initializing of X intercept .\n",
+    "z=2\n",
+    "print \"z intercept = \",z  # initializing of X intercept .\n",
+    "c=2\n",
+    "print \"common factor of all the intercept= \",c  # initializing value of common factor of all the intercepts .\n",
+    "h=[c/x]\n",
+    "print \"miller indices ,h=[c/x] = \",h # calculation\n",
+    "k=[c/y]\n",
+    "print \"k=[c/y] = \",k  # calculation \n",
+    "l=[c/z]\n",
+    "print \"l=[c/z] = \",l #calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_16 pgno: 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Wa =  28.1\n",
+      "D =  2.33 ram/cmˆ3\n",
+      "Na =  6.02e+23 atoms/mole\n",
+      "na =(Na∗D)/(Wa)=  4.99167259786e+22  atoms/cmˆ3\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.16\n",
+    "Wa =28.1\n",
+    "print \"Wa = \",Wa # initializing value of atomic weight .\n",
+    "D=2.33\n",
+    "print \"D = \",D,\"ram/cmˆ3\" # initializing value of density .\n",
+    "Na=6.02*10**23\n",
+    "print \"Na = \",Na,\"atoms/mole\" # initializing value of avagadro number .\n",
+    "na =(Na*D)/(Wa)\n",
+    "print \"na =(Na∗D)/(Wa)= \",na,\" atoms/cmˆ3\" # calculation\n",
+    "# the value of na (number of atoms in 1 cmˆ3 of silicon ) , provided after calculation in the book is wrong."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_17 pgno: 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  5e-08 cm\n",
+      "N=  2\n",
+      "V=aˆ3 =  1.25e-22 cmˆ3\n",
+      "na=(no.of atoms in unit cell/Volume of theunit cell) =(N/(V))=  1.6e+22\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.17\n",
+    "a=5*10**-8\n",
+    "print \"a= \",a,\"cm\" # initializing value of lattice constant .\n",
+    "N=2\n",
+    "print \"N= \",N # initializing value of no. of atoms in unit cell .\n",
+    "V=a**3\n",
+    "print \"V=aˆ3 = \",V,\"cmˆ3\" # initializing value of total Volume of the unit cell.\n",
+    "na =(N/(V))\n",
+    "print \"na=(no.of atoms in unit cell/Volume of theunit cell) =(N/(V))= \",na # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_18 pgno: 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  5.43e-08 cm\n",
+      "N =  8\n",
+      "Number of atom in the cmˆ3,ns =(N/(aˆ3))=  4.99678310227e+22\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.18\n",
+    "a=5.43*10**-8\n",
+    "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n",
+    "N=8\n",
+    "print \"N = \",N # initializing value of no. of atoms in a unit cell .\n",
+    "ns =(N/(a**3))\n",
+    "print \"Number of atom in the cmˆ3,ns =(N/(aˆ3))= \",ns #  calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_19 pgno: 18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  5.43e-08 cm\n",
+      "Wa =  28.1\n",
+      "Na =  6.02e+23\n",
+      "ns =  50000000000000000000000 atoms/cmˆ3\n",
+      "Density of silicon ,D =(ns∗Wa)/(Na)=  2.33388704319  gm/cmˆ2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.19\n",
+    "a=5.43*10**-8\n",
+    "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n",
+    "Wa =28.1\n",
+    "print \"Wa = \",Wa # initializing value of atomic weight .\n",
+    "Na=6.02*10**23\n",
+    "print \"Na = \",Na # initializing value of avagdro number .\n",
+    "ns =5*10**22\n",
+    "print \"ns = \",ns,\"atoms/cmˆ3\" # initializing value of atoms/cmˆ3.\n",
+    "D =(ns*Wa)/(Na)\n",
+    "print \"Density of silicon ,D =(ns∗Wa)/(Na)= \",D,\" gm/cmˆ2\" # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_20 pgno: 19"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  4.75e-08 cm\n",
+      "N =  4\n",
+      "na =(N/(aˆ3))= 3.73232249599e+22\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.20\n",
+    "a=4.75*10**-8\n",
+    "print \"a = \",a,\"cm\" # initializing value of lattice constant .\n",
+    "N=4\n",
+    "print \"N = \",N # initializing value of number of atoms in the unit cell .\n",
+    "na =(N/(a**3))\n",
+    "print \"na =(N/(aˆ3))=\",na  #  calculation"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/DesuSandeep Kumar/Chapter_1_Introduction_to_Radio_Communication_Systems.ipynb b/sample_notebooks/DesuSandeep Kumar/Chapter_1_Introduction_to_Radio_Communication_Systems.ipynb
new file mode 100755
index 00000000..42a985ca
--- /dev/null
+++ b/sample_notebooks/DesuSandeep Kumar/Chapter_1_Introduction_to_Radio_Communication_Systems.ipynb
@@ -0,0 +1,102 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 Introduction to Radio Communication Systems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_3 pgno:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The transfer function has no finite zeros \n",
+      "The poles \n",
+      "[-0.25+0.96824584j -0.25-0.96824584j]\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "#Chapter 1:Introduction to Radio Communication\n",
+    "#example 1.3 page no 3\n",
+    "#given\n",
+    "import numpy\n",
+    "print('The transfer function has no finite zeros ')\n",
+    "p=numpy.array([1, 0.5, 1])\n",
+    "x=numpy.roots(p)\n",
+    "print('The poles ')\n",
+    "print(x)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_4 pgno:8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the image frequency is  MHz 1.91\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "#Chapter 1:Introduction to Radio Communication Systems\n",
+    "#example 1.4 page no 8\n",
+    "#given\n",
+    "fIF=455*10**3#intermediate frequency\n",
+    "fO=1.455*10**6#oscillator frequency\n",
+    "fIM=fIF+fO#image frequency\n",
+    "print'the image frequency is  MHz',fIM*1e-6\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/DivyangGandhi/ch1.ipynb b/sample_notebooks/DivyangGandhi/ch1.ipynb
new file mode 100755
index 00000000..c3414fde
--- /dev/null
+++ b/sample_notebooks/DivyangGandhi/ch1.ipynb
@@ -0,0 +1,532 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a1a2fd08204d20b25cb8797d32a1268bd037e7b99587377263388f4580e76a47"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1  Page No : 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "c = 3*10**8;     #in m/s\n",
+      "f = 1.*10**6;     #in Hz\n",
+      "\n",
+      "# Calculations\n",
+      "lembda = c/f;\n",
+      "\n",
+      "# Results\n",
+      "print 'Wavelength (in m):',lembda\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength (in m): 300.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2  Page No : 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "c = 3*10**8;     #in m/s\n",
+      "f = 100.*10**6;     #in Hz\n",
+      "\n",
+      "# Calculations\n",
+      "lembda = c/f;\n",
+      "\n",
+      "# Results\n",
+      "print 'Wavelength (in m):',lembda\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength (in m): 3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3  Page No : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "G = 175.;     #absolute gain\n",
+      "\n",
+      "# Calculations\n",
+      "Gdb = 10*math.log10(175);   #decibell gain\n",
+      "\n",
+      "# Results\n",
+      "print 'The decibell power gain is:',Gdb,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The decibell power gain is: 22.4303804869 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4  Page No : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Gdb = 28.;   #decibell gain\n",
+      "\n",
+      "# Calculations\n",
+      "G = 10**(Gdb/10);  #Absolute power gain\n",
+      "\n",
+      "# Results\n",
+      "print 'The absolute power gain is:',G\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute power gain is: 630.95734448\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5  Page No : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Gdb = 28.;   #decibell gain\n",
+      "\n",
+      "# Calculations\n",
+      "G = 10**(Gdb/10);  #Absolute power gain\n",
+      "Av = G**0.5;  #Voltage gain\n",
+      "\n",
+      "# Results\n",
+      "print 'The voltage gain is:',Av\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage gain is: 25.1188643151\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6  Page No : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "G = 0.28; #Absolute gain\n",
+      "P1 = 1;  \n",
+      "P2 = .28;  #28 % of input power\n",
+      "\n",
+      "# Calculations and Results\n",
+      "Gdb = 10*math.log10(G);\n",
+      "print 'Decibell gain is',Gdb,'dB'\n",
+      "\n",
+      "Ldb = 10*math.log10(P1/P2); #dB loss\n",
+      "print 'Decibell loss is:',Ldb,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Decibell gain is -5.52841968658 dB\n",
+        "Decibell loss is: 5.52841968658 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7  Page No : 11"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "PmW = 100.; #power in mW\n",
+      "\n",
+      "# Calculations and Results\n",
+      "PdBm = 10*math.log10(PmW/1);  #P in dBm level\n",
+      "print '(a).   Power in dBm level is:',PdBm,'dBm'\n",
+      "\n",
+      "PdBW = PdBm-30;  #P in dBW level\n",
+      "print '(b).   Power in dBW level is:',PdBW,'dBW'\n",
+      "\n",
+      "PdBf = PdBm+120;  #Pin dBf level\n",
+      "print '(c)   Power in dBf level is:',PdBf,'dBf'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).   Power in dBm level is: 20.0 dBm\n",
+        "(b).   Power in dBW level is: -10.0 dBW\n",
+        "(c)   Power in dBf level is: 140.0 dBf\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8  Page No : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "G1 = 5000.;\n",
+      "L = 2000.;\n",
+      "G2 = 400.;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "G = G1*(1/L)*G2;   #Absolute gain\n",
+      "print '(a)  Net absolute gain is:',G\n",
+      "\n",
+      "GdB = 10*math.log10(G);  #System decibell gain\n",
+      "print '(b)  System Decibel gain is:',GdB,'dB'\n",
+      "\n",
+      "G1dB = 10*math.log10(G1);\n",
+      "LdB = 10*math.log10(L);\n",
+      "G2dB = 10*math.log10(G2);\n",
+      "print ('(c)   Individual stage gains are:');\n",
+      "print 'G1dB = ',G1dB\n",
+      "print 'LdB = ',LdB\n",
+      "print 'G2dB = ',G2dB\n",
+      "\n",
+      "GdB = G1dB-LdB+G2dB;\n",
+      "print 'The net dB gain is:',GdB,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)  Net absolute gain is: 1000.0\n",
+        "(b)  System Decibel gain is: 30.0 dB\n",
+        "(c)   Individual stage gains are:\n",
+        "G1dB =  36.9897000434\n",
+        "LdB =  33.0102999566\n",
+        "G2dB =  26.0205999133\n",
+        "The net dB gain is: 30.0 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9  Page No : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "G1 = 5000.;\n",
+      "L = 2000.;\n",
+      "G2 = 400.;\n",
+      "Ps = 0.1;  #in mW\n",
+      "\n",
+      "# Calculations and Results\n",
+      "P1 = G1*Ps;  #in mW\n",
+      "print '(a)  Power level P1 is:',P1,'mW'\n",
+      "\n",
+      "P2 = P1/L;  #in mW\n",
+      "print 'Line output power P2:',P2,'mW'\n",
+      "\n",
+      "Po = G2*P2;  #in mW\n",
+      "print 'System output power Po:',Po,'mW'\n",
+      "\n",
+      "PsdBm = 10*math.log10(Ps/1);\n",
+      "G1dB = 10*math.log10(G1);\n",
+      "LdB = 10*math.log10(L);\n",
+      "G2dB = 10*math.log10(G2);\n",
+      "\n",
+      "print ('(b)  Output power power levels in dBm are');\n",
+      "P1dBm = PsdBm+G1dB;\n",
+      "print 'P1(dBm) = ',P1dBm,'dBm'\n",
+      "\n",
+      "P2dBm = P1dBm-LdB;\n",
+      "print 'P2(dBm) = ',P2dBm,'dBm'\n",
+      "\n",
+      "PodBm = P2dBm+G2dB;\n",
+      "print 'Po(dBm) = ',PodBm,'dBm'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)  Power level P1 is: 500.0 mW\n",
+        "Line output power P2: 0.25 mW\n",
+        "System output power Po: 100.0 mW\n",
+        "(b)  Output power power levels in dBm are\n",
+        "P1(dBm) =  26.9897000434 dBm\n",
+        "P2(dBm) =  -6.02059991328 dBm\n",
+        "Po(dBm) =  20.0 dBm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10  Page No : 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "def voltage(PdBm):\n",
+      "    P = 1*10**(-3)*(10**(PdBm/10));\n",
+      "    return (75*P)**0.5;\n",
+      "\n",
+      "# Variables\n",
+      "S = 10.;  #dBm\n",
+      "G1 = 13.;  #dB\n",
+      "L1 = 26.;  #dB\n",
+      "G2 = 20.;  #dB\n",
+      "L2 = 29.;  #dB\n",
+      "\n",
+      "# Calculations and Results\n",
+      "print '(a)  The output levels are',\n",
+      "PdBm = S;\n",
+      "V = voltage(PdBm);\n",
+      "print PdBm,'1. Signal source in dBm:',PdBm,'in Volts : ',V\n",
+      "\n",
+      "PdBm = S+G1;\n",
+      "V = voltage(PdBm);\n",
+      "print '2. Line Amplifier in dBm:',PdBm,'in Volts : ',V\n",
+      "\n",
+      "PdBm = S+G1-L1;\n",
+      "V = voltage(PdBm);\n",
+      "print '3. Cable section A in dBm:',PdBm,'in Volts : ',V\n",
+      "\n",
+      "PdBm = S+G1-L1+G2;\n",
+      "V = voltage(PdBm);\n",
+      "print '4. Booster amplifier in dBm:',PdBm,'in Volts : ',V\n",
+      "\n",
+      "PdBm = S+G1-L1+G2-L2;\n",
+      "V = voltage(PdBm);\n",
+      "print '5. Cable section B in dBm:',PdBm,'in Volts : ',V\n",
+      "print ('(b).  The output power to get a voltage of 6V'),\n",
+      "V = 6.;   #volts\n",
+      "R = 75.;  #ohm\n",
+      "Po = (V**2)/R;\n",
+      "print Po,'W';\n",
+      "PodBm = 10*math.log10(Po*1000/1);\n",
+      "print 'power in dBm',PodBm,'dBm'\n",
+      "\n",
+      "GrdB = PodBm-PdBm;\n",
+      "print 'The required gain is',GrdB,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)  The output levels are 10.0 1. Signal source in dBm: 10.0 in Volts :  0.866025403784\n",
+        "2. Line Amplifier in dBm: 23.0 in Volts :  3.86839338256\n",
+        "3. Cable section A in dBm: -3.0 in Volts :  0.193878937799\n",
+        "4. Booster amplifier in dBm: 17.0 in Volts :  1.93878937799\n",
+        "5. Cable section B in dBm: -12.0 in Volts :  0.0687908430214\n",
+        "(b).  The output power to get a voltage of 6V 0.48 W\n",
+        "power in dBm 26.8124123738 dBm\n",
+        "The required gain is 38.8124123738 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11  Page No : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P = 5.;   #In mW\n",
+      "N = 100.*10**-6;  #in mW\n",
+      "\n",
+      "# Calculations and Results\n",
+      "S2N = P/N;\n",
+      "print '(a)    Absolute signal to noise ratio :',S2N\n",
+      "\n",
+      "S2NdB = 10*math.log10(S2N);\n",
+      "print '(b)   dB signal to noise ratio is:',S2NdB,'dB'\n",
+      "\n",
+      "PdBm = 10*math.log10(P/1);\n",
+      "print '(c)    Signal Power is',PdBm,'dBm'\n",
+      "\n",
+      "NdBm = 10*math.log10(N/1);\n",
+      "print 'Noise power is',NdBm,'dBm'\n",
+      "\n",
+      "S2NdB = PdBm-NdBm;\n",
+      "print 'Decinel S/N ratio is',S2NdB,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)    Absolute signal to noise ratio : 50000.0\n",
+        "(b)   dB signal to noise ratio is: 46.9897000434 dB\n",
+        "(c)    Signal Power is 6.98970004336 dBm\n",
+        "Noise power is -40.0 dBm\n",
+        "Decinel S/N ratio is 46.9897000434 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/DurgasriInnamuri/Chapter_3_Semoconductor_Devices_Fundamentals.ipynb b/sample_notebooks/DurgasriInnamuri/Chapter_3_Semoconductor_Devices_Fundamentals.ipynb
new file mode 100755
index 00000000..ade5b7fd
--- /dev/null
+++ b/sample_notebooks/DurgasriInnamuri/Chapter_3_Semoconductor_Devices_Fundamentals.ipynb
@@ -0,0 +1,223 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 3 Semoconductor Devices Fundamentals"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.2 page no:35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "resistivity of the si doped with n−dopant is : \n",
+      "0.089 ohm−cm \n"
+     ]
+    }
+   ],
+   "source": [
+    "def resistivity(u,n): #n:doped concentration =10**17 atoms/cubic cm, u: mobility of electrons =700square cm/v−sec .\n",
+    "    q=1.6*10**-19 #q: charge\n",
+    "    Res=1/(q*u*n)# since P is neglegible . \n",
+    "    print \"resistivity of the si doped with n−dopant is : \"\n",
+    "    print \"%0.3f ohm−cm \"%Res \n",
+    "resistivity(10**17,700)\n",
+    "# after executing calling resitivity ( u=700 and n =10ˆ17)i .e. , resistivity (10ˆ17 ,700) ;"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.3 page no:35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "resistivity of intrinsic Ge is : \n",
+      "2595245510.225 ohm−cm \n"
+     ]
+    }
+   ],
+   "source": [
+    "def resistivity(un,np):    # un: electron concentration , up: hole concentration\n",
+    "    q=1.6*10**-19          #in coulumb \n",
+    "    ni=2.5*10*13           # concentration in cmˆ−3 \n",
+    "    Res=1/(q*ni*un*np)     # since n=p=ni  \n",
+    "    print \"resistivity of intrinsic Ge is : \"\n",
+    "    print \"%0.3f ohm−cm \"%Res \n",
+    "resistivity(3900,1900)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.4 page no:37"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "hole concentrartion at 300K is : \n",
+      "2250.000000 per cubic cm \n"
+     ]
+    }
+   ],
+   "source": [
+    "def holeconcentration(ni,Nd): # Nd: donar concentration ; since , Nd>>ni , so Nd=n=10ˆ17 atoms/cmˆ3.\n",
+    "    p=ni**2/Nd\n",
+    "    print \"hole concentrartion at 300K is : \"\n",
+    "    print \"%f per cubic cm \"%p\n",
+    "holeconcentration(1.5*10**10,10**17);"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.5 page no:39"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "resistivity of the copper is : \n",
+      "2.29779411765e-08  ohm−meter\n"
+     ]
+    }
+   ],
+   "source": [
+    "q=1.6*10**-19;\n",
+    "n=8.5*10**28;\n",
+    "u=3.2*10**-3;\n",
+    "p=1/(n*q*u);\n",
+    "print \"resistivity of the copper is : \"\n",
+    "print p,\" ohm−meter\"\n",
+    "# 2.298D−08 means 2.298∗10ˆ −8"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.6 page no:41"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cu is:  0.0570814666846  pF\n",
+      "Ccs is:  0.282102806737 pF\n",
+      "gm is : 7.7519379845 mA/V\n",
+      "C1 is:  3.32558139535 pF\n",
+      "R1 is:  25.8  kilo ohm\n",
+      "R0 is  645.0  kilo Ohm \n",
+      "Ru is:  1290.0 Mega Ohm \n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import sqrt\n",
+    "\n",
+    "Cuo=0.25;        # collector −base depletion region capacitance in pico Farad(pF) for zero bias\n",
+    "Ccso=1.5 ;       # collector −substrate junction capacitance in pico Farad(pF) for zero bias\n",
+    "q=1.6*10**-19 ;  # electron charge in coulomb\n",
+    "Ic=0.2 ;         #collector current in ampere(A)\n",
+    "k=8.6*10**-5;    #in eV/K, where 1eV=1.6∗10ˆ−19\n",
+    "T=300;           # absolute temperature in kelvin (K)\n",
+    "Vcb=10 ;         #forward bias on the junction in volt(v)\n",
+    "Vcs=15 ;         # collector −substrate bias in volt (V)\n",
+    "Cje=1 ;          #depletion region capacitance in pico Farad(pF)\n",
+    "Bo=200;          #small signal current gain\n",
+    "Tf=0.3;          #transit time in forward direction in nano seconds (nS)\n",
+    "n=2*10**-4;      # proportionality constant for Ro and gm\n",
+    "Vo=0.55;         # bias voltage in volt (V)\n",
+    "Cu=Cuo/sqrt(1+(Vcb/Vo));# collector −base capacitance\n",
+    "print \"Cu is: \",Cu,\" pF\"\n",
+    "Ccs=Ccso/sqrt(1+(Vcs/Vo)); # capacitance collector −substrate\n",
+    "print \"Ccs is: \",Ccs,\"pF\"\n",
+    "gm=q*Ic/(k*T*1.6*10**-19);# since k is in eV so converting it in Coulomb/Kelvin\n",
+    "print \"gm is :\",gm,\"mA/V\"# transconductance of the bipolar transistor here\n",
+    "Cb=Tf*gm;# diffusion capacitance in pico Farad(pF)\n",
+    "C1=Cb+Cje;#small signal capacitance of bipolar transistor\n",
+    "print \"C1 is: \",C1,\"pF\"\n",
+    "R1=Bo/gm;# small signal input resistance of bipolar transistor\n",
+    "print \"R1 is: \",R1,\" kilo ohm\"\n",
+    "Ro=1/(n*gm);#small signal output resistance\n",
+    "print \"R0 is \",Ro,\" kilo Ohm \"\n",
+    "Ru=10*Bo*Ro/10**3;# collector −base resistance\n",
+    "print \"Ru is: \",Ru,\"Mega Ohm \""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Gopi KrishnaManchukonda/Chapter_2_Electrostatics_.ipynb b/sample_notebooks/Gopi KrishnaManchukonda/Chapter_2_Electrostatics_.ipynb
new file mode 100755
index 00000000..8ece860c
--- /dev/null
+++ b/sample_notebooks/Gopi KrishnaManchukonda/Chapter_2_Electrostatics_.ipynb
@@ -0,0 +1,281 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Electrostatics "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_1 pgno:13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resultant force acting on charge at C= N 12.72\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,sqrt,cos,sin\n",
+    "\n",
+    "epsilon=8.854e-12\n",
+    "r=sqrt(.1**2+.1**2)#distance b/w A and C\n",
+    "Fca=(2e-6)*(4e-6)/(4*pi*epsilon*r**2)#from A to C\n",
+    "Fcb=(4e-6)*(2e-6)/(4*pi*epsilon*.1**2)#from C to B\n",
+    "Fcd=(4e-6)*(4e-6)/(4*pi*epsilon*.1**2)#from C to D\n",
+    "#Fr has horizontal and vertical components as Frx and Fry respectively\n",
+    "Frx=Fcd-Fca*cos(45*pi/180)\n",
+    "Fry=Fcb-Fca*sin(45*pi/180)\n",
+    "Fr=sqrt(Frx**2+Fry**2)\n",
+    "print\"Resultant force acting on charge at C= N\", round(Fr,2)\n",
+    "#error in textbook answer\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_ pgno:15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resultant intensity on charge at C=*10**4 N/C at angle  eegrees 25.44 37.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,cos,sin,sqrt,atan\n",
+    "epsilon=8.854e-12\n",
+    "E1=(4e-8)/(4*pi*epsilon*.05**2)#fiele intensity eue to charge at A,eirection is from e to A\n",
+    "r=sqrt(2*.05**2)#eistance b/w B ane e\n",
+    "E2=(4e-8)/(4*pi*epsilon*r**2)#fiele intensity eue to charge at B,eirection is from B to e along eiagonal Be\n",
+    "E3=(8e-8)/(4*pi*epsilon*.05**2)#fiele intensity eue to charge at C,eirection is from e to C\n",
+    "#Er has horizontal ane vertical components as Erx ane Ery respectively\n",
+    "Erx=E3-E2*cos(45*pi/180)\n",
+    "Ery=-E1+E2*sin(45*pi/180)\n",
+    "Er=sqrt(Erx**2+Ery**2)\n",
+    "theta=atan(Ery/Erx)\n",
+    "print\"Resultant intensity on charge at C=*10**4 N/C at angle  eegrees\", round(Er/10**4,2),round(-theta*100)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_3 pgno:15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Potential at A eue to charges at B, C ane e= V 3.73\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,sqrt\n",
+    "epsilon=8.854e-12\n",
+    "AB=.05\n",
+    "BC=.07\n",
+    "AC=sqrt(.05**2+.07**2)\n",
+    "V1=2e-10/(4*pi*epsilon*.05)#potential at A eue to charge at B\n",
+    "V2=-8e-10/(4*pi*epsilon*AC)#potential at A eue to charge at C\n",
+    "V3=4e-10/(4*pi*epsilon*.07)#potential at A eue to charge at e\n",
+    "V=V1+V2+V3 \n",
+    "print\"Potential at A eue to charges at B, C ane e= V\", round(V,2)\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_4 pgno:16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Time constant T= sec\n",
+      "0.015\n",
+      "Initial current= A\n",
+      "0.2\n",
+      "Charge on the capacitor after 0.05 sec is  C\n",
+      "0.003\n",
+      "Charging current after 0.05 sec is  A\n",
+      "0.007\n",
+      "Charging current after 0.015 sec is  A\n",
+      "0.074\n",
+      "Voltage across 500 ohm resistor after 0.05 sec is  V 3.567\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import exp\n",
+    "C=30e-6\n",
+    "R=500.\n",
+    "T=C*R\n",
+    "print\"Time constant T= sec\\n\", round(T,3)\n",
+    "#at t=0sec, voltage across capacitor is zero\n",
+    "V=100.#apliee voltage\n",
+    "I=V/R#Ohm's Law\n",
+    "print\"Initial current= A\\n\", round(I,3)\n",
+    "t=.05\n",
+    "Q=C*V\n",
+    "q=Q*(1-exp(-t/T))\n",
+    "print\"Charge on the capacitor after 0.05 sec is  C\\n\", round(q,3)\n",
+    "i1=I*exp(-t/T)\n",
+    "print\"Charging current after 0.05 sec is  A\\n\",round(i1,3)\n",
+    "t=.015\n",
+    "i2=I*exp(-t/T)\n",
+    "print\"Charging current after 0.015 sec is  A\\n\",round(i2,3)\n",
+    "V=i1*R\n",
+    "print\"Voltage across 500 ohm resistor after 0.05 sec is  V\", round(V,3)\n",
+    "#answers vary from the textbook eue to roune off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_5 pgno:17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "P.e. across the combination = V\n",
+      "133.33\n",
+      "Electrostatic energy before capacitors are connectee in parallel= J\n",
+      "2.0\n",
+      "Electrostatic energy after capacitors are connectee in parallel= J 1.33\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "C=100e-6\n",
+    "V=200\n",
+    "Q=C*V\n",
+    "Ct=100e-6+50e-6#total capacitance\n",
+    "Vt=Q/Ct\n",
+    "print\"P.e. across the combination = V\\n\", round(Vt,2)\n",
+    "EE1=100e-6*V**2/2\n",
+    "print\"Electrostatic energy before capacitors are connectee in parallel= J\\n\", EE1\n",
+    "EE2=Ct*Vt**2/2\n",
+    "print\"Electrostatic energy after capacitors are connectee in parallel= J\",round( EE2,2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_6 pgno:18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Three capacitors have capacitances  microF,  microF ane  microF\n",
+      "80.0 100.0 120.0\n",
+      "Voltage across the combination = V 50.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "C1=100e-6  #capacitance of first capacitor which is to be chargee\n",
+    "V=200.  #voltage across C1\n",
+    "Q=C1*V\n",
+    "#Let Q1, Q2, Q3, Q4 be the charges on respective capacitors after connection\n",
+    "Q2=4000e-6\n",
+    "Q3=5000e-6\n",
+    "Q4=6000e-6\n",
+    "Q1=Q-(Q2+Q3+Q4)\n",
+    "C2=C1*(Q2/Q1)\n",
+    "C3=C1*(Q3/Q1)\n",
+    "C4=C1*(Q4/Q1)\n",
+    "print\"Three capacitors have capacitances  microF,  microF ane  microF\\n\", C2*10**6,C3*10**6,C4*10**6\n",
+    "Vt=Q1/C1\n",
+    "print\"Voltage across the combination = V\", Vt\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases.ipynb b/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases.ipynb
new file mode 100755
index 00000000..4ab6ea93
--- /dev/null
+++ b/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases.ipynb
@@ -0,0 +1,656 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "speed of oxygen molecule 473.791093 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.1 speed of air molecules\n",
+    "# Example 1# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=300; # temperature 27 deg C, 27+293=300K\n",
+    "M=32; # oxygen is diatomic\n",
+    "v = sqrt(3*R*(T/M));\n",
+    "print \"speed of oxygen molecule %f m/s\" %v\n",
+    "# Note: Value of R is given wrong in book\n",
+    "# So answer in the book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "volume of gas 1.533151e-03 mˆ3 \n",
+      "\n",
+      "total translational kinetic energy is 154.848250 J \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.2 total translational KE\n",
+    "# Example 2# Ch 3\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=298;#in kelvin\n",
+    "M=32; # oxygen is diatomic\n",
+    "m=2*10**-3; # in kg\n",
+    "p=1.01*10**5; # 1 atm=1.01∗10ˆ5 N/m2\n",
+    "G = (m*R*T)/(M*p);#volume of gas\n",
+    "x=(3/2)*p;#no. of molecules per unit volume where x=N∗0.5∗m∗vˆ2 is given as (3/2)∗p)\n",
+    "print\"volume of gas %e mˆ3 \\n\"%G\n",
+    "KE = x*G;#total translational kinetic energy\n",
+    "print\"total translational kinetic energy is %f J \\n\"%KE\n",
+    "# Note: Value of G is calculated in book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.3 pg.no:88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no . of molecules 2.753546e+19\n",
+      "max pressure in chamber 0.117735 N/m2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.3 max pressure in chamber\n",
+    "# Example 3# Ch 3\n",
+    "from math import pi\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=300; # temperature 27 deg C, 27+293=300K\n",
+    "me=0.10; #mean free path in meters\n",
+    "rm=1.7*10**-10 #molecular radius in angstrom\n",
+    "M=28 #im moleˆ−1\n",
+    "m0=4.8*10**-26 #mass of nitrogen molecule\n",
+    "N = 1/(4*pi*((rm)**2)*me); # no. of molecules in gas\n",
+    "print\"no . of molecules %e\"%N \n",
+    "p = ((N*m0)/M)*R*T; # max pressure in chamber in N/ m2\n",
+    "print\"max pressure in chamber %f N/m2\"%p\n",
+    "# Note: Calculation in the book is wrong So answer in the book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.4 pg.no:88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "temperature 7.729469e+03 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.4 temperature at which avg KE of He atoms in gas become 1 eV\n",
+    "# Example 4# Ch 3\n",
+    "# given data\n",
+    "v = 1.6*10**-19; # avg kinetic energy in j\n",
+    "k = 1.38*10**-23 #boltzmann constant in J/K\n",
+    "T = (2*v)/(3*k);\n",
+    "print \"temperature %e K\"%T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.5 pg.no:89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "volume of 1kg of helium is 11.147071 mˆ3\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.5 volume of 1 kg of He\n",
+    "# Example 6# Ch 3\n",
+    "# given data\n",
+    "m = 1;#in kg\n",
+    "M=2.016;#molecular weight of helium\n",
+    "k =8314# gas constant in J/kg \n",
+    "p = 1.01*10**5;\n",
+    "T = 273; # in kelvin\n",
+    "G = m*k*T/(M*p);#volume of 1kg of helium in mˆ3\n",
+    "print\"volume of 1kg of helium is %f mˆ3\"%G"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.6 pg.no:92"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "density of ions at distance equal to the mean free path 0.367879n0\n",
+      "density of ions at distance equal to five times the mean free path 0.006738n0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.6 density of ions at dist equal to mfp and five times mfp\n",
+    "# Example 6# Ch 3\n",
+    "from math import exp\n",
+    "# given data\n",
+    "z1=-1;#ion at a distance equal to mean free path , −x=mfp\n",
+    "z2=-5;#ion at a distance equal to five times the mean f r e e path , −x=5mfp\n",
+    "#n0 is the density of ions at the origin\n",
+    "n1 = exp(z1);#density of ions at distance equal to the mean free path\n",
+    "n2 = exp(z2);#density of ions at distance equal to five times the mean free path\n",
+    "print\"density of ions at distance equal to the mean free path %fn0\"%n1\n",
+    "print\"density of ions at distance equal to five times the mean free path %fn0\"%n2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.7 pg.no:93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mean square velosity of helium atoms 1304.701955 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.7 mean square velocity of He atoms\n",
+    "# Example 7# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "N = 178*10**-3 #gas density in kg/mˆ3\n",
+    "p = 1.01*10**5 # pressure\n",
+    "v = sqrt((3*p)/N); #mean square velosity of helium atoms\n",
+    "print\"mean square velosity of helium atoms %f m/s\"%v"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "energy of free electron 3.790687 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.8 energy of free electrons\n",
+    "#Example 8# Ch 3\n",
+    "# given data\n",
+    "k =1.38e-21; #boltzmanns constant\n",
+    "T = 293; # temperature in K\n",
+    "e = 1.6*10** -19;\n",
+    "E =(1.5*k*T)/e;\n",
+    "print\"energy of free electron %f eV\"%E"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.9 pg.no:95"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no . of atoms/cmˆ3  4.5168e+22\n",
+      "avg vokume occupied by one atom 2.213957e-23 cmˆ3\n",
+      "avg seperation between atoms 1.000000e+00 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.9 avg separation of atoms and avg vol occupied by one atom\n",
+    "#Example 9# Ch 3\n",
+    "# given data\n",
+    "d = 0.075; #density of solid atomic hydrogen in g/cmˆ3\n",
+    "N_A = 6.0224*10**23; #1g of H consists of NA atoms\n",
+    "N = N_A*d; # number of atoms/cmˆ3\n",
+    "print \"no . of atoms/cmˆ3 \",N\n",
+    "x = 1/N;#avg volume occupied by one atom in cmˆ3\n",
+    "y = (x)**(1/3);#avg seperation between atoms in cm\n",
+    "print \"avg vokume occupied by one atom %e cmˆ3\"%x\n",
+    "print \"avg seperation between atoms %e cm\"%y"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.10 pg.no:96"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "photon enegy 62.250000 eV\n",
+      "kinetic energy of photoelectron 48.650000 ev\n",
+      " velosity of photoelectron 4.133874e+06 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.10 KE in eV and velocity of phototelectron\n",
+    "#Example 10# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "l=200*10**-10;# wavelength in angstrom\n",
+    "h=4.15*10**-15;#planks constant\n",
+    "c=3*10**8;#speed of light\n",
+    "me=9.11*10**-31;\n",
+    "BE=13.6;#binding energy in eV\n",
+    "PE=(h*c)/l;# in eV\n",
+    "print\"photon enegy %f eV\"%PE\n",
+    "KE = PE-BE;#in eV\n",
+    "print\"kinetic energy of photoelectron %f ev\"%KE\n",
+    "ve=sqrt((2*KE*1.6*10**-19)/me);\n",
+    "print\" velosity of photoelectron %e m/s\"%ve"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.11 pg.no:98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "absorption coefficient -0.000000 cmˆ−1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.11 liquid photon absorption coefficient\n",
+    "#Example 11# Ch 3\n",
+    "from math import log\n",
+    "# given data\n",
+    "I = 1.;\n",
+    "I0 = 6.;\n",
+    "x=20;#in cm\n",
+    "u = -(1/x)*log(I/I0);\n",
+    "print\"absorption coefficient %f cmˆ−1\"%u"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.12 pg.no:98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "binding energy of gas 12.450000 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.12 binding energy of the gas\n",
+    "#Example 12# Ch 3\n",
+    "# given data\n",
+    "c=3*10**8;\n",
+    "h=4.15*10**-15;\n",
+    "lmax =1000*10** -10;\n",
+    "We=(c*h)/lmax;\n",
+    "print\"binding energy of gas %f eV\"%We"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.14 pg.no:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no of gas molecules 3.527492e+22 atoms/mˆ3\n",
+      "diameter of argon atom 2.769501e-10 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.14 diameter of the argon atom\n",
+    "#Example 14# Ch 3\n",
+    "from math import sqrt,pi\n",
+    "# given data\n",
+    "p=1.01*10**5/760;# 1 torr in N/m2\n",
+    "k=1.38*10**-23;\n",
+    "T=273; # in Kelvin\n",
+    "n=85*10**2;#no of collisions per meter\n",
+    "N=p/(k*T);\n",
+    "print \"no of gas molecules %e atoms/mˆ3\"%N\n",
+    "r_a=sqrt(n/(pi*N*1));\n",
+    "print \"diameter of argon atom %e m\"%r_a"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.15 pg.no:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mobility of electrons 9375.000000 mˆ2/sV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.15 mobility of electrons\n",
+    "#Example 15# Ch 3\n",
+    "# given data\n",
+    "Ie=3;# current flow in amperes\n",
+    "A=8*10**-4;#area of the electrodes in mˆ2\n",
+    "V=20;#voltage across the electrodes\n",
+    "d=0.8;#spacing between the electrodes in meters\n",
+    "n_e=1*10**17;#electron density in mˆ−3\n",
+    "e=1.6*10**-19;\n",
+    "ke=(Ie*d)/(A*V*n_e*e);\n",
+    "print\"mobility of electrons %f mˆ2/sV\"%ke"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.17 pg.no:108"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ion density 0.02m away 1.911913e+09 ions/mˆ3 \n",
+      "\n",
+      "ion density −0.02m away 5.230363e+12 ions/mˆ3 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.17 ion density point 02 m away in both directions at 25 deg C amperes\n",
+    "#Example 17# Ch 3\n",
+    "from math import exp\n",
+    "# given data\n",
+    "E = 5; #electric field in V/m\n",
+    "n_o = 10**11; #ion density in ions/m3\n",
+    "T = 293; # in kelvin\n",
+    "z = 0.02; #distance in meters\n",
+    "e = 1.6*10**-19; #in couloumb\n",
+    "k = 1.38*10**-23; # in m2 kg s−2 K−1\n",
+    "n1 = n_o*exp((-e*E*z)/(k*T));# ion density away \n",
+    "n2 = n_o*exp((e*E*z)/(k*T));# ion density away −0.02m\n",
+    "print\"ion density 0.02m away %e ions/mˆ3 \\n\"%n1\n",
+    "print\"ion density −0.02m away %e ions/mˆ3 \\n\"%n2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.18 pg.no:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "diameter before drift 3.032550e-05 m \n",
+      "\n",
+      "diameter after drift 5.515025e-03 m \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.18 diameter of cloud after drifting a distance of point 05\n",
+    "#Example 18# Ch 3 2 clc;\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "E = 250; #electric field in V/m\n",
+    "r1 = 0.3*10**-3#intial diameter of cloud in meters \n",
+    "k = 1.38*10**-23;#in m2 kg s−2 K−1\n",
+    "T = 293; #in kelvin\n",
+    "e = 1.6*10**-19;# in couloumb\n",
+    "z = 0.05;#drift distance in meters\n",
+    "r = (6*k*T*z)/(e*E);#diameter before drift\n",
+    "print\"diameter before drift %e m \\n\"%r\n",
+    "r2 = sqrt (r1**2 + r );#diamter after drifting a distance\n",
+    "print\"diameter after drift %e m \\n\"%r2 \n",
+    "# round off value calculated for r and r2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.19 pg.no:111"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mean free path of electron in nitrogen 2.221482e-04 m\n",
+      "ionization potential of nitrogen 28.000000 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.19 a mean free path of electrons in nitrogen and b ionization potential of nitrogen\n",
+    "#Example 19# Ch 3 \n",
+    "# given data\n",
+    "a = 9003;#constant in m−1kPa−1 \n",
+    "B = 256584;#in V/m.kPa\n",
+    "p = 0.5;#in kPa\n",
+    "M = 1/(a*p);#mean free path in meters\n",
+    "print\"mean free path of electron in nitrogen %e m\"%M\n",
+    "Vi = B/a; #ionization potential of nitrogen\n",
+    "print\"ionization potential of nitrogen %f V\"%Vi"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases_1.ipynb b/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases_1.ipynb
new file mode 100755
index 00000000..6237f1aa
--- /dev/null
+++ b/sample_notebooks/GundaChaitnaya rani/Chapter_3_Ionization_and_Deionization_Processes_in_gases_1.ipynb
@@ -0,0 +1,663 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 3 Ionization and Deionization Processes in gases"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "speed of oxygen molecule 473.791093 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.1 speed of air molecules\n",
+    "# Example 1# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=300; # temperature 27 deg C, 27+293=300K\n",
+    "M=32; # oxygen is diatomic\n",
+    "v = sqrt(3*R*(T/M));\n",
+    "print \"speed of oxygen molecule %f m/s\" %v\n",
+    "# Note: Value of R is given wrong in book\n",
+    "# So answer in the book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "volume of gas 1.533151e-03 mˆ3 \n",
+      "\n",
+      "total translational kinetic energy is 154.848250 J \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.2 total translational KE\n",
+    "# Example 2# Ch 3\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=298;#in kelvin\n",
+    "M=32; # oxygen is diatomic\n",
+    "m=2*10**-3; # in kg\n",
+    "p=1.01*10**5; # 1 atm=1.01∗10ˆ5 N/m2\n",
+    "G = (m*R*T)/(M*p);#volume of gas\n",
+    "x=(3/2)*p;#no. of molecules per unit volume where x=N∗0.5∗m∗vˆ2 is given as (3/2)∗p)\n",
+    "print\"volume of gas %e mˆ3 \\n\"%G\n",
+    "KE = x*G;#total translational kinetic energy\n",
+    "print\"total translational kinetic energy is %f J \\n\"%KE\n",
+    "# Note: Value of G is calculated in book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.3 pg.no:88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no . of molecules 2.753546e+19\n",
+      "max pressure in chamber 0.117735 N/m2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.3 max pressure in chamber\n",
+    "# Example 3# Ch 3\n",
+    "from math import pi\n",
+    "# given data\n",
+    "R=8314; # gas constant in J/kg . mol .K\n",
+    "T=300; # temperature 27 deg C, 27+293=300K\n",
+    "me=0.10; #mean free path in meters\n",
+    "rm=1.7*10**-10 #molecular radius in angstrom\n",
+    "M=28 #im moleˆ−1\n",
+    "m0=4.8*10**-26 #mass of nitrogen molecule\n",
+    "N = 1/(4*pi*((rm)**2)*me); # no. of molecules in gas\n",
+    "print\"no . of molecules %e\"%N \n",
+    "p = ((N*m0)/M)*R*T; # max pressure in chamber in N/ m2\n",
+    "print\"max pressure in chamber %f N/m2\"%p\n",
+    "# Note: Calculation in the book is wrong So answer in the book is wrong"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.4 pg.no:88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "temperature 7.729469e+03 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.4 temperature at which avg KE of He atoms in gas become 1 eV\n",
+    "# Example 4# Ch 3\n",
+    "# given data\n",
+    "v = 1.6*10**-19; # avg kinetic energy in j\n",
+    "k = 1.38*10**-23 #boltzmann constant in J/K\n",
+    "T = (2*v)/(3*k);\n",
+    "print \"temperature %e K\"%T"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.5 pg.no:89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "volume of 1kg of helium is 11.147071 mˆ3\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.5 volume of 1 kg of He\n",
+    "# Example 6# Ch 3\n",
+    "# given data\n",
+    "m = 1;#in kg\n",
+    "M=2.016;#molecular weight of helium\n",
+    "k =8314# gas constant in J/kg \n",
+    "p = 1.01*10**5;\n",
+    "T = 273; # in kelvin\n",
+    "G = m*k*T/(M*p);#volume of 1kg of helium in mˆ3\n",
+    "print\"volume of 1kg of helium is %f mˆ3\"%G"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.6 pg.no:92"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "density of ions at distance equal to the mean free path 0.367879n0\n",
+      "density of ions at distance equal to five times the mean free path 0.006738n0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.6 density of ions at dist equal to mfp and five times mfp\n",
+    "# Example 6# Ch 3\n",
+    "from math import exp\n",
+    "# given data\n",
+    "z1=-1;#ion at a distance equal to mean free path , −x=mfp\n",
+    "z2=-5;#ion at a distance equal to five times the mean f r e e path , −x=5mfp\n",
+    "#n0 is the density of ions at the origin\n",
+    "n1 = exp(z1);#density of ions at distance equal to the mean free path\n",
+    "n2 = exp(z2);#density of ions at distance equal to five times the mean free path\n",
+    "print\"density of ions at distance equal to the mean free path %fn0\"%n1\n",
+    "print\"density of ions at distance equal to five times the mean free path %fn0\"%n2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.7 pg.no:93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mean square velosity of helium atoms 1304.701955 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.7 mean square velocity of He atoms\n",
+    "# Example 7# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "N = 178*10**-3 #gas density in kg/mˆ3\n",
+    "p = 1.01*10**5 # pressure\n",
+    "v = sqrt((3*p)/N); #mean square velosity of helium atoms\n",
+    "print\"mean square velosity of helium atoms %f m/s\"%v"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.1 pg.no:93"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "energy of free electron 3.790687 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.8 energy of free electrons\n",
+    "#Example 8# Ch 3\n",
+    "# given data\n",
+    "k =1.38e-21; #boltzmanns constant\n",
+    "T = 293; # temperature in K\n",
+    "e = 1.6*10** -19;\n",
+    "E =(1.5*k*T)/e;\n",
+    "print\"energy of free electron %f eV\"%E"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.9 pg.no:95"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no . of atoms/cmˆ3  4.5168e+22\n",
+      "avg vokume occupied by one atom 2.213957e-23 cmˆ3\n",
+      "avg seperation between atoms 1.000000e+00 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.9 avg separation of atoms and avg vol occupied by one atom\n",
+    "#Example 9# Ch 3\n",
+    "# given data\n",
+    "d = 0.075; #density of solid atomic hydrogen in g/cmˆ3\n",
+    "N_A = 6.0224*10**23; #1g of H consists of NA atoms\n",
+    "N = N_A*d; # number of atoms/cmˆ3\n",
+    "print \"no . of atoms/cmˆ3 \",N\n",
+    "x = 1/N;#avg volume occupied by one atom in cmˆ3\n",
+    "y = (x)**(1/3);#avg seperation between atoms in cm\n",
+    "print \"avg vokume occupied by one atom %e cmˆ3\"%x\n",
+    "print \"avg seperation between atoms %e cm\"%y"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.10 pg.no:96"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "photon enegy 62.250000 eV\n",
+      "kinetic energy of photoelectron 48.650000 ev\n",
+      " velosity of photoelectron 4.133874e+06 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.10 KE in eV and velocity of phototelectron\n",
+    "#Example 10# Ch 3\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "l=200*10**-10;# wavelength in angstrom\n",
+    "h=4.15*10**-15;#planks constant\n",
+    "c=3*10**8;#speed of light\n",
+    "me=9.11*10**-31;\n",
+    "BE=13.6;#binding energy in eV\n",
+    "PE=(h*c)/l;# in eV\n",
+    "print\"photon enegy %f eV\"%PE\n",
+    "KE = PE-BE;#in eV\n",
+    "print\"kinetic energy of photoelectron %f ev\"%KE\n",
+    "ve=sqrt((2*KE*1.6*10**-19)/me);\n",
+    "print\" velosity of photoelectron %e m/s\"%ve"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.11 pg.no:98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "absorption coefficient -0.000000 cmˆ−1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.11 liquid photon absorption coefficient\n",
+    "#Example 11# Ch 3\n",
+    "from math import log\n",
+    "# given data\n",
+    "I = 1.;\n",
+    "I0 = 6.;\n",
+    "x=20;#in cm\n",
+    "u = -(1/x)*log(I/I0);\n",
+    "print\"absorption coefficient %f cmˆ−1\"%u"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.12 pg.no:98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "binding energy of gas 12.450000 eV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.12 binding energy of the gas\n",
+    "#Example 12# Ch 3\n",
+    "# given data\n",
+    "c=3*10**8;\n",
+    "h=4.15*10**-15;\n",
+    "lmax =1000*10** -10;\n",
+    "We=(c*h)/lmax;\n",
+    "print\"binding energy of gas %f eV\"%We"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.14 pg.no:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "no of gas molecules 3.527492e+22 atoms/mˆ3\n",
+      "diameter of argon atom 2.769501e-10 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.14 diameter of the argon atom\n",
+    "#Example 14# Ch 3\n",
+    "from math import sqrt,pi\n",
+    "# given data\n",
+    "p=1.01*10**5/760;# 1 torr in N/m2\n",
+    "k=1.38*10**-23;\n",
+    "T=273; # in Kelvin\n",
+    "n=85*10**2;#no of collisions per meter\n",
+    "N=p/(k*T);\n",
+    "print \"no of gas molecules %e atoms/mˆ3\"%N\n",
+    "r_a=sqrt(n/(pi*N*1));\n",
+    "print \"diameter of argon atom %e m\"%r_a"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.15 pg.no:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mobility of electrons 9375.000000 mˆ2/sV\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.15 mobility of electrons\n",
+    "#Example 15# Ch 3\n",
+    "# given data\n",
+    "Ie=3;# current flow in amperes\n",
+    "A=8*10**-4;#area of the electrodes in mˆ2\n",
+    "V=20;#voltage across the electrodes\n",
+    "d=0.8;#spacing between the electrodes in meters\n",
+    "n_e=1*10**17;#electron density in mˆ−3\n",
+    "e=1.6*10**-19;\n",
+    "ke=(Ie*d)/(A*V*n_e*e);\n",
+    "print\"mobility of electrons %f mˆ2/sV\"%ke"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.17 pg.no:108"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ion density 0.02m away 1.911913e+09 ions/mˆ3 \n",
+      "\n",
+      "ion density −0.02m away 5.230363e+12 ions/mˆ3 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.17 ion density point 02 m away in both directions at 25 deg C amperes\n",
+    "#Example 17# Ch 3\n",
+    "from math import exp\n",
+    "# given data\n",
+    "E = 5; #electric field in V/m\n",
+    "n_o = 10**11; #ion density in ions/m3\n",
+    "T = 293; # in kelvin\n",
+    "z = 0.02; #distance in meters\n",
+    "e = 1.6*10**-19; #in couloumb\n",
+    "k = 1.38*10**-23; # in m2 kg s−2 K−1\n",
+    "n1 = n_o*exp((-e*E*z)/(k*T));# ion density away \n",
+    "n2 = n_o*exp((e*E*z)/(k*T));# ion density away −0.02m\n",
+    "print\"ion density 0.02m away %e ions/mˆ3 \\n\"%n1\n",
+    "print\"ion density −0.02m away %e ions/mˆ3 \\n\"%n2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.18 pg.no:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "diameter before drift 3.032550e-05 m \n",
+      "\n",
+      "diameter after drift 5.515025e-03 m \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.18 diameter of cloud after drifting a distance of point 05\n",
+    "#Example 18# Ch 3 2 clc;\n",
+    "from math import sqrt\n",
+    "# given data\n",
+    "E = 250; #electric field in V/m\n",
+    "r1 = 0.3*10**-3#intial diameter of cloud in meters \n",
+    "k = 1.38*10**-23;#in m2 kg s−2 K−1\n",
+    "T = 293; #in kelvin\n",
+    "e = 1.6*10**-19;# in couloumb\n",
+    "z = 0.05;#drift distance in meters\n",
+    "r = (6*k*T*z)/(e*E);#diameter before drift\n",
+    "print\"diameter before drift %e m \\n\"%r\n",
+    "r2 = sqrt (r1**2 + r );#diamter after drifting a distance\n",
+    "print\"diameter after drift %e m \\n\"%r2 \n",
+    "# round off value calculated for r and r2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 3.7.19 pg.no:111"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "mean free path of electron in nitrogen 2.221482e-04 m\n",
+      "ionization potential of nitrogen 28.000000 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 3.7.19 a mean free path of electrons in nitrogen and b ionization potential of nitrogen\n",
+    "#Example 19# Ch 3 \n",
+    "# given data\n",
+    "a = 9003;#constant in m−1kPa−1 \n",
+    "B = 256584;#in V/m.kPa\n",
+    "p = 0.5;#in kPa\n",
+    "M = 1/(a*p);#mean free path in meters\n",
+    "print\"mean free path of electron in nitrogen %e m\"%M\n",
+    "Vi = B/a; #ionization potential of nitrogen\n",
+    "print\"ionization potential of nitrogen %f V\"%Vi"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/HirenShah/ch1.ipynb b/sample_notebooks/HirenShah/ch1.ipynb
new file mode 100755
index 00000000..e4371040
--- /dev/null
+++ b/sample_notebooks/HirenShah/ch1.ipynb
@@ -0,0 +1,377 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3303c10fbdf0badf34a5e53239631d65156a6198f60e9dc97e75e274501b7ad0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1  Page No : 3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Given:\n",
+      "n  =  4.      #Number of cylinders\n",
+      "d  =  68./10      #Bore in cm\n",
+      "l  =  75./10      #Stroke in cm\n",
+      "r  =  8.      #Compression ratio\n",
+      "\n",
+      "#Solution:\n",
+      "V_s  =  (math.pi/4)*d**2*l      #Swept volume of one cylinder in cm**3\n",
+      "cubic_capacity  =  n*V_s      #Cubic capacity in cm**3\n",
+      "#Since, r   =   (V_c + V_s)/V_c\n",
+      "V_c  =  V_s/(r-1)      #Clearance volume in cm**3\n",
+      "\n",
+      "#Results:\n",
+      "print \" The cubic capacity of the engine   =   %.1f cm**3\"%(cubic_capacity)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The cubic capacity of the engine   =   1089.5 cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2  Page No : 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given:\n",
+      "ip  =  10.      #Indicated power in kW\n",
+      "eta_m  =  80.      #Mechanical efficiency in percent\n",
+      "\n",
+      "#Solution:\n",
+      "#Since, eta_m   =   bp/ip\n",
+      "bp  =  (eta_m/100)*ip      #Brake power in kW\n",
+      "fp  =  ip-bp      #Friction power in kW\n",
+      "\n",
+      "#Results:\n",
+      "print \" The brake power delivered, bp   =   %d kW\"%(bp)\n",
+      "print \" The friction power, fp   =   %d kW\"%(fp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The brake power delivered, bp   =   8 kW\n",
+        " The friction power, fp   =   2 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3  Page No : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#Given:\n",
+      "bp  =  100.      #Brake power at full load in kW\n",
+      "fp  =  25.      #Frictional power in kW (printing error)\n",
+      "\n",
+      "#Solution:\n",
+      "eta_m  =  bp/(bp+fp)      #Mechanical efficiency at full load\n",
+      "#(a)At half load\n",
+      "bp  =  bp/2      #Brake power at half load in kW\n",
+      "eta_m1  =  bp/(bp+fp)      #Mechanical efficiency at half load\n",
+      "#(b)At quarter load\n",
+      "bp  =  bp/2      #Brake power at quarter load in kW\n",
+      "eta_m2  =  bp/(bp+fp)      #Mechanical efficiency at quarter load\n",
+      "\n",
+      "#Results:\n",
+      "print \" The mechanical efficiency at full load, eta_m   =   %d percent\"%(eta_m*100)\n",
+      "print \" The mechanical efficiency, \\\n",
+      "\\na)At half load, eta_m   =   %.1f percent \\\n",
+      "\\nb)At quarter load, eta_m   =   %d percent\"%(eta_m1*100,eta_m2*100)\n",
+      "\n",
+      "#Data in the book is printed wrong\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The mechanical efficiency at full load, eta_m   =   80 percent\n",
+        " The mechanical efficiency, \n",
+        "a)At half load, eta_m   =   66.7 percent \n",
+        "b)At quarter load, eta_m   =   50 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4  Page No : 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#Calculations on four stroke petrol engine\n",
+      "#Given:\n",
+      "bp  =  35.      #Brake power in kW\n",
+      "eta_m  =  80.      #Mechanical efficiency in percent\n",
+      "bsfc  =  0.4      #Brake specific fuel consumption in kg/kWh\n",
+      "A_F  =  14./1      #Air-fuel ratio\n",
+      "CV  =  43000.      #Calorific value in kJ/kg\n",
+      "\n",
+      "#Solution:\n",
+      "#(a)\n",
+      "ip  =  bp*100/eta_m      #Indicated power in kW\n",
+      "#(b)\n",
+      "fp  =  ip-bp      #Frictional power in kW\n",
+      "#(c)\n",
+      "#Since, 1 kWh   =   3600 kJ\n",
+      "eta_bt  =  1/(bsfc*CV/3600)      #Brake thermal efficiency\n",
+      "#(d)\n",
+      "eta_it  =  eta_bt/eta_m*100      #Indicated thermal efficiency\n",
+      "#(e)\n",
+      "m_f  =  bsfc*bp      #Fuel consumption in kg/hr\n",
+      "#(f)\n",
+      "m_a  =  A_F*m_f      #Air consumption in kg/hr\n",
+      "\n",
+      "\n",
+      "#Results:\n",
+      "print \" a)The indicated power, ip   =   %.2f kW \\\n",
+      "\\nb)The friction power, fp   =   %.2f kW\"%(ip,fp)\n",
+      "print \" c)The brake thermal efficiency, eta_bt   =   %.1f percent \\\n",
+      "\\nd)The indicated thermal efficiency, eta_it   =   %.1f percent\"%(eta_bt*100,eta_it*100)\n",
+      "print \" e)The fuel consumption per hour, m_f   =   %.1f kg/hr \\\n",
+      "\\nf)The air consumption per hour, m_a   =   %d kg/hr\"%(m_f,m_a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a)The indicated power, ip   =   43.75 kW \n",
+        "b)The friction power, fp   =   8.75 kW\n",
+        " c)The brake thermal efficiency, eta_bt   =   20.9 percent \n",
+        "d)The indicated thermal efficiency, eta_it   =   26.2 percent\n",
+        " e)The fuel consumption per hour, m_f   =   14.0 kg/hr \n",
+        "f)The air consumption per hour, m_a   =   196 kg/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5  Page No : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given:\n",
+      "F_A  =  0.07/1      #Fuel-air ratio\n",
+      "bp  =  75.      #Brake power in kW\n",
+      "eta_bt  =  20.      #Brake thermal efficiency in percent\n",
+      "rho_a  =  1.2      #Density of air in kg/m**3\n",
+      "rho_f  =  4*rho_a      #Density of fuel vapour in kg/m**3\n",
+      "CV  =  43700.      #Calorific value of fuel in kJ/kg\n",
+      "     \n",
+      "#Solution:\n",
+      "m_f  =  bp*3600/(eta_bt*CV/100)      #Fuel consumption in kg/hr\n",
+      "m_a  =  m_f/F_A      #Air consumption in kg/hr\n",
+      "V_a  =  m_a/rho_a      #Volume of air in m**3/hr\n",
+      "V_f  =  m_f/rho_f      #Volume of fuel in m**3/hr\n",
+      "V_mixture  =  V_f+V_a      #Mixture volume in m**3/hr\n",
+      "     \n",
+      "#Results:\n",
+      "print \" The air consumption, m_a   =   %.1f kg/hr\"%(m_a)\n",
+      "print \" The volume of air required, V_a   =   %.1f m**3/hr\"%(V_a)\n",
+      "print \" The volume of mixture required   =   %.1f m**3/hr\"%(V_mixture)      #printing error)\n",
+      "     #Answer in the book is printed wrong\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The air consumption, m_a   =   441.3 kg/hr\n",
+        " The volume of air required, V_a   =   367.8 m**3/hr\n",
+        " The volume of mixture required   =   374.2 m**3/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6  Page No : 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Given:\n",
+      "bp  =  5.      #Brake power in kW\n",
+      "eta_it  =  30.      #Indicated thermal efficiency in percent\n",
+      "eta_m  =  75.      #Mechanical efficiency in percent (printing error)\n",
+      "\n",
+      "#Solution:\n",
+      "ip  =  bp*100/eta_m      #Indicated power in kW\n",
+      "CV  =  42000.      #Calorific value of diesel(fuel) in kJ/kg\n",
+      "m_f  =  ip*3600/(eta_it*CV/100)      #Fuel consumption in kg/hr\n",
+      "#Density of diesel(fuel)   =   0.87 kg/l\n",
+      "rho_f  =  0.87      #Density of fuel in kg/l\n",
+      "V_f  =  m_f/rho_f      #Fuel consumption in l/hr\n",
+      "isfc  =  m_f/ip      #Indicated specific fuel consumption in kg/kWh\n",
+      "bsfc  =  m_f/bp      #Brake specific fuel consumption in kg/kWh\n",
+      "\n",
+      "#Results:\n",
+      "print \" The fuel consumption of engine, m_f in, \\\n",
+      "\\na)kg/hr   =   %.3f kg/hr \\\n",
+      "\\nb)litres/hr   =   %.2f l/hr\"%(m_f,V_f)\n",
+      "print \" c)Indicated specific fuel consumption, isfc   =   %.3f kg/kWh\"%(isfc)\n",
+      "print \" d)Brake specific fuel consumption, bsfc   =   %.3f kg/kWh\"%(bsfc)\n",
+      "#Data in the book is printed wrong\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fuel consumption of engine, m_f in, \n",
+        "a)kg/hr   =   1.905 kg/hr \n",
+        "b)litres/hr   =   2.19 l/hr\n",
+        " c)Indicated specific fuel consumption, isfc   =   0.286 kg/kWh\n",
+        " d)Brake specific fuel consumption, bsfc   =   0.381 kg/kWh\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7  Page No : 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Given:\n",
+      "bp  =  5000.      #Brake power in kW\n",
+      "fp  =  1000.      #Friction power in kW\n",
+      "m_f  =  2300.      #Fuel consumption in kg/hr\n",
+      "A_F  =  20./1      #Air-fuel ratio\n",
+      "CV  =  42000.      #Calorific value of fuel in kJ/kg\n",
+      "\n",
+      "#Solution:\n",
+      "#(a)\n",
+      "ip  =  bp+fp      #Indicated power in kW\n",
+      "#(b)\n",
+      "eta_m  =  bp/ip      #Mechanical efficiency\n",
+      "#(c)\n",
+      "m_a  =  A_F*m_f      #Air consumption in kg/hr\n",
+      "#(d)\n",
+      "eta_it  =  ip*3600/(m_f*CV)      #Indicated thermal efficiency\n",
+      "#(e)\n",
+      "eta_bt  =  eta_it*eta_m      #Brake thermal efficiency\n",
+      "\n",
+      "#Results:\n",
+      "print \" a)The indicated power, ip   =   %d kW\"%(ip)\n",
+      "print \" b)The mechanical efficiency, eta_m   =   %d percent\"%(eta_m*100)\n",
+      "print \" c)The air consumption, m_a   =   %d kg/hr\"%(m_a)\n",
+      "print \" d)The indicated thermal efficiency, eta_it   =   %.1f percent \\\n",
+      "\\ne)The brake thermal efficiency, eta_bt   =   %.1f percent\"%(eta_it*100,eta_bt*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a)The indicated power, ip   =   6000 kW\n",
+        " b)The mechanical efficiency, eta_m   =   83 percent\n",
+        " c)The air consumption, m_a   =   46000 kg/hr\n",
+        " d)The indicated thermal efficiency, eta_it   =   22.4 percent \n",
+        "e)The brake thermal efficiency, eta_bt   =   18.6 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/InnamuriBhavitha/Chapter_1_CRYSTAL_STRUCTURES.ipynb b/sample_notebooks/InnamuriBhavitha/Chapter_1_CRYSTAL_STRUCTURES.ipynb
new file mode 100755
index 00000000..3369137f
--- /dev/null
+++ b/sample_notebooks/InnamuriBhavitha/Chapter_1_CRYSTAL_STRUCTURES.ipynb
@@ -0,0 +1,234 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 CRYSTAL STRUCTURES"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_4 pgno:22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1.0\n",
+      "r=a/2 =  0.5\n",
+      "Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)=  0.523598775598\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.0\n",
+      "Fp(S.C)=(v∗100/V)=  52.3598775598\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.4\n",
+    "from math import pi\n",
+    "a=1.\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=a/2.\n",
+    "print \"r=a/2 = \",r # initializing value of radius of atom for simple cubic .\n",
+    "v=((4*pi*(r**3))/3)\n",
+    "print \"Volume of one atom ,v=((4∗%pi∗(rˆ3))/3)= \",v # calcuation . \n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(S.C)=(v∗100/V)= \",Fp,# calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1_5 pgno:24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1.0\n",
+      "Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) =  0.433012701892\n",
+      "Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 =  0.680174761588\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.0\n",
+      "Fp(B.C.C)=(v∗100/V)=  68.0174761588 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.5\n",
+    "from math import sqrt\n",
+    "a=1.\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=(sqrt(3)*(a**2/4))\n",
+    "print \"Radius of the atoms,r=(sqrt(3)∗(aˆ2/4)) = \",r # initializing value of radius of atom for BCC.\n",
+    "v=((4*pi*(r**3))/3)*2\n",
+    "print \"Volume of two atom,v=((4∗pi∗(rˆ3))/3)∗2 = \",v # calcuation \n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(B.C.C)=(v∗100/V)= \",Fp,\"%\" # calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_6 pgno:25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1\n",
+      "Radius of the atom,r=(a/(2∗sqrt(2)))=  0.353553390593\n",
+      "Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)=  0.740480489693\n",
+      "Total volume of the cube ,V=aˆ3=  2\n",
+      "Fp(F.C.C)=(v∗100/V)=  37.0240244847 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.6\n",
+    "a=1\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=(a/(2*sqrt(2)))\n",
+    "print \"Radius of the atom,r=(a/(2∗sqrt(2)))= \",r  # initializing value of radius of atom for FCC .\n",
+    "v=(((4*pi*(r**3))/3)*4)\n",
+    "print \"Volume of the four atom,v=(((4∗pi∗(rˆ3))/3)∗4)= \",v # calcuation \n",
+    "V=a^3\n",
+    "print \"Total volume of the cube ,V=aˆ3= \",V  #  calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(F.C.C)=(v∗100/V)= \",Fp,\"%\"  # calculation\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_8 pgno:26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a=  1\n",
+      "Radius of the atom , r=(sqrt (3)∗a/8))=  0.216506350946\n",
+      "v=(((4∗pi∗(rˆ3))/3)∗8) =  0.340087380794\n",
+      "V=aˆ3=  2\n",
+      "Fp(Diamond)=(v∗100/V) =  17.0043690397 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 1.8 \n",
+    "a=1\n",
+    "print \"a= \",a # initializing value of lattice constant(a)=1.\n",
+    "r=((sqrt(3)*a/8))\n",
+    "print \"Radius of the atom , r=(sqrt (3)∗a/8))= \",r # initializing value of radius of atom for diamond .\n",
+    "v=(((4*pi*(r**3))/3)*8)\n",
+    "print \"v=(((4∗pi∗(rˆ3))/3)∗8) = \",v # calcuation .\n",
+    "V=a^3\n",
+    "print \"V=aˆ3= \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(Diamond)=(v∗100/V) = \",Fp,\"%\" # calculation\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1_9 pgno:28"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a =  5e-08  cm\n",
+      "Radius of the atom,r=(sqrt(3)∗(a/4))=  2.16506350946e-08\n",
+      "Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2=  8.50218451985e-23\n",
+      "Total Volume of the cube ,V=aˆ3 =  1.25e-22\n",
+      "Fp(B.C.C)=(v∗100/V) =  68.0174761588 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "#exa 1.9\n",
+    "a=5*10**-8\n",
+    "print \"a = \",a,\" cm\" # initializing value of lattice constant .\n",
+    "r=(sqrt(3)*(a/4))\n",
+    "print \"Radius of the atom,r=(sqrt(3)∗(a/4))= \",r  # initializing value of radius of atom for BCC.\n",
+    "v=((4*pi*(r**3))/3)*2\n",
+    "print \"Volume of the two atoms ,v=((4∗pi∗(rˆ3))/3)∗2= \",v # calcuation .\n",
+    "V=a**3\n",
+    "print \"Total Volume of the cube ,V=aˆ3 = \",V # calcuation .\n",
+    "Fp=(v*100/V)\n",
+    "print \"Fp(B.C.C)=(v∗100/V) = \",Fp,\"%\"  #  calculation"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/KavinkumarD/Chapter_11__Impulse_and_Reaction_Turbines.ipynb b/sample_notebooks/KavinkumarD/Chapter_11__Impulse_and_Reaction_Turbines.ipynb
new file mode 100644
index 00000000..5415ad01
--- /dev/null
+++ b/sample_notebooks/KavinkumarD/Chapter_11__Impulse_and_Reaction_Turbines.ipynb
@@ -0,0 +1,345 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:38f9fe4fd8a5c174c9e1dd9b5dc21976f4cdd814f7eb8fcfe0c266e278f9a77b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Impulse and Reaction Turbines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 and Page No:454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "p02=6; # Inlet pressure in bar\n",
+      "T02=900; # Inlet temperature in kelvin\n",
+      "p0fs=1; # Outlet pressure in bar\n",
+      "eff_isenT=0.85; # insentropic efficiency of turbine\n",
+      "alpha_2=math.radians(75); # Nozzle outlet angle in degree and conversion to radians\n",
+      "u=250; # Mean blade velocity in m/s\n",
+      "Cp=1.15*10**3; # Specific heat in J/ kg K\n",
+      "r=1.333; # Specific heat ratio\n",
+      "\n",
+      "#Calculations\n",
+      "T0fs=T02/(p02/p0fs)**((r-1)/r); # Isentropic temperature at the exit of the final stage\n",
+      "Del_Toverall=eff_isenT*(T02-T0fs); # Actual overall temperature drop\n",
+      "c2=2*u/math.sin (alpha_2); # absolute velocity\n",
+      "c3= c2*math.cos (alpha_2);# absolute velocity\n",
+      "c1=c3; # From velocity triangles\n",
+      "Del_Tstage=(c2**2-c1**2)/(2*Cp); # Stage temperature drop\n",
+      "n=Del_Toverall/Del_Tstage; # Number of stages\n",
+      "\n",
+      "#Results\n",
+      "print \"Number of stages n =\",round (n,0);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of stages n = 3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 and Page No:455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "N=10000; # Speed of gas turbine in rpm\n",
+      "T01=700+273.15; # Total head temperature at nozzle entry in kelvin\n",
+      "P01=4.5; #Total head pressure at nozzle entry in bar\n",
+      "P02=2.6; # Outlet pressure from nozzle in bar\n",
+      "p3=1.5;# Pressure at trbine outlet annulus in bar\n",
+      "M=0.5; # Mach number at outlet\n",
+      "alpha_2=math.radians(70); # outlet nozzle angle in degrees and conversion to radians\n",
+      "D=64; # Blade mean diameter in cm\n",
+      "m=22.5; # Mass flow rate in kg/s\n",
+      "eff_T=0.99; # turbine mechanical efficiency\n",
+      "Cp=1.147; # Specific heat in kJ/kg K\n",
+      "r=1.33; # Specific heat ratio\n",
+      "fl=0.03; # frictional loss\n",
+      "R=284.6; # characteristic gas constant in J/kg K\n",
+      "\n",
+      "#Calculations\n",
+      "eff_N=1-fl; # Nozzle efficiency\n",
+      "T_02=(P02/P01)**((r-1)/r)*T01; # Isentropic temperature after expansion\n",
+      "T02=T01-eff_N*(T01-T_02); # Actual temperature after expansion\n",
+      "c2=math.sqrt (2*Cp*10**3*(T01-T02)); # Absolute velocity\n",
+      "u=(3.14*D*10**-2*N)/60; # Mean blade velocity\n",
+      "# From velocity triangles\n",
+      "wt2=c2*math.sin( (alpha_2))-u;\n",
+      "ca=c2*math.cos( (alpha_2));\n",
+      "beta_2=(math.atan((wt2)/ca));\n",
+      "T3=T02/(P02/p3)**((r-1)/r); # Assuming rotor losses are negligible\n",
+      "c3=M*math.sqrt (r*R*T3); # Absolute velocity\n",
+      "beta_3=(math.atan(u/c3));\n",
+      "ct2=c2*math.sin((alpha_2));\n",
+      "P=eff_T*m*(ct2)*u/1000; # Power developed\n",
+      "\n",
+      "#Results\n",
+      "print \"(i).\"\n",
+      "print \"\\tGas angle at entry = \",round (math.degrees(beta_2),3),\"degree\"\n",
+      "print \"\\tGas angle at exit = \",round (math.degrees(beta_3),3),\"degree\"\n",
+      "print \"(ii).\"\n",
+      "print \"\\tPower developed = \",round(P,3),\"kW   (roundoff error)\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i).\n",
+        "\tGas angle at entry =  41.411 degree\n",
+        "\tGas angle at exit =  51.609 degree\n",
+        "(ii).\n",
+        "\tPower developed =  3680.184 kW   (roundoff error)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 and Page No:457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "alpha_2=math.radians(65); # Nozzle discharge angle in degree and conversion to radians\n",
+      "c3=300; # Absolute velocity in m/s\n",
+      "alpha_3=math.radians(30); # in degrees and conversion to radians\n",
+      "\n",
+      "#Calculations\n",
+      "ca2=c3*math.cos (alpha_3); # Axial velocity\n",
+      "c2=ca2/math.cos(alpha_2); #  Absolute velocity\n",
+      "# ca3=ca2=ca and equal blade angles then\n",
+      "ca=ca2;\n",
+      "beta_2=math.atan((c2*math.sin(alpha_2)+c3*math.sin(alpha_3))/(2*ca)); # Blade angle\n",
+      "beta_3=beta_2; # equal blade angles\n",
+      "u=c2*math.sin(alpha_2)-ca2*math.tan(beta_2); # Mean blade velocity\n",
+      "# From velocity triangles\n",
+      "ct2=c2*math.sin(alpha_2);\n",
+      "ct3=c3*math.sin(alpha_3);\n",
+      "WT=u*(ct2+ct3)/1000; # Work done\n",
+      "sigma=u/c2; # optimum speed ratio\n",
+      "eff_B=4*(sigma*math.sin(alpha_2)-sigma**2);\n",
+      "\n",
+      "#Results\n",
+      "print \"Blade angle = beta_2= beta_3 = \",round (math.degrees(beta_2),3),\"degree\"\n",
+      "print \"Power Produced = \",round(WT,3),\"kJ/kg   (roundoff error)\"\n",
+      "print \"Blade efficiency = \",round(eff_B*100,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Blade angle = beta_2= beta_3 =  53.692 degree\n",
+        "Power Produced =  143.963 kJ/kg   (roundoff error)\n",
+        "Blade efficiency =  76.19 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 and Page No:458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "P01=7; # Pressure at inlet in bar\n",
+      "T01=300+273.15; # Temperature at inlet in kelvin\n",
+      "P02=3; # Pressure at outlet in bar\n",
+      "alpha_2=math.radians(70); # Nozzle angle in degree and conversion to radians\n",
+      "eff_N=0.9; # Isentropic efficiency of nozzle\n",
+      "WT=75; # Power Produced in kW\n",
+      "Cp=1.15; # Specific heat in kJ/kg K\n",
+      "r=1.33; # Specific heat ratio\n",
+      "\n",
+      "#Calculations\n",
+      "T_02=T01*(P02/P01)**((r-1)/r); # Isentropic temperature after expansion\n",
+      "T02=T01-eff_N*(T01-T_02); # Actual temperature after expansion\n",
+      "c2=math.sqrt (2*Cp*10**3*(T01-T02)); # Absolute velocity\n",
+      "# For optimum blade speed ratio\n",
+      "u=(c2*math.sin (alpha_2)/2); # Mean blade velocity\n",
+      "beta_2=math.atan((c2*math.sin(alpha_2)-u)/(c2*math.cos(alpha_2))); # Blade angle\n",
+      "# From velocity triangles\n",
+      "ct2=c2*math.sin(alpha_2);\n",
+      "w2=c2*math.cos(alpha_2)/math.cos(beta_2);\n",
+      "w3=w2; # Equal inlet and outlet angles\n",
+      "beta_3=54; # in degrees\n",
+      "ct3=w3*math.sin(beta_3)-u;\n",
+      "m=(WT*10**3)/(u*(ct2+ct3)); # Gas mass flow rate\n",
+      "\n",
+      "#Results\n",
+      "print \"Blade angle = \",round(math.degrees(beta_2),3),\"degree\"\n",
+      "print \"Gas Mass Flow Rate = \",round(m,3),\"kg/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Blade angle =  53.948 degree\n",
+        "Gas Mass Flow Rate =  4.89 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 and Page No:460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from __future__ import division\n",
+      "\n",
+      "#Variable declaration\n",
+      "P01=4.6; # Total head inlet pressure in bar\n",
+      "T01=700+273.15; # Total head inlet temperature in kelvin\n",
+      "P2=1.6; # Static head pressure at mean radius in bar\n",
+      "Dm_h=10; # Mean blade diameter/blade height\n",
+      "lc=0.1; # Nozzle losses coefficient\n",
+      "alpha_2=math.radians(60); # Nozzle outlet angle in degree and conversion to radians\n",
+      "Cp=1.147; # Specific heat in kJ/kg K\n",
+      "r=1.33; # Specific heat ratio\n",
+      "m=20; # Mass flow rate in kg/s\n",
+      "R=284.6; # characteristic gas constant in J/kg K\n",
+      "\n",
+      "#Calculations\n",
+      "T_2=T01*(P2/P01)**((r-1)/r); # Isentropic temperature after expansion\n",
+      "T2=(lc*T01+T_2)/(1+lc); # Actual temperature after expansion\n",
+      "c2=math.sqrt(2*Cp*10**3*(T01-T2)); # Absolute velocity\n",
+      "# From velocity triangles\n",
+      "ca=c2*math.cos(alpha_2);\n",
+      "row=P2*10**5/(R*T2); # Density of gas\n",
+      "A=m/(ca*row); # Area\n",
+      "Dm=math.sqrt (A*Dm_h/3.14); # Mean Diameter\n",
+      "h=Dm/10; # Blade height\n",
+      "rm=Dm/2; # Mean radius\n",
+      "# At root\n",
+      "r_root=(Dm-h)/2;\n",
+      "#At the tip\n",
+      "r_tip=(Dm+h)/2;\n",
+      "# Free vorte flow\n",
+      "ct_mean=c2*math.sin (alpha_2);\n",
+      "# At the root\n",
+      "ct2_root=(ct_mean*rm)/r_root;\n",
+      "alpha2_root=math.atan(ct2_root/ca);\n",
+      "c2_root=ct2_root/math.sin (alpha2_root);\n",
+      "T2_root=T01-c2_root**2/(2*Cp*10**3);\n",
+      "# At the tip\n",
+      "ct2_tip=ct_mean*rm/r_tip;\n",
+      "alpha2_tip = math.atan (ct2_tip/ca);\n",
+      "c2_tip=ct2_tip/math.sin(alpha2_tip);\n",
+      "T2_tip=T01-c2_tip**2/(2*Cp*10**3);\n",
+      "\n",
+      "#Results\n",
+      "print \"A the Root\"\n",
+      "print \"\\tGas Temperature at the root = \",round(T2_root,3),\"K\"\n",
+      "print \"\\tGas velocity at the root = \",round(c2_root,3),\"m/s\"\n",
+      "print \"\\tDischarge angle at the root = \",round(math.degrees(alpha2_root),3),\"degree\"\n",
+      "print \"\\nA the Tip\"\n",
+      "print \"\\tGas Temperature at the tip = \",round(T2_tip,3),\"K\"\n",
+      "print \"\\tGas velocity at the tip = \",round(c2_tip,3),\"m/s\"\n",
+      "print \"\\tDischarge angle at the tip = \",round(math.degrees(alpha2_tip),3),\"degree\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A the Root\n",
+        "\tGas Temperature at the root =  733.345 K\n",
+        "\tGas velocity at the root =  741.696 m/s\n",
+        "\tDischarge angle at the root =  62.543 degree\n",
+        "\n",
+        "A the Tip\n",
+        "\tGas Temperature at the tip =  795.766 K\n",
+        "\tGas velocity at the tip =  637.902 m/s\n",
+        "\tDischarge angle at the tip =  57.581 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/KhushbuPattani/chapter1.ipynb b/sample_notebooks/KhushbuPattani/chapter1.ipynb
new file mode 100644
index 00000000..f014e773
--- /dev/null
+++ b/sample_notebooks/KhushbuPattani/chapter1.ipynb
@@ -0,0 +1,53 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Solution of Equation & Curve Fitting"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1.1, page no. 12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "import numpy\n",
+    "\n",
+    "p = 2*(x^3)+x^2-13*x+6\n",
+    "print \"The roots of above equation are: \", numpy.roots([2,1,-13,6])"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/ManchukondaMaruthi Naga Vijaya Durga/Chapter_2_Generalized_Configurations_and_Functional_Descriptions_of_Measuring_Instruments.ipynb b/sample_notebooks/ManchukondaMaruthi Naga Vijaya Durga/Chapter_2_Generalized_Configurations_and_Functional_Descriptions_of_Measuring_Instruments.ipynb
new file mode 100755
index 00000000..2cb57529
--- /dev/null
+++ b/sample_notebooks/ManchukondaMaruthi Naga Vijaya Durga/Chapter_2_Generalized_Configurations_and_Functional_Descriptions_of_Measuring_Instruments.ipynb
@@ -0,0 +1,109 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Generalized Configurations and Functional Descriptions of Measuring Instruments"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_1 pgno:22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "ts=0.1\n",
+      "ps=2.5\n",
+      "dT=20\n",
+      "the error in measurement is d percent\n",
+      "0.007\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Caption_Error in measurement\n",
+    "#Ex_1 part_2 #page 22\n",
+    "print (\"ts=0.1\")\n",
+    "print (\"ps=2.5\")\n",
+    "print (\"dT=20\")\n",
+    "\n",
+    "ts=0.1   #('enter the temperature sensitivity=:')\n",
+    "ps=2.5   #('enter the pressure sensitivity(in units/MPa)=:')\n",
+    "dT=20   #('enter the temperature change during pressure measurement=:')\n",
+    "P=120   #('enter the pressure to be measured (in MPa)=:')\n",
+    "error=(ts*dT)/(ps*P);\n",
+    "print'the error in measurement is d percent\\n',round(error,3) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_2 pgno:23"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the error in measurement is d percent\n",
+      "0.017\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Caption_Error in measurement\n",
+    "#Ex_2 part_2 #page 23\n",
+    "\n",
+    "\n",
+    "ts=0.5   #('enter the temperature sensitivity=:')\n",
+    "ps=7.5   #('enter the pressure sensitivity(in units/MPa)=:')\n",
+    "dT=40   #('enter the temperature change during pressure measurement=:')\n",
+    "P=160   #('enter the pressure to be measured (in MPa)=:')\n",
+    "error=(ts*dT)/(ps*P);\n",
+    "print'the error in measurement is d percent\\n',round(error,3) "
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/MandalaManoj  pruthvi/Chapter_4_Radian_Measure.ipynb b/sample_notebooks/MandalaManoj  pruthvi/Chapter_4_Radian_Measure.ipynb
new file mode 100755
index 00000000..212743ac
--- /dev/null
+++ b/sample_notebooks/MandalaManoj  pruthvi/Chapter_4_Radian_Measure.ipynb
@@ -0,0 +1,658 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Radian Measure"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.1 page.no:96"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Radian measure is 0.314159 rad\n",
+      "(or)\n",
+      "Radian measure is (pi/10)rad\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To convert a degree measure to radians\n",
+    "from math import pi\n",
+    "\n",
+    "deg=18 # degree measure\n",
+    "radian=deg*(pi/180) # radian measure\n",
+    "print \"Radian measure is %f rad\\n(or)\"%radian\n",
+    "print \"Radian measure is (pi/%.0f)rad\"%(1/(radian/pi))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.2 page.no:96"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Degree measure is 20 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To convert a radian meeasure to degree\n",
+    "from math import pi\n",
+    "\n",
+    "radian=pi/9 # radian measure\n",
+    "deg=radian/(pi/180) # degree measure\n",
+    "print \"Degree measure is %.0f degree\"%deg"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.3 page.no:99"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of arc intercepted =2.4 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine length of the intercepted arc\n",
+    "r=2. #radius of circle\n",
+    "theta=1.2 # central angle in radian\n",
+    "s=r*theta # length of arc\n",
+    "print \"Length of arc intercepted =%.1f cm\"%s"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.4 page.no:99"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of arc intercepted = 7.16 ft \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine length of the arc intercepted\n",
+    "from math import pi\n",
+    "\n",
+    "r=10 #radius of circle\n",
+    "theta=41*(pi/180) # central angle in radian\n",
+    "s=r*theta # length of arc\n",
+    "print \"Length of arc intercepted = %.2f ft \"%s"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.5 page.no:100"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Measure of central angle = 0.40 rad\n",
+      " \n",
+      "Measure of central angle =22.92 degree\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine angle in radians and degrees\n",
+    "from math import pi\n",
+    "\n",
+    "r=5. #radius of circle\n",
+    "s=2. #length of arc\n",
+    "theta = s/r #central angle in radian\n",
+    "print \"Measure of central angle = %.2f rad\\n \"%theta\n",
+    "print \"Measure of central angle =%.2f degree\"%(theta*(180/pi))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.6 page.no:100"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of the rope =13.4 ft\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine the length of the rope\n",
+    "from math import sqrt,pi,atan,acos\n",
+    "\n",
+    "d=8. #distance between places in feet\n",
+    "r=2. #radius of cylinder in feet\n",
+    "#from the figure\n",
+    "DA=d/2\n",
+    "BE=r\n",
+    "DE=3 #distance from centre of container to wall\n",
+    "AE=sqrt(DE**2 + DA**2) # pythagoras theorem\n",
+    "AB=sqrt(AE**2 - BE**2) # pythagoras theorem\n",
+    "#all angles below are in radians\n",
+    "angle_AED = atan((d/2)/DE)\n",
+    "angle_AEB = acos(BE/AE)\n",
+    "angle_BEC = pi - (angle_AED + angle_AEB)\n",
+    "arc_BC = BE*angle_BEC #length of arc BC\n",
+    "L = 2*(AB + arc_BC) #length of rope\n",
+    "print \"Length of the rope =%.1f ft\"%L"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.7 page.no:101"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of belt around pulley = 71.4 cm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine the length of the belt around the pulleys\n",
+    "from math import pi,sqrt,asin\n",
+    "\n",
+    "AE= 5. #radius of first pulley in cm\n",
+    "BF= 8. #radius of second pulley in cm\n",
+    "AB=15. #distance between centre of pulleys in cm\n",
+    "#from the figure\n",
+    "CF=AE #parallel side of rectangle ACFE\n",
+    "BC= BF- CF\n",
+    "AC = sqrt(AB**2 - BC**2) #by pythagoras theorem\n",
+    "EF=AC# parallel side of rectangle ACFE 14\n",
+    "angle_EAC = pi/2\n",
+    "angle_BAC = asin(BC/AB)\n",
+    "angle_DAE = pi - angle_EAC - angle_BAC\n",
+    "angle_ABC = angle_DAE #AE and BF are parallel\n",
+    "angle_GBF= pi - angle_ABC\n",
+    "arc_DE=AE*angle_ABC # length of arc DE\n",
+    "arc_FG=BF*angle_GBF # length of arc FG\n",
+    "L=2*(arc_DE + EF + arc_FG) #length of belt\n",
+    "print \"Length of belt around pulley = %.1f cm\"%L"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.8 page.no:103"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area of sector = 1.6∗pi cmˆ2\n",
+      "(or)\n",
+      "Area of sector = 5.026548 cmˆ2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find the area of sector of circle\n",
+    "from math import pi\n",
+    "\n",
+    "theta= pi/5 # angle in radian\n",
+    "r=4. #radius in cm\n",
+    "A=r*r*theta/2 #Area of sector\n",
+    "print \"Area of sector = %.1f∗pi cmˆ2\\n(or)\"%(A/pi)\n",
+    "print \"Area of sector = %f cmˆ2\"%A"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.9 page.no:103"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area of sector =12.51 mˆ2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine area of sector of a circle\n",
+    "from math import pi\n",
+    "\n",
+    "theta= 117*(pi/180) # angle in radian\n",
+    "r=3.5 #radius in m\n",
+    "A=r*r*theta/2 #Area of sector\n",
+    "print \"Area of sector =%.2f mˆ2\"%A"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.10 page.no:104"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area of sector =27 cmˆ2\n",
+      "\n",
+      "Note: Angle subtended by arc = 0.666667 rad\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine area of sector of circle\n",
+    "\n",
+    "s=6. #arc length in cm\n",
+    "r=9. #radius in cm\n",
+    "A=r*s/2 #Area of sector\n",
+    "print \"Area of sector =%.0f cmˆ2\\n\"%A\n",
+    "print \"Note: Angle subtended by arc = %f rad\"%(s/r)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.11 page.no:104"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area enclosed by belt pulley system = 338.71 cmˆ2 \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine area insude belt pulley system\n",
+    "from math import pi,sqrt,asin\n",
+    "\n",
+    "AE= 5. #radius of first pulley\n",
+    "BF= 8. #radius of second pulley\n",
+    "AB=15. #distance between centre of pulleys\n",
+    "#from the figure\n",
+    "CF=AE\n",
+    "BC= BF- CF\n",
+    "AC = sqrt(AB**2 - BC**2)\n",
+    "#from the figure\n",
+    "angle_EAC = pi/2\n",
+    "angle_BAC = asin(BC/AB)\n",
+    "angle_DAE = pi - angle_EAC - angle_BAC\n",
+    "angle_ABC = angle_DAE #AE and BF are parallel\n",
+    "angle_GBF= pi - angle_ABC\n",
+    "area_DAE = AE**2*angle_DAE/2 #area of sector DAE\n",
+    "area_GBF = BF**2*angle_GBF/2 #area of sector GBF\n",
+    "area_AEFC = AE*AC #area of rectangle AEFC\n",
+    "area_ABC = AC*BC/2 #area of triangle ABC\n",
+    "area_K =2*( area_DAE + area_AEFC + area_ABC +area_GBF)\n",
+    "print \"Area enclosed by belt pulley system = %.2f cmˆ2 \"%area_K\n",
+    "#Note: answer differs from book due to approximations by them "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.12 page.no:105"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Required area of segment = 1.408 square units\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine area of segment formed by a chord in circle\n",
+    "from math import acos,sin\n",
+    "\n",
+    "radius = 2.\n",
+    "chord = 3.\n",
+    "#Use law of cosines\n",
+    "cos_theta = (radius**2+radius**2-chord**2)/(2*radius*radius)\n",
+    "theta=acos(cos_theta) #subtended central angle in radians\n",
+    "area_K=radius**2*(theta-sin(theta))/2\n",
+    "print \"Required area of segment = %.3f square units\"%area_K"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.13 page.no:106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area of intersection of 2 circles =7.66 cm ˆ2 \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To determine area of intersection of 2 circles\n",
+    "from math import acos\n",
+    "\n",
+    "d=7. #distance between centres in cm\n",
+    "r1= 5. #radius of first circle in cm\n",
+    "r2= 4. #radius of second circle in cm\n",
+    "#use law of cosines\n",
+    "cos_alpha=(d**2+ r1**2 - r2**2 ) /(2*d*r1)\n",
+    "cos_beeta=(d**2+ r2**2 - r1**2 ) /(2*d*r2)\n",
+    "#from the geometry of the figure\n",
+    "#all the angles below are in radians\n",
+    "alpha= acos(cos_alpha)\n",
+    "beeta= acos(cos_beeta)\n",
+    "angle_BAC = alpha\n",
+    "angle_ABC = beeta\n",
+    "angle_CAD =2* angle_BAC\n",
+    "angle_CBD =2* angle_ABC\n",
+    "#required area = area at segment CD in circle at A and at B\n",
+    "area_K = r1**2*(angle_CAD-sin(angle_CAD))/2 + r2 **2*(angle_CBD-sin(angle_CBD))/2\n",
+    "print \"Area of intersection of 2 circles =%.2f cm ˆ2 \"%area_K"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.14 page.no:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Angular speed= 2.094395 radian/sec\n",
+      "\n",
+      "Linear speed=6.283185m/sec\n",
+      "\n",
+      "(or)\n",
+      "\n",
+      "Angular speed= 0.666667∗pi radian/sec\n",
+      " \n",
+      "Linear speed = 2.000000∗pi m/sec \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find linear and angular speed of a moving object\n",
+    "from math import pi\n",
+    "t=0.5 #time in second\n",
+    "r= 3 #radius in m of the circle\n",
+    "theta = pi/3 # central angle in radian\n",
+    "w = theta/t #angular speed in rad /sec\n",
+    "v=w*r#linear speed in m/sec\n",
+    "print \"Angular speed= %f radian/sec\\n\"%w\n",
+    "print \"Linear speed=%fm/sec\"%v\n",
+    "print \"\\n(or)\\n\\nAngular speed= %f∗pi radian/sec\\n \"%(w/pi)\n",
+    "print \"Linear speed = %f∗pi m/sec \"%(v/pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.15 page.no:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Linear speed = 12.96 ft/sec\n",
+      "\n",
+      "Angular speed= 6.48 radian/sec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find linear and angular speed of a moving object\n",
+    "\n",
+    "t=2.7 #time in second\n",
+    "r= 2. #radius in ft of the circle\n",
+    "s=35. #distance in feet\n",
+    "v=s/t #linear speed in ft/sec\n",
+    "w=v/r #angular speed in rad /sec\n",
+    "print \"Linear speed = %.2f ft/sec\\n\"%v\n",
+    "print \"Angular speed= %.2f radian/sec\"%w"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.16 page.no:109"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "central angle swept = 7.75 radian \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find the central angle swept by a moving object\n",
+    "t=3.1 #time in second\n",
+    "v= 10 #linear speed in m/sec\n",
+    "r= 4 #radius in m of the circle\n",
+    "s=v*t # distance in m\n",
+    "theta = s/r #central angle swept\n",
+    "print \"central angle swept = %.2f radian \"%theta"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.17 page.no:110"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Angular speed of larger gear=20 rpm \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find the angular speed of larger gear interlocked with smaller gear\n",
+    "r1=5 #radius of larger gear\n",
+    "r2=4 #radius smaller gear\n",
+    "w2=25 #angular speed of smaller gear\n",
+    "# Imagine a particle on outer radii of each gear\n",
+    "#At any time , for every rotation , circular displacement of each particle is same\n",
+    "# (or) s1=s2 implies v1∗t=v2∗t\n",
+    "#v1= v2 implies w1∗r1=w2∗r2\n",
+    "w1=(w2*r2)/r1 #angular speed of larger gear\n",
+    "print \"Angular speed of larger gear=%.0f rpm \"%w1"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/MayurSabban/Chapter02.ipynb b/sample_notebooks/MayurSabban/Chapter02.ipynb
new file mode 100755
index 00000000..ada20973
--- /dev/null
+++ b/sample_notebooks/MayurSabban/Chapter02.ipynb
@@ -0,0 +1,502 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 02 : Magnetic Circuits and Induction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "\n",
+      "from pylab import *\n",
+      "import math\n",
+      "#initialisation of variables\n",
+      "U_o=4*math.pi*10**-7\n",
+      "U_r=6000\n",
+      "l_g=0.0006\n",
+      "l_c=.40\n",
+      "A_c=.04*.04\n",
+      "B_c=1.2\n",
+      "N=600\n",
+      "\n",
+      "#Calculataions\n",
+      "i=(1/(U_o*N))*(((B_c*l_c)/U_r)+(B_c*l_g))\n",
+      "phi=B_c*A_c\n",
+      "lmda=N*phi\n",
+      "A_g=(.04+l_g)**2\n",
+      "B_g=phi/A_g\n",
+      "\n",
+      "#Results\n",
+      "print(\"Neglecting fringing,current(A)=%.2f ohm\" %i)\n",
+      "print(\"Flux(Wb)=%.4f \" %phi)\n",
+      "print(\"Flux linkages(Wb-turns)=%.2f \" %lmda)\n",
+      "print(\"Fringing taken into account,current(A)=%.2f \" %B_g)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Neglecting fringing,current(A)=1.06 ohm\n",
+        "Flux(Wb)=0.0019 \n",
+        "Flux linkages(Wb-turns)=1.15 \n",
+        "Fringing taken into account,current(A)=1.16 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "import math\n",
+      "#Calculation of current reqd to produce flux in the given magnetic circuit.\n",
+      "\n",
+      "U_o=4*math.pi*10**-7\n",
+      "U_r=4000\n",
+      "N=600\n",
+      "l_c=.30\n",
+      "l_g=0\n",
+      "dia=.02\n",
+      "phi=.5*10**-3        #flux\n",
+      "\n",
+      "#Calculations\n",
+      "A=(math.pi/4)*dia**2\n",
+      "i=0;\n",
+      "R=((l_c/U_r)+l_g)/(U_o*A)\n",
+      "i=(phi*R)/N\n",
+      "\n",
+      "print(\"no air gap current(A) =%.4f \" %i)\n",
+      "#l_g=0.001\n",
+      "R=((l_c/U_r)+l_g)/(U_o*A)\n",
+      "i=(phi*R)/N\n",
+      "print(\"no air gap current(A) =%.4f \" %i)\n",
+      "\n",
+      "B_g=phi/A\n",
+      "print(\"B(T) =%.4f \" %B_g)\n",
+      "H_g=B_g/U_o\n",
+      "\n",
+      "AT_g=H_g*0.001\n",
+      "\n",
+      "print(\"AT_g =%.4f \" %AT_g)\n",
+      "\n",
+      "H_c=3000\n",
+      "AT_c=H_c*0.30\n",
+      "print(\"AT_c =%.4f \" %AT_c)\n",
+      "\n",
+      "i=(AT_g+AT_c)/N\n",
+      "\n",
+      "#Results\n",
+      "print(\"from magnetisation data, current(A) =%.4f \" %i)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "no air gap current(A) =0.1583 \n",
+        "no air gap current(A) =0.1583 \n",
+        "B(T) =1.5915 \n",
+        "AT_g =1266.5148 \n",
+        "AT_c =900.0000 \n",
+        "from magnetisation data, current(A) =3.6109 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page No 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "#Determination of mmf of the exciting coil\n",
+      "\n",
+      "U_o=4*math.pi*10**-7\n",
+      "A1=.0001\n",
+      "A2=.0002\n",
+      "l1=.025*10**-2\n",
+      "l2=.02*10**-2\n",
+      "phi=.75*10**-3\n",
+      "\n",
+      "#Calculations\n",
+      "def reluctance(l,U_r,A):\n",
+      "\tRe=l/(U_o*U_r*A)\n",
+      "\treturn Re\n",
+      "\n",
+      "def mmf(R1,R2,R3):\n",
+      "\tNi=phi*(R3+((R1*R2)/(R1+R2)))\n",
+      "\treturn Ni\n",
+      "\n",
+      "R_g1=reluctance(l1,1,A1)\n",
+      "R_g2=reluctance(l2,1,A1)\n",
+      "R_g3=reluctance(l2,1,A2)\n",
+      "print(\"when U_r=1,mmf(AT) =%.4f \" %mmf(R_g1,R_g2,R_g3))\n",
+      "L1=l1*2*10**3\n",
+      "L2=l2*10**3\n",
+      "R_c1=reluctance(L1,5000,A1)\n",
+      "R_c2=reluctance(L1,5000,A1)\n",
+      "R_c3=reluctance(L2,5000,A2)\n",
+      "\n",
+      "#Results\n",
+      "print(\"when U_r=5000,mmf(AT) =%.4f \" %mmf(R_c1+R_g1,R_c2+R_g2,R_c3+R_g3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when U_r=1,mmf(AT) =1259.9766 \n",
+        "when U_r=5000,mmf(AT) =1680.3089 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variablesimport math\n",
+      "# Exciting current calculation needed to setup reqd flux\n",
+      "\n",
+      "U_o=4*math.pi*10**-7\n",
+      "A1=800*10**-6\n",
+      "A2=600*10**-6\n",
+      "l1=1*10**-3            #air gap length\n",
+      "l2=160*10**-3          #length of central limb\n",
+      "l3=400*10**-3          #length of side limb\n",
+      "phi=.8*10**-3\n",
+      "N=500\n",
+      "\n",
+      "#Calculations\n",
+      "def fd(A):\n",
+      "\tB=phi/A\n",
+      "\treturn B\n",
+      "\n",
+      "def mmf(l,B):\n",
+      "\tF=l/B\n",
+      "\treturn F\n",
+      "\n",
+      "#air gap\n",
+      "B_g=fd(A1)\n",
+      "F_g=mmf(l1,B_g)/U_o\n",
+      "print(\"F_g(AT) =%.4f \" %F_g)\n",
+      "#central limb\n",
+      "B_c=B_g\n",
+      "F_c=mmf(l2,B_c)/10**-3\n",
+      "print(\"F_c(AT)=%.4f \" %F_c)\n",
+      "#outer limb                flux is divided into half\n",
+      "B_o=fd(A2)/2\n",
+      "F_o=mmf(l3,B_o)/(4*10**-3)\n",
+      "print(\"F_o(AT)=%.4f \" %F_o)\n",
+      "i=(F_g+F_c+F_o)/N            #    total mmf/no of turns\n",
+      "\n",
+      "#Results\n",
+      "print(\"exciting current(A)=%.4f \" %i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "F_g(AT) =795.7747 \n",
+        "F_c(AT)=160.0000 \n",
+        "F_o(AT)=150.0000 \n",
+        "exciting current(A)=2.2115 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "# determination of excitation coil mmf\n",
+      "U_o=4*math.pi*10**-7 \n",
+      "A1=25*10**-4 \n",
+      "A2=12.5*10**-4 \n",
+      "A3=25*10**-4 \n",
+      "l1=.5             #length of side limb(ab+cd)\n",
+      "l2=.2           #length of central limb(ad)\n",
+      "l3=.5             #length of side limb(dea)\n",
+      "l4=.25*10**-3           #length of air gap\n",
+      "phi=.75*10**-3 \n",
+      "N=500 \n",
+      "\n",
+      "#Calculations\n",
+      "def fd(A):\n",
+      "\tB=phi/A\n",
+      "\treturn B\n",
+      "\t\n",
+      "def flux(B,l):\n",
+      "\tF=B*l/(U_o)\n",
+      "\treturn F\n",
+      "\t\n",
+      "def fl(H,l):\n",
+      "\tf=H*l\n",
+      "\treturn f\n",
+      "\n",
+      "B_abcd=fd(A1) \n",
+      "F_bc=flux(B_abcd,l4) \n",
+      "print(\"B_abcd(T) =%.4f \" %B_abcd)\n",
+      "H_ab=200                         #for cast iron for B=0.3\n",
+      "F_abcd=fl(H_ab,l1) \n",
+      "F_ad=F_abcd+F_bc \n",
+      "H_ad=F_ad/l2 \n",
+      "print(\"H_ad(AT/m) =%.4f \" %H_ad)\n",
+      "B_ad=1.04                        #for cast iron for H=800\n",
+      "phi_ad=B_ad*A2 \n",
+      "phi_dea=phi+phi_ad \n",
+      "B_dea=phi_dea/A3 \n",
+      "H_dea=500                         #for cast iron for B=.82\n",
+      "F_dea=H_dea*l3 \n",
+      "F=F_dea+F_ad \n",
+      "\n",
+      "#Results\n",
+      "print(\"reqd mmf(AT) =%.4f \" %F)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B_abcd(T) =0.3000 \n",
+        "H_ad(AT/m) =798.4155 \n",
+        "reqd mmf(AT) =409.6831 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7, Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "#determination of self and mutual inductance b/w 2 coils\n",
+      "\n",
+      "U_o=4*math.pi*10**-7 \n",
+      "U_r=1600 \n",
+      "A1=4*10**-4 \n",
+      "A2=4*10**-4 \n",
+      "A0=2*10**-4 \n",
+      "N1=500 \n",
+      "N2=1000 \n",
+      "\n",
+      "#Calculations\n",
+      "l1=.01*((6+0.5+1)*2+(4+2)) \n",
+      "l2=.01*((3+0.5+1)*2+(4+2)) \n",
+      "l0=.01*(4+2) \n",
+      "\n",
+      "def reluc(l,A):\n",
+      "\tR=l/(U_o*U_r*A)\n",
+      "\treturn R\n",
+      "\t\n",
+      "R1=reluc(l1,A1) \n",
+      "R2=reluc(l2,A2) \n",
+      "R0=reluc(l0,A0) \n",
+      "\n",
+      "def re(r0,r1,r2):\n",
+      "\tre=r0+((r1*r2)/(r1+r2)) \n",
+      "\treturn re\n",
+      "\n",
+      "print('coil 1 excited with 1A') \n",
+      "R_1=re(R1,R0,R2) \n",
+      "phi1=N1/R_1 \n",
+      "phi2=phi1*R0/(R0+R2) \n",
+      "L11=N1*phi1 \n",
+      "print(\"self inductance(H) =%.4f \" %L11)\n",
+      "M21=N2*phi2 \n",
+      "print(\"mutual inductance(H) =%.4f \" %M21)\n",
+      "print('coil 2 excited with 1A') \n",
+      "R_2=re(R2,R0,R1) \n",
+      "phi2=N2/R_2 \n",
+      "L22=N2*phi2 \n",
+      "print(\"self inductance(H) =%.4f \" %L22)\n",
+      "M12=M21 \n",
+      "\n",
+      "#Results\n",
+      "print(\"mutual inductance(H) =%.4f \" %M12)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coil 1 excited with 1A\n",
+        "self inductance(H) =0.7267 \n",
+        "mutual inductance(H) =0.6460 \n",
+        "coil 2 excited with 1A\n",
+        "self inductance(H) =3.5529 \n",
+        "mutual inductance(H) =0.6460 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#initialisation of variables\n",
+      "#determination of R_c,R_g,L,W_f\n",
+      "\n",
+      "U_o=4*math.pi*10**-7 \n",
+      "U_r=6000 \n",
+      "l_g=0.0006 \n",
+      "l_c=.40 \n",
+      "A_c=.04*.04 \n",
+      "B_c=1.2 \n",
+      "N=600 \n",
+      "\n",
+      "#Calculations\n",
+      "def current(B_g):\n",
+      "\ti=(1/(U_o*N))*(((B_c*l_c)/U_r)+(B_g*l_g))\n",
+      "\treturn i\n",
+      "\n",
+      "print(\"neglecting fringing,current(A)= %.4f \" %current(B_c))\n",
+      "\n",
+      "phi=B_c*A_c \n",
+      "print(\"flux(Wb)=%.4f \" %phi)\n",
+      "\n",
+      "def flux_linkage(phi):\n",
+      "\tlmda=N*phi\n",
+      "\treturn lmda\n",
+      "\n",
+      "print(\"flux linkages(Wb-turns)= %.4f \" %flux_linkage(phi))\n",
+      "\n",
+      "def reluc(l,U,A):\n",
+      "\tR=l/(U_o*U*A)\n",
+      "\treturn R\n",
+      "R_c=reluc(l_c,U_r,A_c) \n",
+      "print(\"R_c=%.4f \" %R_c)\n",
+      "R_g=reluc(l_g,1,A_c) \n",
+      "print(\"R_g=%.4f \" %R_g)\n",
+      "\n",
+      "L=N**2/(R_c+R_g) \n",
+      "print(\"coil inductance(H)=%.4f \" %L)\n",
+      "W_f=(N*phi)**2/(2*L) \n",
+      "\n",
+      "#Results\n",
+      "print(\"energy stored in the magnetic field(J)=%.4f \" %W_f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "neglecting fringing,current(A)= 1.0610 \n",
+        "flux(Wb)=0.0019 \n",
+        "flux linkages(Wb-turns)= 1.1520 \n",
+        "R_c=33157.2798 \n",
+        "R_g=298415.5183 \n",
+        "coil inductance(H)=1.0857 \n",
+        "energy stored in the magnetic field(J)=0.6112 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits.ipynb b/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits.ipynb
new file mode 100755
index 00000000..5503c007
--- /dev/null
+++ b/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits.ipynb
@@ -0,0 +1,289 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1,Page number 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The current in circuit is  0.3672\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "import math   \n",
+    "# given\n",
+    "Bc=0.8\n",
+    "Hc=510\n",
+    "Bg=0.8\n",
+    "A=12.566 \n",
+    "lg=0.0015\n",
+    "lc=0.36\n",
+    "N=500\n",
+    "# calculations\n",
+    "Fg=Bg/A*(2*lg)\n",
+    "Fc=Hc*lc\n",
+    "F=Fc+Fg\n",
+    "i=F/N\n",
+    "Pre=Bc/Hc\n",
+    "RelPre=Pre/A\n",
+    "F=Hc*lc\n",
+    "i=F /N #current\n",
+    "print 'The current in circuit is ',i\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2,Page number 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The flux density is 5\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "A=12.566 \n",
+    "lc=360\n",
+    "N=500\n",
+    "i=4\n",
+    "lg=2*10**-3\n",
+    "m=-A*(lc/lg)\n",
+    "c=(N*i*A)/(lg)\n",
+    "Hc=(N*i)/(lc)  #flux density\n",
+    "print 'The flux density is',Hc"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3,Page number 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The airgap flux value is -7.47688567997e-07\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#given\n",
+    "N1=500\n",
+    "I1=10\n",
+    "N2=500\n",
+    "I2=10\n",
+    "Ibafe=3*52*10**-2\n",
+    "A=12.566\n",
+    "b=1200\n",
+    "Ag=4*10^-4\n",
+    "Ac=4*10^-4\n",
+    "lg=5*10^-3\n",
+    "Ibecore=0.515\n",
+    "c=0.0002067\n",
+    "d=0.0004134\n",
+    "#calculations\n",
+    "F1=N1*I1\n",
+    "F2=N2*I2\n",
+    "Pre=1200*A\n",
+    "Rbafe=(Ibafe)/(Pre*Ac)\n",
+    "Rg=lg/(A*Ag)\n",
+    "Rbecore=Ibecore/(Pre*Ac)\n",
+    "Bg=d/(Ag)\n",
+    "Hg=Bg/A    # airgap ﬂux\n",
+    "print 'The airgap flux value is',Hg\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4,Page number 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The magnetic flux is 0.153938040026\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# given \n",
+    "Irad=20\n",
+    "Orad=25\n",
+    "Dia=22.5\n",
+    "N=250\n",
+    "i=2.5\n",
+    "B=1.225\n",
+    "# calculations\n",
+    "l=2*pi*Dia*10**-2\n",
+    "radius=1/2*(Irad+Orad)\n",
+    "H=(N*i)/l\n",
+    "A=pi*((Orad -Irad)/2)**2*10**-4\n",
+    "z=(1.225)*(pi*6.25*10**-4)\n",
+    "y=(N*z)\n",
+    "L=(y/i)\n",
+    "core=(B/H)\n",
+    "l=(2*pi*22.5*10**-2)\n",
+    "Rcore=(l)/(core*A)\n",
+    "L=(N**2)/(Rcore)\n",
+    "print 'The magnetic flux is',L"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 5,Page number 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "flux density 144.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# given\n",
+    "n=500\n",
+    "E=100\n",
+    "A=0.001\n",
+    "b=1/120\n",
+    "f=1.2\n",
+    "#calculations\n",
+    "max1=(E/1000)*(b)\n",
+    "max2=(f*A)\n",
+    "E=(120*n*max2*2) # result\n",
+    "print 'flux density',E\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 6,Page number 9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The dimension Am is 0.000210526315789\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "#given\n",
+    "lg=0.4*10**-2\n",
+    "Bg=0.8\n",
+    "Hm=42*10**3\n",
+    "A=4*pi*10**-7\n",
+    "Ag=2.5*10**-4\n",
+    "Bm=0.95\n",
+    "#calculations\n",
+    "Hg=Bg/A\n",
+    "lm=(lg/Hm)*Hg\n",
+    "Am=(Bg*Ag)/(Bm)\n",
+    "print 'The dimension Am is',Am\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits_1.ipynb b/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits_1.ipynb
new file mode 100755
index 00000000..8781ef4e
--- /dev/null
+++ b/sample_notebooks/MeenaChandrupatla/Chapter_1_Magnetic_Circuits_1.ipynb
@@ -0,0 +1,296 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1 Magnetic circuits"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1,Page number 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The current in circuit is  0.3672\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "import math   \n",
+    "# given\n",
+    "Bc=0.8\n",
+    "Hc=510\n",
+    "Bg=0.8\n",
+    "A=12.566 \n",
+    "lg=0.0015\n",
+    "lc=0.36\n",
+    "N=500\n",
+    "# calculations\n",
+    "Fg=Bg/A*(2*lg)\n",
+    "Fc=Hc*lc\n",
+    "F=Fc+Fg\n",
+    "i=F/N\n",
+    "Pre=Bc/Hc\n",
+    "RelPre=Pre/A\n",
+    "F=Hc*lc\n",
+    "i=F /N #current\n",
+    "print 'The current in circuit is ',i\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2,Page number 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The flux density is 5\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "A=12.566 \n",
+    "lc=360\n",
+    "N=500\n",
+    "i=4\n",
+    "lg=2*10**-3\n",
+    "m=-A*(lc/lg)\n",
+    "c=(N*i*A)/(lg)\n",
+    "Hc=(N*i)/(lc)  #flux density\n",
+    "print 'The flux density is',Hc"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3,Page number 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The airgap flux value is -7.47688567997e-07\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "#given\n",
+    "N1=500\n",
+    "I1=10\n",
+    "N2=500\n",
+    "I2=10\n",
+    "Ibafe=3*52*10**-2\n",
+    "A=12.566\n",
+    "b=1200\n",
+    "Ag=4*10^-4\n",
+    "Ac=4*10^-4\n",
+    "lg=5*10^-3\n",
+    "Ibecore=0.515\n",
+    "c=0.0002067\n",
+    "d=0.0004134\n",
+    "#calculations\n",
+    "F1=N1*I1\n",
+    "F2=N2*I2\n",
+    "Pre=1200*A\n",
+    "Rbafe=(Ibafe)/(Pre*Ac)\n",
+    "Rg=lg/(A*Ag)\n",
+    "Rbecore=Ibecore/(Pre*Ac)\n",
+    "Bg=d/(Ag)\n",
+    "Hg=Bg/A    # airgap ﬂux\n",
+    "print 'The airgap flux value is',Hg\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4,Page number 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The magnetic flux is 0.153938040026\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "# given \n",
+    "Irad=20\n",
+    "Orad=25\n",
+    "Dia=22.5\n",
+    "N=250\n",
+    "i=2.5\n",
+    "B=1.225\n",
+    "# calculations\n",
+    "l=2*pi*Dia*10**-2\n",
+    "radius=1/2*(Irad+Orad)\n",
+    "H=(N*i)/l\n",
+    "A=pi*((Orad -Irad)/2)**2*10**-4\n",
+    "z=(1.225)*(pi*6.25*10**-4)\n",
+    "y=(N*z)\n",
+    "L=(y/i)\n",
+    "core=(B/H)\n",
+    "l=(2*pi*22.5*10**-2)\n",
+    "Rcore=(l)/(core*A)\n",
+    "L=(N**2)/(Rcore)\n",
+    "print 'The magnetic flux is',L"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 5,Page number 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "flux density 144.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "# given\n",
+    "n=500\n",
+    "E=100\n",
+    "A=0.001\n",
+    "b=1/120\n",
+    "f=1.2\n",
+    "#calculations\n",
+    "max1=(E/1000)*(b)\n",
+    "max2=(f*A)\n",
+    "E=(120*n*max2*2) # result\n",
+    "print 'flux density',E\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 6,Page number 9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The dimension Am is 0.000210526315789\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "#given\n",
+    "lg=0.4*10**-2\n",
+    "Bg=0.8\n",
+    "Hm=42*10**3\n",
+    "A=4*pi*10**-7\n",
+    "Ag=2.5*10**-4\n",
+    "Bm=0.95\n",
+    "#calculations\n",
+    "Hg=Bg/A\n",
+    "lm=(lg/Hm)*Hg\n",
+    "Am=(Bg*Ag)/(Bm)\n",
+    "print 'The dimension Am is',Am\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/MohdAsif/chapter2.ipynb b/sample_notebooks/MohdAsif/chapter2.ipynb
new file mode 100644
index 00000000..5856d478
--- /dev/null
+++ b/sample_notebooks/MohdAsif/chapter2.ipynb
@@ -0,0 +1,349 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7252e18079dddfe390b8ba1aeebeccc0ce63f41488e78d7d5f7b5e2c14af82fb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2 - Semiconductor Devices"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex-2.1 Pg-2.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Vf=0.2  #voltage in volts\n",
+      "Vr=60  #voltage in volts\n",
+      "If=60*10**(-3)  #current in  ampere\n",
+      "I0=0.025*10**(-3)  #current in  ampere\n",
+      "Rf=Vf/If #forward resistance\n",
+      "Rr=Vr/I0 #reverse resistance\n",
+      "Rr=Rr*1e-6\n",
+      "print \"the equivalent resistance are Rf=%.3f ohm and Rr=%-.1f M ohm\"%(Rf,Rr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the equivalent resistance are Rf=3.333 ohm and Rr=2.4 M ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Chapter-2 Ex-2.2 Pg-2.18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Vf=0.7 \n",
+      "If=0.06 \n",
+      "Rf=Vf/If #DC forward resistance\n",
+      "print \"\\n DC forward resistance Rf : %.2f ohm\\n\"%(Rf)\n",
+      "print \"as the forward voltage changes from P to Q\"\n",
+      "delta_Vf=0.77-.7 \n",
+      "delta_If=(120-60)*10**(-3) \n",
+      "rf=delta_Vf/delta_If #dynamic forward resistance\n",
+      "print \"\\n Dynamic forward resistance rf : %.3f ohm\"%(rf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " DC forward resistance Rf : 11.67 ohm\n",
+        "\n",
+        "as the forward voltage changes from P to Q\n",
+        "\n",
+        " Dynamic forward resistance rf : 1.167 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2_3 Pg-2-22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "V=10 #supply voltage\n",
+      "Rf=0 #forward resistance\n",
+      "Rl=1 #load resistance in k ohm\n",
+      "Vin=0.7 #cut in voltage\n",
+      "Il=(V-Vin)/Rl #applying KVL to the loop\n",
+      "If=Il \n",
+      "print \"\\n \\n current through the resistance Il=If = is %.1f mA\"%(If)\n",
+      "Vl=Il*Rl \n",
+      "print \"\\n \\n voltage across Rl is %.1f V\"%(Vl)\n",
+      "Pd=If*Vin \n",
+      "print \"\\n \\n diode power Pd = %.2f mW\"%(Pd)\n",
+      "Pl=Il*Vl \n",
+      "print \"\\n \\n load power Pl = %.2f mW\"%(Pl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " \n",
+        " current through the resistance Il=If = is 9.3 mA\n",
+        "\n",
+        " \n",
+        " voltage across Rl is 9.3 V\n",
+        "\n",
+        " \n",
+        " diode power Pd = 6.51 mW\n",
+        "\n",
+        " \n",
+        " load power Pl = 86.49 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EX2_4 PG-2.23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Vf=0.7 #cut-in voltage\n",
+      "V=10 #supply voltage\n",
+      "Rl=500 #load resistance\n",
+      "If=(V-Vf)/Rl #applying KVL to the circuit\n",
+      "If=If*1e3\n",
+      "print \"\\n Forward current is %.2f mA\\n\"%(If)\n",
+      "print \"When forward resistance is Rf is 3.2 Ohm then\"\n",
+      "print \"the equivalent circuit is as shown in fig-2.25(b)\"\n",
+      "Rf=3.2 \n",
+      "If=(V-Vf)/(Rl+Rf) #applying KVL to the circuit\n",
+      "If=If*1e3\n",
+      "print \"\\n therefore Forward current is %.4f mA\"%(If)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Forward current is 18.60 mA\n",
+        "\n",
+        "When forward resistance is Rf is 3.2 Ohm then\n",
+        "the equivalent circuit is as shown in fig-2.25(b)\n",
+        "\n",
+        " therefore Forward current is 18.4817 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EX2_5 PG-2.23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "%matplotlib inline\n",
+      "from matplotlib.pyplot import plot, xlabel, ylabel, show, grid, title\n",
+      "If=80e-3 #maximum forward current\n",
+      "Rf=0.4 #dynamic resistance\n",
+      "Vin=0.3 #cut-in voltage for germanium\n",
+      "print \"when forward current is zero then\"\n",
+      "Vf=Vin #voltage across the diode\n",
+      "print \" voltage across the diode is %1.1f V\\n\"%(Vf)\n",
+      "print \"when forward current is 80mA then\"\n",
+      "Vf=Vin+If*Rf \n",
+      "print \" voltage across the diode is %1.3f V\"%(Vf)\n",
+      "x=np.array([0,.1, .2, .3, .332]) #x-coordinate\n",
+      "y=np.array([0,  0,  0,  0,  80]) #y-coordinate\n",
+      "plot(x,y)\n",
+      "xlabel('voltage across the diode (V) ') \n",
+      "ylabel('current (mA)') \n",
+      "title('Piecewise linear characteristic')\n",
+      "grid()\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "when forward current is zero then\n",
+        " voltage across the diode is 0.3 V\n",
+        "\n",
+        "when forward current is 80mA then\n",
+        " voltage across the diode is 0.332 V\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ZAjvs0Pidwkhxx2BmNoBWKjyDh5LMzAb03e/CPffAD35QdJKBeSjJzKwOWm2P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\nZmbWjTsGMzPr5v8Ds1Mla6CkMg8AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f0aca60fe50>"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EX2_6 PG-2.28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy as np\n",
+      "%matplotlib inline\n",
+      "from matplotlib.pyplot import plot, xlabel, ylabel, show, grid, title, subplot\n",
+      "If=25e-3 #current at Q-point\n",
+      "x=np.array([ 0, 0.5, 0.6,    1,   1.1 ]) #x-coordinate\n",
+      "y=np.array([ 0 , 1 ,  5,     25,   30 ]) #y-coordinate\n",
+      "subplot(1,3,1)\n",
+      "plot(x,y)\n",
+      "x1=np.array([0.5,   1 ,   3]) #x-coordinate\n",
+      "y1=np.array([ 31,   25,   0]) #y-coordinate\n",
+      "subplot(1,3,2)\n",
+      "plot(x1,y1)\n",
+      "x2=np.array([ 0, 1] )\n",
+      "y2=np.array([25 ,25] )\n",
+      "subplot(1,3,3)\n",
+      "plot(x2,y2)\n",
+      "xlabel('Vf (volts)') \n",
+      "ylabel('If (mA)') \n",
+      "title(\"Piece-wise linear characteristic\")\n",
+      "grid()\n",
+      "show()\n",
+      "print \"Q-point is denoted by the intersection of two lines as shown in the plot\"\n",
+      "delta_If=10e-3 #from the graph plotted\n",
+      "delta_Vf=0.9 #from the graph plotted\n",
+      "s=delta_If/delta_Vf #slope\n",
+      "print \"Therefore load resistance is the reciprocal of the slope \"\n",
+      "Rl=1/s #load resistance\n",
+      "print \"\\n required load resistance is %.0f ohm\"%(Rl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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2Qfv28Kc/wWmn+a1NfHbu3Mns2bOZOXMmc+bMYfDgwVxyySVcdNFFKcsKe7kW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\nCv8fFIeXAyYn0X4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f0ac7557350>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q-point is denoted by the intersection of two lines as shown in the plot\n",
+        "Therefore load resistance is the reciprocal of the slope \n",
+        "\n",
+        " required load resistance is 90 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "EX2_7 PG-2.29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "V=0.22 #forward bias voltage\n",
+      "T=25+273 #room temperature in degree kelvin\n",
+      "I0=2e-3 #reverse saturation current\n",
+      "n=1 #for germanium diode\n",
+      "k=8.62e-5#Boltzmann's constant\n",
+      "Vt=k*T \n",
+      "I=I0*(exp(V/(n*Vt))) # diode current\n",
+      "print \"therefore the P-N junction diode current is %f A\"%(I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "therefore the P-N junction diode current is 10.483844 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/MonicaVenkatesh/Chapter_1.ipynb b/sample_notebooks/MonicaVenkatesh/Chapter_1.ipynb
new file mode 100644
index 00000000..a8371dbd
--- /dev/null
+++ b/sample_notebooks/MonicaVenkatesh/Chapter_1.ipynb
@@ -0,0 +1,49 @@
+{
+ "metadata": {
+  "name": "Chapter 1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 1:Properties of Fluids"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.1,Page number 5"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#variable initialisation\n\nw=500        #mass of liquid in N\n\ng=9.806      #gravity in m/s*2\n\ng1=3.5       #gravity in m/s*2\n\ng2=18.0      #gravity in m/s*2\n\n\n#Solution\n\nm=w/g        #mass\n\nprint \"a)m=\",round(m,2),\"kg\"\n\nw1=m*g1      #weight\n\nw2=m*g2\n\nprint \"b)w1=\",round(w1,2),\"N\"\n\nprint \"w2=\",round(w2,1),\"N\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "a)m= 50.99 kg\nb)w1= 178.46 N\nw2= 917.8 N\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/MukteshChaudhary/ch1_1.ipynb b/sample_notebooks/MukteshChaudhary/ch1_1.ipynb
new file mode 100644
index 00000000..079c7b6e
--- /dev/null
+++ b/sample_notebooks/MukteshChaudhary/ch1_1.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Basic Ideas: Energy Bands in Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1, Page 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable decalaration\n",
+      "pd = 100   #potential difference, V\n",
+      "m0=9.11*(10**-31);#m0=rest mass of the electron in kg\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#solving final velocity of the electron\n",
+      "Ek=1.6*(10**-19)*pd;#Ek=final kinetic energy of electron in Joules\n",
+      "print \"Final kinetic energy = %.1e J,%.f eV\"%(Ek,Ek*6.242*10**18)\n",
+      "v=math.sqrt((2*Ek)/m0)#v=final velocity of the electron\n",
+      "print \"Final velocity = %.3e m/s\"%v"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Final kinetic energy = 1.6e-17 J,100 eV\n",
+        "Final velocity = 5.927e+06 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2, Page 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Variable declaration\n",
+      "m=7360*9.11*(10**-31);#m=mass of the ion in kg\n",
+      "q=2*1.6*(10**-19);#q=charge of the ion in Coulomb\n",
+      "V=2000;#V=potential difference in Volt\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#solving velocity & kinetic energy of the ion\n",
+      "v=math.sqrt((2*q*V)/m)#v=velocity of the ion\n",
+      "print \"Velocity acquired by the ion = %.3e m/s\"%v\n",
+      "Ek=(1./2)*m*(v**2)#Ek=kinetic energy of the ion\n",
+      "print \"Kinetic energy of ion = %.1e J = %.f eV\"%(Ek,Ek*6.242*10**18)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity acquired by the ion = 4.369e+05 m/s\n",
+        "Kinetic energy of ion = 6.4e-16 J = 3995 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/NarasimhaMamidala/Chapter_4_BJT_Fundamentals.ipynb b/sample_notebooks/NarasimhaMamidala/Chapter_4_BJT_Fundamentals.ipynb
new file mode 100755
index 00000000..c7db6367
--- /dev/null
+++ b/sample_notebooks/NarasimhaMamidala/Chapter_4_BJT_Fundamentals.ipynb
@@ -0,0 +1,544 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 BJT Fundamentals"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−1 in page 208"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)The value of the Base Current is 3.85e-04 A\n",
+      "\n",
+      "(b)The value of the Collector Current is  3.615e-03 A \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculate Base and Collector Currents\n",
+    "# Given Data\n",
+    "alpha=0.90; # Current Gain in CB mode\n",
+    "Ico=15*10**-6; # Reverse saturation Current in micro−A\n",
+    "Ie=4*10**-3; # Emitter Current in mA\n",
+    "# Calculations\n",
+    "Ic=Ico+(alpha*Ie);\n",
+    "Ib=Ie-Ic;\n",
+    "print \"(a)The value of the Base Current is %0.2e A\\n\" %Ib;\n",
+    "print \"(b)The value of the Collector Current is %0 .3e A \\n\" %Ic"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−2 in page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)The Current gain alpha for BJT is 0.989 \n",
+      "\n",
+      "(b)The value of the base Current is 4.44e-05 A\n",
+      "\n",
+      "(c)The value of the Emitter Current is 4.04e-03 A \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculate alpha using beta\n",
+    "# Given Data\n",
+    "\n",
+    "beta_bjt=90.; # beta gain for the BJT\n",
+    "Ic=4*10**-3; # Collector Current in mA\n",
+    "# Calculations\n",
+    "alpha=beta_bjt/(1.+beta_bjt);\n",
+    "Ib=Ic/beta_bjt;\n",
+    "Ie=Ic+Ib;\n",
+    "print \"(a)The Current gain alpha for BJT is %0.3f \\n\"%alpha\n",
+    "print \"(b)The value of the base Current is %0.2e A\\n\"%Ib\n",
+    "print \"(c)The value of the Emitter Current is %0.2e A \\n\"%Ie"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−3 in page 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)The value of Current gain beta for BJT is 9 \n",
+      "\n",
+      "(b)The value of the Collector Current is 4.65e-03 A \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Collector Current in C E mode\n",
+    "# Given Data\n",
+    "alpha=0.90; # Current Gain of BJT\n",
+    "Ico=15*10**-6; # Reverse Saturation Current of BJT in micro−A\n",
+    "Ib=0.5*10**-3; # Base Current in C−E mode in mA\n",
+    "# Calculations\n",
+    "beta_bjt=alpha/(1-alpha);\n",
+    "Ic=(beta_bjt*Ib)+(beta_bjt+1)*Ico;\n",
+    "print \"(a)The value of Current gain beta for BJT is %0.0f \\n\"%beta_bjt\n",
+    "print \"(b)The value of the Collector Current is %0.2e A \\n\"%Ic"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−4 in page 20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Current gain beta for the Device is 250 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculate beta for the BJT\n",
+    "Ib=20*10**-6; # Base current in micro−A\n",
+    "Ic=5*10**-3; # Collector Current in mA\n",
+    "# Calculations\n",
+    "beta_bjt=Ic/Ib;\n",
+    "print \"The Current gain beta for the Device is %0.0f \\n\"%beta_bjt;\n",
+    "# The Current Gain beta for the Device is 250"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−5 in page 209"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)The value of the Emitter Current is 5.05e-03A \n",
+      "\n",
+      "(b)The value of beta gain of the BJT is 100 \n",
+      "\n",
+      "(c)The value of alpha gain of the BJT is 0.990 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To Compute Alpha Beta and Emitter Current\n",
+    "# Given Data\n",
+    "Ib=50*10**-6; # Base Current in mu−A\n",
+    "Ic=5*10**-3; # Collector Current in mA\n",
+    "# Calculations\n",
+    "Ie=Ic+Ib;\n",
+    "beta_bjt=Ic/Ib;\n",
+    "alpha=Ic/Ie;\n",
+    "print \"(a)The value of the Emitter Current is %0.2eA \\n\"%Ie\n",
+    "print \"(b)The value of beta gain of the BJT is %0.0f \\n\"%beta_bjt\n",
+    "print \"(c)The value of alpha gain of the BJT is %0.3f \\n\"%alpha"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−6 in page 210"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The value of inverse beta of the BJT is 1 \n",
+      "\n",
+      "The value of inverse alpha of the BJT is 2 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculate alpha reverse and beta reverse\n",
+    "# Given Data\n",
+    "Ie=10.*10**-3; # Emitter Current in mA\n",
+    "Ib=5*10**-3; # Base Current in mu−A\n",
+    "# Calculations\n",
+    "Ic=Ie-Ib;\n",
+    "beta_reverse=Ib/Ic;\n",
+    "alpha_reverse=Ie/Ic;\n",
+    "print \"The value of inverse beta of the BJT is %0.0f \\n\"%beta_reverse\n",
+    "print \"The value of inverse alpha of the BJT is %0.0f \\n\"%alpha_reverse"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−7 in page 210"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Circuit 1:\n",
+      "(a)Emitter Current=9.30e-04 A\n",
+      "(b)Base Current=9.21e-06 A\n",
+      "(c)Collector Voltage=0.792 V\n",
+      "\n",
+      "\n",
+      "Circuit 2:\n",
+      "(a)Emitter Current=1.86e-03 A\n",
+      "(b) Collector Current=1.842e-03 A\n",
+      "(c)Collector Voltage=-5.700 V\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Calculate Labeled Currents and Voltages\n",
+    "# Given Data\n",
+    "beta_bjt=100.; # beta gain of BJT\n",
+    "Vbe=0.7; # Base−Emitter voltage of BJT in V\n",
+    "#Calculation\n",
+    "Vcc1=10.;\n",
+    "Vee1=-10.;\n",
+    "Ve1=-0.7;\n",
+    "R1=10*10**3;\n",
+    "Ie1=(Vcc1-Vbe)/R1;\n",
+    "Ib1=Ie1/(beta_bjt+1);\n",
+    "Vc1=Vcc1-R1*(Ie1-Ib1);\n",
+    "Vcc2=10.;\n",
+    "Vee2=-15.;\n",
+    "Ve2=-0.7;\n",
+    "R2 =5*10**3;\n",
+    "Ie2=(Vcc2-Vbe)/R2;\n",
+    "Ic2=(beta_bjt/(beta_bjt+1.))*Ie2;\n",
+    "Vc2=Vee2+R2*(Ie2);\n",
+    "print \"Circuit 1:\\n(a)Emitter Current=%0.2e A\\n(b)Base Current=%0.2e A\\n(c)Collector Voltage=%0.3f V\\n\\n\"%(Ie1,Ib1,Vc1);\n",
+    "print \"Circuit 2:\\n(a)Emitter Current=%0.2e A\\n(b) Collector Current=%0.3e A\\n(c)Collector Voltage=%0.3f V\\n\"%(Ie2,Ic2,Vc2);"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−8 in page 211"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Circuit 1:\n",
+      "(a)Base Voltage = 0.0 V\n",
+      "(b)Emitter Voltage = -0.7 V\n",
+      "\n",
+      "Circuit 2:\n",
+      "(a)Emitter Voltage = 0.7 V\n",
+      "(b) Collector Voltage = -5.7 V\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculate labeled Voltages\n",
+    "# Given Data\n",
+    "Vbe=0.7; # Base−Emitter voltage of BJT in V\n",
+    "Vcc2=10; # DC voltage across Collector in V\n",
+    "Vee2=-15; # DC voltage across Emitter in V\n",
+    "Rc2=5*10**3; # Collector Resistance in K−ohms\n",
+    "# Beta Current Gain of BJT is Infinity\n",
+    "# Calculations\n",
+    "Vb1=0;\n",
+    "Ve1=-0.7;\n",
+    "Ve2=0.7;\n",
+    "Vc2=Vee2+Rc2*((Vcc2-Vbe)/Rc2);\n",
+    "print \"Circuit 1:\\n(a)Base Voltage = %0.1f V\\n(b)Emitter Voltage = %0.1f V\\n\"%(Vb1,Ve1);\n",
+    "print \"Circuit 2:\\n(a)Emitter Voltage = %0.1f V\\n(b) Collector Voltage = %0.1f V\\n\"%(Ve2,Vc2);"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−9 in page 211"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Circuit Parameters:\n",
+      "(a)Base Voltage = 0.300V\n",
+      "(b)Base Current = 1.500e-05 A\n",
+      "(c)Emitter Current= 8.000e-04 A\n",
+      "(d)Collector Current = 7.850e-04 A\n",
+      "(e) Collector Voltage = -1.075 V\n",
+      "(f) beta gain = 52.333\n",
+      "(g)alpha gain = 0.981\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Calculating BJT parameters assuming Vbe\n",
+    "# Given Data\n",
+    "Ve=1.; # Emitter Voltage of BJT in V\n",
+    "Vbe=0.7; # Base−Emitter Voltage of BJT in V\n",
+    "Rb=20*10**3; # Base Resistance of Circuit in K−ohms\n",
+    "Rc=5*10**3; # Collector Resistance of Circuit in K−ohms\n",
+    "Re=5*10**3; # Emitter Resistance of Circuit in K−ohms\n",
+    "Vcc=5.; # DC voltage across Collector in V\n",
+    "Vee=-5; # DC voltage across Emitter in V\n",
+    "# Calculations\n",
+    "Vb=Ve-Vbe;\n",
+    "Ib=Vb/Rb;\n",
+    "Ie=(Vcc -1)/Re;\n",
+    "Ic=Ie-Ib;\n",
+    "Vc=(Rc*Ic)-Vcc;\n",
+    "beta_bjt=Ic/Ib;\n",
+    "alpha=Ic/Ie;\n",
+    "print \"Circuit Parameters:\\n(a)Base Voltage = %0.3fV\\n(b)Base Current = %0.3e A\\n(c)Emitter Current= %0.3e A\\n(d)Collector Current = %0.3e A\\n(e) Collector Voltage = %0.3f V\\n(f) beta gain = %0.3f\\n(g)alpha gain = %0.3f\\n\"%(Vb,Ib,Ie,Ic,Vc, beta_bjt ,alpha);"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−10 in page 211"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)Change in Emitter voltage is +0.40 V\n",
+      "\n",
+      "(b)Change in Collector Voltage is 0.00 V\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Measurement of Circuit Voltage changes\n",
+    "# Given Data\n",
+    "Vb=-5; # Base Voltage of BJT in V\n",
+    "Rc=1*10**3; # Collector Resistance in K−ohms\n",
+    "Ie=2*10**-3; # Emitter Current of BJT in mA\n",
+    "delB=+0.4; # Change in Base Voltage\n",
+    "# Calculations\n",
+    "delE =+0.4;\n",
+    "delC=0;\n",
+    "print \"(a)Change in Emitter voltage is +%0.2f V\\n\"%delE\n",
+    "print \"(b)Change in Collector Voltage is %0.2f V\\n\"%delC"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4−11 in page 212"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Assume active mode for circuit 1\n",
+      "(a)Ve = 1.30 V\n",
+      "(b)Ic = 0.00e+00 A\n",
+      "(c)Ve = 3.03 V\n",
+      "\n",
+      "Thus the circuit operates in an active mode\n",
+      "\n",
+      "\n",
+      "For circuit 2,assume active mode\n",
+      "\n",
+      "(a)Ve = 1.7 V\n",
+      "(b)Ie = 4.30e-04 A\n",
+      "(c)Vc = 4.30 V\n",
+      "\n",
+      "This circuit operates in a saturated mode\n",
+      "\n",
+      "\n",
+      "For circuit 3,assume active mode\n",
+      "\n",
+      "(a)Ve = -4.3 V\n",
+      "(b)Ie = 6.9000e-05 A\n",
+      "(c)Ic = 0.000e+00 A\n",
+      "(d)Vc = -40.2 V\n",
+      "\n",
+      "The circuit operates in an active mode\n",
+      "\n",
+      "\n",
+      "For circuit 4,assume active mode\n",
+      "\n",
+      "(a)Ie = 1.86e-03 A\n",
+      "(b)Vc = -10.00 V\n",
+      "\n",
+      "The circuit operates in an active mode\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Determine mode of operation of BJT\n",
+    "# Given Data\n",
+    "Vbe=0.7; # Base−Emitter Voltage in V\n",
+    "beta_bjt=100; # beta gain of BJ\n",
+    "# Calculation\n",
+    "print \"Assume active mode for circuit 1\"\n",
+    "Vb1=2;\n",
+    "Ve_1=Vb1-Vbe;\n",
+    "Ie1 =1*10** -3;\n",
+    "Ic1=Ie1*(beta_bjt/(1+beta_bjt));\n",
+    "Ve1=6-(3*0.99);\n",
+    "print \"(a)Ve = %0.2f V\\n(b)Ic = %0.2e A\\n(c)Ve = %0.2f V\\n\"%(Ve_1,Ic1,Ve1);\n",
+    "print \"Thus the circuit operates in an active mode\\n\\n\"\n",
+    "print \"For circuit 2,assume active mode\\n\"\n",
+    "Vcc=1;\n",
+    "Ve2=Vcc+Vbe;\n",
+    "Ie2=(6-Ve2)/(10*10**3);\n",
+    "Vc=0+(10*0.43);\n",
+    "print \"(a)Ve = %0.1f V\\n(b)Ie = %0.2e A\\n(c)Vc = %0.2f V\\n\"%(Ve2,Ie2,Vc);\n",
+    "print \"This circuit operates in a saturated mode\\n\\n\"\n",
+    "print \"For circuit 3,assume active mode\\n\"\n",
+    "Ve3=-5+Vbe;\n",
+    "Ie3=(9.5-Ve3)/(200*10**3);\n",
+    "Ic=Ie3*(beta_bjt/(1+beta_bjt));\n",
+    "Vc3=-50+(0.492*20);\n",
+    "print \"(a)Ve = %0.1f V\\n(b)Ie = %0.4e A\\n(c)Ic = %0.3e A\\n(d)Vc = %0.1f V\\n\"%(Ve3,Ie3,Ic,Vc3);\n",
+    "print \"The circuit operates in an active mode\\n\\n\"\n",
+    "print \"For circuit 4,assume active mode\\n\"\n",
+    "Ve4 = -20.7;\n",
+    "Ie4=(30+Ve4)/(5*10**3);\n",
+    "Vc4=(-Ie4*(beta_bjt/(1+beta_bjt))*(2*10**3))-10;\n",
+    "print \"(a)Ie = %0.2e A\\n(b)Vc = %0.2f V\\n\"%(Ie4,Vc4)\n",
+    "print \"The circuit operates in an active mode\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/PADMAVATHITHIYAGARAJAN/Chapter_5.ipynb b/sample_notebooks/PADMAVATHITHIYAGARAJAN/Chapter_5.ipynb
new file mode 100644
index 00000000..3159e1a4
--- /dev/null
+++ b/sample_notebooks/PADMAVATHITHIYAGARAJAN/Chapter_5.ipynb
@@ -0,0 +1,49 @@
+{
+ "metadata": {
+  "name": "Chapter 5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 5 , Resistance and Wire Size"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Problem 5.5, Page number: 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "from __future__ import division\n\n#Comparing the Resistances of wires\n\n#Initialization\n\nA10=10380                         #Area in cmil\n\nA14=4106                          #Area in cmil\n\nfrom decimal import *\n\n#Calculation\n\nR14_R10=A10/A14\n\nprint \"The no. 14 wire has\",round(R14_R10,2),\"times as much resistance as the no. 10 wire\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The no. 14 wire has 2.53 times as much resistance as the no. 10 wire\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/PADMAVATHITHIYAGARAJAN/chapter5.ipynb b/sample_notebooks/PADMAVATHITHIYAGARAJAN/chapter5.ipynb
new file mode 100644
index 00000000..7ed0f4e9
--- /dev/null
+++ b/sample_notebooks/PADMAVATHITHIYAGARAJAN/chapter5.ipynb
@@ -0,0 +1,50 @@
+{
+ "metadata": {
+  "name": "chapter5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": "Chapter 5 , Resistance and Wire Size"
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Problem 5.5, Page number: 48"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "from __future__ import division\n\n#Comparing the Resistances of wires\n\n#Initialization\n\nA10=10380                         #Area in cmil\n\nA14=4106                          #Area in cmil\n\n#Calculation\n\nR14_R10=A10/A14\n\nprint \"The no. 14 wire has\",round(R14_R10,2),\"times as much resistance as the no. 10 wire\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "The no. 14 wire has 2.53 times as much resistance as the no. 10 wire\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/ParitoshMehta/ch4.ipynb b/sample_notebooks/ParitoshMehta/ch4.ipynb
new file mode 100755
index 00000000..c94b7f1e
--- /dev/null
+++ b/sample_notebooks/ParitoshMehta/ch4.ipynb
@@ -0,0 +1,147 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:6400e06363df718a0205b64ca5c3fc7d61a14e362b1dd10172150d22b3db1e89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Communication Filters and Signal Transmission"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5  Page No : 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array,log10,sqrt\n",
+      "\n",
+      "# Variables\n",
+      "f = array([500., 2000., 10000.]);    #frequency in Hz\n",
+      "\n",
+      "# Calculations\n",
+      "Af = 1/sqrt(1+(f/1000)**8);   #Linear amplitude response\n",
+      "AdBf = 20*log10(Af);\n",
+      "\n",
+      "# Results\n",
+      "print '   f,Hz     (Af)        (AdBf)'\n",
+      "for i in range(3):\n",
+      "    print ' %5i Hz   %.5f     %.3f dB'%(f[i],Af[i],AdBf[i])\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "   f,Hz     (Af)        (AdBf)\n",
+        "   500 Hz   0.99805     -0.017 dB\n",
+        "  2000 Hz   0.06238     -24.099 dB\n",
+        " 10000 Hz   0.00010     -80.000 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6  Page No : 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "L = 4.*10**-6;   #Henry\n",
+      "C = 9.*10**-12;  #Farad\n",
+      "R = 20.*10**3;    #ohm\n",
+      "\n",
+      "# Calculations and Results\n",
+      "f0 = 1/(2*math.pi*math.sqrt(L*C));    #frequency in Hz\n",
+      "print 'a) The resonant frequency is f0 = %.2f  MHz'%(f0*10**-6)\n",
+      "Q = R*math.sqrt(C/L)\n",
+      "print ' b) The Q is %i'%(Q);\n",
+      "B = f0/Q;\n",
+      "print ' c) The 3-dB bandwidth is B = %i KHz'%(B*10**-3);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The resonant frequency is f0 = 26.53  MHz\n",
+        " b) The Q is 30\n",
+        " c) The 3-dB bandwidth is B = 884 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7  Page No : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#misprinted example number\n",
+      "pulse_width = 2*10**-6;    #second\n",
+      "rise_time = 10*10**-9;      #second\n",
+      "\n",
+      "# Calculations and Results\n",
+      "B = .5/pulse_width;       #in Hz\n",
+      "print 'a) The aproximate bandwidth for coarse reproduction is B = %i  KHz'%(B*10**-3)\n",
+      "B = .5/rise_time;\n",
+      "print ' b) The aproximate bandwidth for fine reproduction is B = %i  MHz'%(B*10**-6)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The aproximate bandwidth for coarse reproduction is B = 250  KHz\n",
+        " b) The aproximate bandwidth for fine reproduction is B = 50  MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/PrashantSahu/Chapter_2_Molecular_Diffusion.ipynb b/sample_notebooks/PrashantSahu/Chapter_2_Molecular_Diffusion.ipynb
new file mode 100644
index 00000000..7df6880b
--- /dev/null
+++ b/sample_notebooks/PrashantSahu/Chapter_2_Molecular_Diffusion.ipynb
@@ -0,0 +1,657 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Chapter 2 : Molecular Diffusion\n",
+    "##Example 2.1 "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Molar average velocity of gas mixture is: 0.0303\n",
+      "Mass average velocity of gas mixture is: 0.029\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "N2 = 0.05   #mole fraction of Nitrogen denoted as 1\n",
+    "H2 = 0.15   #mole fraction of Hydrogen denoted as 2\n",
+    "NH3 = 0.76  #mole fraction of Ammonia denoted as 3\n",
+    "Ar = 0.04   #mole fraction of Argon denoted as 4\n",
+    "u1 = 0.03\n",
+    "u2 = 0.035\n",
+    "u3 = 0.03\n",
+    "u4 = 0.02\n",
+    "#Calculating molar average velocity\n",
+    "U = N2*u1 + H2*u2 + NH3*u3 + Ar*u4\n",
+    "print 'Molar average velocity of gas mixture is: %.4f'%U\n",
+    "#Calculating of mass average velocity\n",
+    "M1 = 28\n",
+    "M2 = 2\n",
+    "M3 = 17\n",
+    "M4 = 40\n",
+    "M = N2*M1 + H2*M2 + NH3*M3 + Ar*M4\n",
+    "u = (1/M)*(N2*M1*u1 + H2*M2*u2 + NH3*M3*u3 + Ar*M4*u4)\n",
+    "print 'Mass average velocity of gas mixture is: %.3f'%u"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    " ##Example 2.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false,
+    "scrolled": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Time for complete evaporation is: 15.93 hours\n",
+      "(b) Time for disappearance of water is: 8.87 hours\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "import numpy as np\n",
+    "\n",
+    "#Calcualtion for (a) part\n",
+    "#calculating vapor pressure of water at 301K\n",
+    "pv = math.exp(13.8573 - (5160.2/301))   #in bar\n",
+    "#wet-bulb temperature is 22.5 degree centigrade\n",
+    "#calculating mean air-film temperature\n",
+    "Tm = ((28+22.5)/2)+273  #in kelvin\n",
+    "#calculating diffusion coefficient\n",
+    "Dab = ((0.853*(30.48**2))*((298.2/273)**1.75))/(3600*10000)   #in m^2/s\n",
+    "l = 2.5e-3     #in m\n",
+    "P = 1.013      #in bar\n",
+    "R = 0.08317    #Gas constant\n",
+    "pAo = math.exp(13.8573 - (5160.2/295.2))    #vapor pressure of water at the wet-bulb temperature, 22.2C\n",
+    "pAl = 0.6*round(pv,4)\n",
+    "Na = (((round(Dab,7)*P)/(R*298.2*l))*math.log((P-pAl)/(P-round(pAo,3))))*18     #in kg/m^2s\n",
+    "#amount of water per m^2 of floor area is\n",
+    "thickness = 2e-3\n",
+    "Amount = thickness*1    #in m^3 \n",
+    "#density of water is 1000kg/m^3\n",
+    "#therefore in kg it is\n",
+    "amount = Amount*1000\n",
+    "Time_for_completion = amount/Na   #in seconds\n",
+    "Time_for_completion_hours = Time_for_completion/3600\n",
+    "print '(a) Time for complete evaporation is: %.2f'%Time_for_completion_hours,'hours'\n",
+    "\n",
+    "#Calculation for (b) part\n",
+    "water_loss = 0.1   #in kg/m^2.h\n",
+    "water_loss_by_evaporation = Na*3600\n",
+    "total_water_loss = water_loss + water_loss_by_evaporation\n",
+    "time_for_disappearance = amount/total_water_loss\n",
+    "print '(b) Time for disappearance of water is: %0.2f'%time_for_disappearance,'hours'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) The molar flux of Ammonia is:1.922E-05 gmol/cm^2.s\n",
+      "(b) and (c)\n",
+      "Velocity of A is 0.522 cm/s\n",
+      "Velocity of B is 0.000 cm/s\n",
+      "Mass average velocity of A is 0.439 cm/s\n",
+      "Molar average velocity of A is 0.47 cm/s\n",
+      "(d) Molar flux of NH3 is 3.062E-06 gmol/cm^2.s\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x3705ac8>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "%matplotlib inline\n",
+    "import matplotlib\n",
+    "import math\n",
+    "import numpy as np\n",
+    "from matplotlib import pyplot as plt\n",
+    "#calculation for (a) part\n",
+    "l = 1         #thickness of air in cm\n",
+    "pAo = 0.9     #in atm\n",
+    "pAl = 0.1     #in atm\n",
+    "Dab = 0.214   #in cm^2/s\n",
+    "T = 298       #in K\n",
+    "P = 1         #in atm\n",
+    "R = 82.1      #in (cm^3)(atm)/(K)(gmol)\n",
+    "#calculating molar flux of ammonia\n",
+    "Na = ((Dab*P)/(R*T*l))*math.log((P-pAl)/(P-pAo))\n",
+    "print '(a) The molar flux of Ammonia is:%0.3E'%Na,'gmol/cm^2.s'\n",
+    "\n",
+    "#calculation for (b) and (c) part\n",
+    "Nb = 0        #air is non-diffusing\n",
+    "U = (Na/(P/(R*T)))     #molar average velocity\n",
+    "yA = pAo/P\n",
+    "yB = pAl/P\n",
+    "uA = U/yA           #\n",
+    "uB = 0              #since Nb=0\n",
+    "Ma = 17\n",
+    "Mb = 29\n",
+    "M = Ma*yA + Mb*yB\n",
+    "u = uA*yA*Ma/M      #since u =(uA*phoA + uB*phoB)/pho\n",
+    "print '(b) and (c)'\n",
+    "print 'Velocity of A is %0.3f'%uA,'cm/s'\n",
+    "print 'Velocity of B is %0.3f'%uB,'cm/s'\n",
+    "print 'Mass average velocity of A is %0.3f'%u,'cm/s'\n",
+    "print 'Molar average velocity of A is %0.2f'%U,'cm/s'\n",
+    "\n",
+    "#calculation for (d) part\n",
+    "Ca = pAo/(R*T)\n",
+    "Ia = Ca*(uA - u)      #molar flux of NH3 relative to an observer moving\n",
+    "                      #with the mass average velocity \n",
+    "print '(d) Molar flux of NH3 is %0.3E'%Ia,'gmol/cm^2.s'\n",
+    "\n",
+    "z = []\n",
+    "pa =[]\n",
+    "for i in np.arange(0,1,0.01):\n",
+    "    z.append(i)\n",
+    "    \n",
+    "for i in range(0,len(z)):\n",
+    "    pa.append(1-(0.1*math.exp(2.197*z[i])))\n",
+    "    \n",
+    "from matplotlib.pyplot import*\n",
+    "plot(z,pa);\n",
+    "plt.xlabel('z(cm)');\n",
+    "plt.ylabel('pA(atm)');"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "##Example 2.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) Rate of diffusion of oxygen 7.151E-10 kmol/s\n",
+      "(b) The partial pressure gradient of oxygen at midway in diffusion path is: -4.25 bar/m\n",
+      "(c)\n",
+      "Molar average velocity and diffusion velocities at \"midway\"\n",
+      "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+      "The diffusion velocity of oxygen 7.9E-04 m/s\n",
+      "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+      "Molar average velocity and diffusion velocities at \"top of tube\"\n",
+      "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+      "The diffusion velocity of oxygen 3.72E-04 m/s\n",
+      "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+      "Molar average velocity and diffusion velocities at \"bottom of tube\"\n",
+      "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+      "The diffusion velocity of oxygen is not infinity\n",
+      "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+      "(d)\n",
+      "New molar flux of (A) 4.95E-06 kmol/m^2.s\n",
+      "New molar flux of (B) 7.19E-06 kmol/m^2.s\n"
+     ]
+    },
+    {
+     "data": {
+      "image/png": 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ogmzw96am8BARkYyS3PQkIiIJoEQhIiIZKVGIiEhGShQiIpKREoWIiGSkRCEi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\nAABJRU5ErkJggg==\n",
+      "text/plain": [
+       "<matplotlib.figure.Figure at 0x9dd6748>"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "#calculation of (a) part\n",
+    "#given data\n",
+    "import math\n",
+    "T = 298              #in kelvin\n",
+    "P = 1.013            #in bar\n",
+    "pAl = 0              #partial pressure of oxygen(A) at liquid surface\n",
+    "pAo = 0.21*1.013     #partial pressure of oxygen at open mouth\n",
+    "l = 0.05             #length of diffusion path in m\n",
+    "Dab = 2.1e-5         #diffusivity in m^2/s\n",
+    "R = 0.08317          #in m^3.bar.kmol.K\n",
+    "Na = Dab*P*math.log((P-pAl)/(P-pAo))/(R*T*l)       #in kmol/m^2.s\n",
+    "area = (math.pi/4)*(0.015)**2\n",
+    "rate = area*Na\n",
+    "print '(a) Rate of diffusion of oxygen %0.3E'%rate,'kmol/s'\n",
+    "z = []\n",
+    "pa =[]\n",
+    "for i in np.arange(0,1,0.01):\n",
+    "    z.append(i)\n",
+    "    \n",
+    "for i in range(0,len(z)):\n",
+    "    pa.append(P-(P-pAo)*math.exp((R*T*Na*z[i])/(Dab*P)))\n",
+    "    \n",
+    "from matplotlib.pyplot import*\n",
+    "plot(z,pa);\n",
+    "plt.xlabel('z(cm)');\n",
+    "plt.ylabel('pA(atm)');\n",
+    "\n",
+    "#calculation of (b) part\n",
+    "z = 0.025           #diffusion path\n",
+    "pA = 0.113          #in bar\n",
+    "#we have to find partial pressure gradient of oxygen at mid way of diffusion path\n",
+    "#let dpA/dz = ppd\n",
+    "ppd = -(R*T*round(Na,8)*(P-pA))/(Dab*P)\n",
+    "print '(b) The partial pressure gradient of oxygen at midway in diffusion path is: %0.2f'%ppd,'bar/m'\n",
+    "\n",
+    "#calculation of (c) part\n",
+    "uA = Na*(R*T/pA)     #velocity of oxygen\n",
+    "uB = 0               #since nitrogen is non-diffusing hence Nb = 0\n",
+    "U = pA*uA/P          #since U=1/C*(uA*Ca + uB*Cb)\n",
+    "vAd = uA - U         #diffusion velocity of oxygen\n",
+    "vBd = uB - U         #diffusion velocity of nitrogen\n",
+    "print '(c)'\n",
+    "print 'Molar average velocity and diffusion velocities at \"midway\"'\n",
+    "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+    "print 'The diffusion velocity of oxygen %0.1E'%vAd,'m/s'\n",
+    "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+    "#at z=0(at top of tube)\n",
+    "uA = Na*(R*T/pAo)\n",
+    "uB = 0\n",
+    "U = pAo*uA/P\n",
+    "vAd = uA - U\n",
+    "vBd = uB - U\n",
+    "print 'Molar average velocity and diffusion velocities at \"top of tube\"'\n",
+    "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+    "print 'The diffusion velocity of oxygen %0.2E'%vAd,'m/s'\n",
+    "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+    "#at z=0.05(at bottom of tube)\n",
+    "#uA = inf\n",
+    "uB = 0\n",
+    "U = pAo*uA/P\n",
+    "vAd = uA - U\n",
+    "vBd = uB - U\n",
+    "print 'Molar average velocity and diffusion velocities at \"bottom of tube\"'\n",
+    "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+    "print 'The diffusion velocity of oxygen is not infinity'\n",
+    "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+    "\n",
+    "#calculation of (d) part\n",
+    "V = -2*U\n",
+    "pA = 0.113\n",
+    "Nad = round(Na,8) - V*(pA/(R*T))\n",
+    "Nbd = 0 - (P - pA)*V/(R*T)\n",
+    "print '(d)'\n",
+    "print 'New molar flux of (A) %0.2E'%Nad,'kmol/m^2.s'\n",
+    "print 'New molar flux of (B) %0.2E'%Nbd,'kmol/m^2.s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a) The air-film thickness is :0.00193 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given data\n",
+    "area = 3*4           #in m^2\n",
+    "mperarea = 3.0/12    #in kg/m^2\n",
+    "#part (a)\n",
+    "P = 1.013            #in bar\n",
+    "Dab = 9.95e-6        #in m^2/s\n",
+    "R = 0.08317          #in m^3.bar./K.kmol\n",
+    "T = 273+27           #in K\n",
+    "#let d=1\n",
+    "d = 1                #in m\n",
+    "pAo = 0.065          #partial pressure of alcohol on liquid surface\n",
+    "pAd = 0              #partial pressure over d length of stagnant film of air\n",
+    "Na = (Dab*P*math.log((P-pAd)/(P-pAo)))/(R*T*d)     #in kmol/m^2.s\n",
+    "Na = Na*60           #in kg/m^2.s\n",
+    "flux = mperarea/(5*60)   #since the liquid evaporates completely in 5 minutes\n",
+    "#now we have to find the value of d\n",
+    "d = Na/flux\n",
+    "print '(a) The air-film thickness is :%0.5f'%d,'m'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "(a)\n",
+      "The steady-state flux is: 3.35E-06 kmol/m^2.s\n",
+      "The rate of transport of N2 from vessel 1 to 2: 6.6E-09 kmol/s\n",
+      "(b)\n",
+      "The flux and the rate of transport of oxygen is: -3.35E-06 kmol/m^2.s\n",
+      "(c)\n",
+      "Partial pressure at a point 0.05m from vessel 1 is: 1.2 atm\n",
+      "(d)\n",
+      "Net or total mass flux: -1.340E-05 kmol/m^2.s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given data\n",
+    "#part (a)\n",
+    "Dab = 0.23e-4*0.5*(293.0/316)**1.75     #in m^2/s\n",
+    "pA1 = 2*0.8           #in atm\n",
+    "pA2 = 2*0.2           #in atm\n",
+    "l = 0.15              #in m\n",
+    "R = 0.0821            #in m^3.atm./K.kmol\n",
+    "T = 293               #in K\n",
+    "Ma = 28\n",
+    "Mb = 32\n",
+    "Na = Dab*(pA1-pA2)/(R*T*l)   #in kmol/m^2.s\n",
+    "area = math.pi/4*(0.05)**2     #in m^2\n",
+    "rate = area*Na\n",
+    "print '(a)'\n",
+    "print 'The steady-state flux is: %0.2E'%Na,'kmol/m^2.s'\n",
+    "print 'The rate of transport of N2 from vessel 1 to 2: %0.1E'%rate,'kmol/s'\n",
+    "\n",
+    "#part (b)\n",
+    "Nb = -Na\n",
+    "print '(b)'\n",
+    "print 'The flux and the rate of transport of oxygen is: %0.2E'%Nb,'kmol/m^2.s'\n",
+    "\n",
+    "#part (c)\n",
+    "#let dpA/dz = ppg\n",
+    "dz = 0.05               #in m\n",
+    "ppg = (pA2 - pA1)/l     #in atm/m\n",
+    "pA = pA1 + (ppg)*dz     #in atm\n",
+    "print '(c)'\n",
+    "print 'Partial pressure at a point 0.05m from vessel 1 is: %0.1f'%pA,'atm'\n",
+    "\n",
+    "#part (d)\n",
+    "nt = Ma*Na + Mb*Nb\n",
+    "print '(d)'\n",
+    "print 'Net or total mass flux: %0.3E'%nt,'kmol/m^2.s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Methanol flux: 4.64e-05 kmol/m^2.s\n",
+      "Water flux: -4.06e-05 kmol/m^2.s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given data\n",
+    "Ha = 274.6*32     #molar latent heat of methanol(a)\n",
+    "Hb = 557.7*18     #molar latent heat of water(b)\n",
+    "yAl = 0.76        #mole fraction of methanol in the vapour\n",
+    "yAo = 0.825       #mole fraction of methanol in the vapour at the liquid-vapour interface\n",
+    "P = 1             #in atm\n",
+    "l = 1e-3          #in m\n",
+    "T =344.2          #in K\n",
+    "R = 0.0821        #m^3.atm./K.kmol\n",
+    "Dab = 1.816e-5    #in m^2/s\n",
+    "Na = Dab*P*math.log((1-0.1247*yAl)/(1-0.1247*yAo))/(0.1247*R*T*l)\n",
+    "print 'Methanol flux: %0.2e'%Na,'kmol/m^2.s'\n",
+    "Nb = -(Ha/Hb)*Na\n",
+    "print 'Water flux: %0.2e'%Nb,'kmol/m^2.s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Value of pA1 is 0.937 atm\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given values\n",
+    "import math\n",
+    "V1 = 3000      #in cm^3\n",
+    "V2 = 4000      #in cm^3\n",
+    "Dab = 0.23     #in cm^2/s\n",
+    "Dba = 0.23     #in cm^2/s\n",
+    "l1 = 4         #in cm\n",
+    "d1 = 0.5       #in cm\n",
+    "l2 = 2         #in cm\n",
+    "d2 = 0.3       #in cm\n",
+    "pA3 = 1        #in atm\n",
+    "#unknowns\n",
+    "# pA1   and   pA2\n",
+    "# dpA1bydt = (Dab/V1*l1)*((pA1)-(pA2))*((math.pi*(d1**2))/4)\n",
+    "#on integrating using Laplace trandformation\n",
+    "# initial conditions\n",
+    "t=18000       #in seconds\n",
+    "pA1 = 1-0.57*(math.exp((-1.005)*(10**(-6))*t)-math.exp((-7.615)*(10**(-6))*t))\n",
+    "print 'Value of pA1 is %0.3f'%pA1,'atm'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "##Example 2.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Value of flux of water vapour: 2.96E-05 kmol/m^2.s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given values\n",
+    "import math\n",
+    "y1l = 0           #mol fraction of dry air\n",
+    "y10 = (17.53/760) #mol fraction of water\n",
+    "l = 1.5           #in mm\n",
+    "C = 0.0409        #in kmol/m^3 : calculated by P/RT\n",
+    "D12 = 0.923       #Diffusivity of hydrogen over water\n",
+    "D13 = 0.267       #Diffusivity of oxygen over water\n",
+    "y2 = 0.6          #mole fraction of hydrogen\n",
+    "y3 = 0.4          #mole fraction of oxygen\n",
+    "D1m = 1/((y2/D12)+(y3/D13))       #calculating mean diffusivity\n",
+    "Ni = (D1m*C*1000/(l*10000))*math.log((1-y1l)/(1-y10))\n",
+    "print 'Value of flux of water vapour: %0.2E'%Ni,'kmol/m^2.s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Flux of ethane 4.804E-05 gmol/cm^2.s\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given data\n",
+    "y1 = 0.4          #mole fraction of ethane(1)\n",
+    "y2 = 0.3          #mole fraction of ethylene(2)\n",
+    "y3 = 0.3          #mole fraction of hydrogen(3)\n",
+    "#calculating D13\n",
+    "#The Lennard-Jones parameters are\n",
+    "sigma1 = 4.443    #in angstrom\n",
+    "sigma2 = 4.163    #in angstrom\n",
+    "sigma3 = 2.827    #in angstrom\n",
+    "e1byk = 215.7\n",
+    "e2byk = 224.7\n",
+    "e3byk = 59.7\n",
+    "sigma13 = (sigma1 + sigma3)/2     #in angstrom\n",
+    "e13byk = (e1byk*e3byk)**0.5\n",
+    "kTbye13 = 993/113.5\n",
+    "ohmD13 = 0.76      #from collision integral table\n",
+    "D13 = ((0.001858)*(993**1.5)*((1.0/30)+(1.0/2))**0.5)/((2)*(sigma13**2)*(ohmD13))\n",
+    "#calculating D23\n",
+    "sigma23 = (sigma2+sigma3)/2\n",
+    "kTbye23 = ((993/224.7)*(993/59.7))*0.5\n",
+    "ohmD23 = 0.762\n",
+    "D23 = (0.001858*(993**1.5)*((1.0/28)+(1.0/2))**0.5)/(2*(sigma23**2)*ohmD23)\n",
+    "D = (D13+D23)/2   #in cm^2/s\n",
+    "l = 0.15          #in cm\n",
+    "#at z=0 (bulk gas)\n",
+    "y10 = 0.6\n",
+    "y20 = 0.2\n",
+    "y30 = 0.2\n",
+    "#at z=l (catalyst surface)\n",
+    "y1l = 0.4\n",
+    "y2l = 0.3\n",
+    "y3l = 0.3\n",
+    "C = 2.0/(82.1*993)   #calculated by P/RT\n",
+    "N1 = (D*C/l)*math.log((y10+y20)/(y1l+y2l))\n",
+    "print 'Flux of ethane %0.3E'%N1,'gmol/cm^2.s'"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Raj Phani/chapter1.ipynb b/sample_notebooks/Raj Phani/chapter1.ipynb
new file mode 100755
index 00000000..af42a9cd
--- /dev/null
+++ b/sample_notebooks/Raj Phani/chapter1.ipynb
@@ -0,0 +1,993 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# chapter 1: Atomic Nucleus"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.1;pg no:2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.1, Page:2  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The electric field in V/m is = 20000.0\n",
+      "\n",
+      " The force in N/C is = 20000.0\n",
+      "\n",
+      " The force on metal sphere in N is = 7.6e-05\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of elelectric field and force\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.1, Page:2  \\n \\n\"\n",
+    "#Given:\n",
+    "v=1000# potential\n",
+    "d=0.05# distance\n",
+    "q=3.8*10**-9# charge\n",
+    "#solution:\n",
+    "e=v/d;#electric field\n",
+    "f=e;# force\n",
+    "f1=f*q;# force on metal sphere\n",
+    "print\"\\n The electric field in V/m is =\",e\n",
+    "print\"\\n The force in N/C is =\",f\n",
+    "print\"\\n The force on metal sphere in N is =\",f1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.2;pg no:2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.2, Page:2  \n",
+      " \n",
+      "\n",
+      "The potential in V is = 80.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of potential\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.2, Page:2  \\n \\n\"\n",
+    "#Given:\n",
+    "energy=2*10**-6\n",
+    "c=2.5*10**-8# velocity of light\n",
+    "#solution:\n",
+    "v=energy/c# potential\n",
+    "print\"The potential in V is =\",v"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.3;pg no:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.3, Page:3 \n",
+      "\n",
+      "The wavelength in Angstroms is = 3.88\n",
+      "The photon wavelength in Angstroms is = 1242.38\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of elecrtron and photon wavelength\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.3, Page:3 \\n\"\n",
+    "#Given:\n",
+    "energy=10 #in electron volts\n",
+    "m=9.1*10**-31# mass of electron in kg\n",
+    "h=6.626*10**-34# planck's constant J.s\n",
+    "c=3*10**8# speed of light in m/s\n",
+    "#solution (a):\n",
+    "energy1=energy*1.6*10**-19# energy in J\n",
+    "p=(2*m*energy1)**0.5# momentum\n",
+    "wavelength=h/p*(10)**10\n",
+    "print\"The wavelength in Angstroms is =\",round(wavelength,2)\n",
+    "#solution (b):\n",
+    "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n",
+    "print\"The photon wavelength in Angstroms is =\",round(wavelength1,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.4;pg no:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.4, Page:3  \n",
+      " \n",
+      "\n",
+      "The energy in eV is = 150.77\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of kinetic energy of an electron\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.4, Page:3  \\n \\n\"\n",
+    "#Given:\n",
+    "wavelength=10**-10\n",
+    "m=9.1*10**-31\n",
+    "h=6.626*10**-34\n",
+    "#solution:\n",
+    "p=h/wavelength\n",
+    "e=p*p/(2*m) # energy in J\n",
+    "e1=e/(1.6*10**-19)# energy in eV\n",
+    "print\"The energy in eV is =\",round(e1,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.5;pg no:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.5, Page:3  \n",
+      " \n",
+      "\n",
+      "The wavelength in 10^-5 Angstroms is = 0.66\n",
+      "The wavelength in 10^-5 Angstroms is = 0.65\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of wavelength of oxygen and nitrogen nucleus\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.5, Page:3  \\n \\n\"\n",
+    "#Given:\n",
+    "m=1.66*10**-27# 1u=1.66*10^-27 kg\n",
+    "h=6.6262*10**-34#planck's constant in J.s\n",
+    "energy1=120# in Mev for oxygen\n",
+    "energy2=140# in MeV for nitrogen\n",
+    "#solution(a):\n",
+    "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n",
+    "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+    "print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength1,2)\n",
+    "#solution (b):\n",
+    "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n",
+    "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+    "print\"The wavelength in 10^-5 Angstroms is =\",round(wavelength2,2)\n",
+    "# 1 Angstrom = 10^-10 m"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.6;pg no:3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.6, Page:3  \n",
+      " \n",
+      "\n",
+      "The energy in eV is = 8275.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of energy of a gamma photon\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.6, Page:3  \\n \\n\"\n",
+    "#Given:\n",
+    "wavelength=1.5*10**-10\n",
+    "h=6.62*10**-34\n",
+    "c=3*10**8\n",
+    "#solution:\n",
+    "e=(h*c)/wavelength# energy in J\n",
+    "e1=e/(1.6*10**-19)# energy in eV\n",
+    "print\"The energy in eV is =\",e1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.7;pg no:4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.7, Page:4  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The threshold frequency in s^-1 is = 1.23634168427e+15\n",
+      "\n",
+      " The threshold wavelength in Angstroms is = 2426.51\n",
+      "\n",
+      " The energy of photoelectrone in eV is = 3.91\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of threshold frequency,wavelength,energy of photoelectrone\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.7, Page:4  \\n \\n\"\n",
+    "#Given:\n",
+    "E=5.12*1.6*10**-19# energy in J\n",
+    "h=6.626*10**-34\n",
+    "c=3*10**8\n",
+    "wavelength=200*10**-9\n",
+    "w=2.3# in eV\n",
+    "#solution:\n",
+    "tf=E/h# (part a)\n",
+    "print\"\\n The threshold frequency in s^-1 is =\",round(tf,2)\n",
+    "tl=c/tf*10**10# (part b)\n",
+    "print\"\\n The threshold wavelength in Angstroms is =\",round(tl,2)\n",
+    "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n",
+    "pe=e-w\n",
+    "print\"\\n The energy of photoelectrone in eV is =\",round(pe,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.8;pg no:4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.8, Page:4  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The velocity of alpha particles for 1 MeV in mega m/s is = 6.94\n",
+      "\n",
+      " The velocity of alpha particles for 2 MeV in mega m/s is = 9.82\n",
+      "\n",
+      " The velocity of deuteron particles for 1 MeV in mega m/s is = 9.82\n",
+      "\n",
+      " The velocity of deuteron particles for 2 MeV in mega m/s is = 13.88\n",
+      "\n",
+      " The velocity of proton particles for 1 MeV in mega m/s is = 13.88\n",
+      "\n",
+      " The velocity of proton particles for 2 MeV in mega m/s is = 19.63\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of velocity of alpha particles,deuteron,proton\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.8, Page:4  \\n \\n\"\n",
+    "#Given:\n",
+    "e1=1 # in MeV\n",
+    "e2=2 # in MeV\n",
+    "ma=4 # in u(amu)\n",
+    "md=2 # in u(amu)\n",
+    "mp=1 # in u(amu)\n",
+    "# 1u = 1.6*10^-27 Kg\n",
+    "#solution: part a)For alpha particles\n",
+    "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of alpha particles for 1 MeV in mega m/s is =\",round(v1a/10**6,2)# For 1 MeV\n",
+    "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of alpha particles for 2 MeV in mega m/s is =\",round(v2a/10**6,2)# For 2 MeV\n",
+    "#solution: part b)For deuteron particles\n",
+    "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of deuteron particles for 1 MeV in mega m/s is =\",round(v1b/10**6,2) # For 1 MeV\n",
+    "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of deuteron particles for 2 MeV in mega m/s is =\",round(v2b/10**6,2) # For 2 MeV\n",
+    "#solution: part c)For proton particles\n",
+    "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of proton particles for 1 MeV in mega m/s is =\",round(v1p/10**6,2) # For 1 MeV\n",
+    "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+    "print\"\\n The velocity of proton particles for 2 MeV in mega m/s is =\",round(v2p/10**6,2) # For 2 MeV"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.9;pg no:5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.9, Page:5  \n",
+      " \n",
+      "\n",
+      "The energy in MeV is = 934.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The energy equivalence\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.9, Page:5  \\n \\n\"\n",
+    "#Given:\n",
+    "m=1./(6.023*10**23)#mass of 1 atom in g\n",
+    "m1=m*10**-3#mass of 1 atom in Kg\n",
+    "c=3.*10**8# velocity in m/s\n",
+    "#solution:\n",
+    "e=m1*c*c; # energy in J\n",
+    "e1=e/(1.6*10**-13)# energy in MeV\n",
+    "print\"The energy in MeV is =\",round(e1)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.10;pg no:5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.10, Page:5  \n",
+      " \n",
+      "\n",
+      "The energy in eV is = 13.26\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The energy of formation\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.10, Page:5  \\n \\n\"\n",
+    "#Given:\n",
+    "enthalpy=1278 # enthalpy of combustion in kJ/mol\n",
+    "#solution:\n",
+    "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n",
+    "print\"The energy in eV is =\",round(energy,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.11;pg no:5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.11, Page:5  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The mean binding energy of helium atom in MeV is = 7.07\n",
+      "\n",
+      " The mean binding energy of oxygen atom in MeV is = 7.98\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of mean binding energy of helium and oxygen\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.11, Page:5  \\n \\n\"\n",
+    "#Given:\n",
+    "mh=1.0078\n",
+    "mn=1.0087\n",
+    "ma=4.0026\n",
+    "mo=15.9949\n",
+    "Ah=4.0026 # atomic mass of helium\n",
+    "Ao=15.9949 # atomic mass of oxygen\n",
+    "#solution:\n",
+    "# part (a)\n",
+    "B1=(2*mh+2*mn-ma)*931 # in MeV\n",
+    "Bh=B1/Ah\n",
+    "print\"\\n The mean binding energy of helium atom in MeV is =\",round(Bh,2)\n",
+    "# part (b)\n",
+    "B2=(8*mh+8*mn-mo)*931 # in MeV\n",
+    "Bo=B2/Ao\n",
+    "print\"\\n The mean binding energy of oxygen atom in MeV is =\",round(Bo,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.12;pg no:6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.12, Page:6 \n",
+      "\n",
+      "\n",
+      " The mean binding energy of Be atom in MeV is = 7.059\n",
+      "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of mean binding energy of Be atom\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.12, Page:6 \\n\"\n",
+    "#Given:\n",
+    "mh=1.0078;\n",
+    "mn=1.0087;\n",
+    "ABe=8.0053; # atomic mass of beryllium\n",
+    "#solution:\n",
+    "B1=(4*mh+4*mn-ABe)*931; # in MeV\n",
+    "Bh=B1/ABe;\n",
+    "print\"\\n The mean binding energy of Be atom in MeV is =\",round(Bh,3)\n",
+    "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.13;pg no:6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.13, Page:6 \n",
+      "\n",
+      "The amount of coal required in Kg is = 2.5\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of amount of coal\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.13, Page:6 \\n\"\n",
+    "#Given:\n",
+    "e=200; # in Mev\n",
+    "m=0.235; # weight of uranium atom in Kg\n",
+    "enthalpy=393.5; # in KJ/mol\n",
+    "Na=6.02*10**23;\n",
+    "#solution:\n",
+    "e1=e*1.6*10**-19*10**6;\n",
+    "atoms=Na/m;\n",
+    "e2=atoms*e1;#energy released in J\n",
+    "m1=(e2*12)/(393.5*1000*1000);# in Kg\n",
+    "m2=m1/1000;# in tons\n",
+    "print\"The amount of coal required in Kg is =\", round(m2/1000,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.14;pg no:7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.14, Page:7  \n",
+      " \n",
+      "\n",
+      "The energy release in part (a) in eV/molecule is = 2.51\n",
+      "The energy release in part (b) in eV/molecule is = 9.23\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The energy releases\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.14, Page:7  \\n \\n\"\n",
+    "#Given:\n",
+    "H1=241.8; # in KJ/mol\n",
+    "H2=887.2; # in KJ/mol\n",
+    "# 1 KJ/mol = 0.0104 eV/atom\n",
+    "#solution: part (a)\n",
+    "e1=H1*0.0104;\n",
+    "print\"The energy release in part (a) in eV/molecule is =\",round(e1,2)\n",
+    "#solution: part (b)\n",
+    "e2=H2*0.0104;\n",
+    "print\"The energy release in part (b) in eV/molecule is =\",round(e2,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.15;pg no:7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.15, Page:7 \n",
+      "\n",
+      "The energy release in part (a) in KJ/mol of carbondioxide is = 394.9\n",
+      "The energy release in part (b) in KJ/mol of alumina is = 1676.0\n",
+      "The energy release in part (c) in MJ/atom of U(235) is = 19.264\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The energy releases\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.15, Page:7 \\n\"\n",
+    "#Given:\n",
+    "H1=4.1; # in eV/molecule\n",
+    "H2=17.4; # in eV/molecule\n",
+    "H3=200;# in MeV/atom of U\n",
+    "# 1 eV/atom  = 96.32 KJ/mol\n",
+    "#solution: part (a)\n",
+    "e1=H1*96.32;\n",
+    "print\"The energy release in part (a) in KJ/mol of carbondioxide is =\",round(e1,1)\n",
+    "#solution: part (b)\n",
+    "e2=H2*96.32;\n",
+    "print\"The energy release in part (b) in KJ/mol of alumina is =\",round(e2,1)\n",
+    "#solution: part (c)\n",
+    "e3=H3*1000*96.32;# in MJ/atom of U(235)\n",
+    "print\"The energy release in part (c) in MJ/atom of U(235) is =\",round(e3/10**6,3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.16;pg no:7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.16, Page:7  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The rate of energy release in MW is= 949.25\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The rate of energy release\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.16, Page:7  \\n \\n\"\n",
+    "#Given:\n",
+    "e=200.; #MeV/ atom of U\n",
+    "# 1 eV = 1.6*10^-19 J\n",
+    "Na=6.023*10**23;\n",
+    "M=0.235; # mass in Kg\n",
+    "#solution:\n",
+    "e1=e*1.6*10**-19*10**6;\n",
+    "A=Na/M;\n",
+    "e2=A*e1; # energy released in MJ/day\n",
+    "e3=e2/(24.*3600.);\n",
+    "print\"\\n The rate of energy release in MW is=\",round(e3/10**6,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.17;pg no:8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.17, Page:8  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The mass loss in 10^-27 Kg/He formed is = 0.0464\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The mass loss\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.17, Page:8  \\n \\n\"\n",
+    "#Given:\n",
+    "e=26.03; # in MeV\n",
+    "#solution:\n",
+    "loss=e/931; #in atomic mass units (u)\n",
+    "# 1 u = 1.66*10^-27 Kg\n",
+    "m=(loss*1.66*10**-27)/(1*10**-27);\n",
+    "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",round(m,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.18;pg no:8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.18, Page:8  \n",
+      "\n",
+      "\n",
+      " The energy loss in MeV is = -4.0312\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of The energy loss\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.18, Page:8  \\n\"\n",
+    "#Given:\n",
+    "mh=1.007825;\n",
+    "mt=3.016049;\n",
+    "md=2.014102;\n",
+    "#solution:\n",
+    "m1=(mh+mt-2*md);\n",
+    "e=(-m1)*931; # in MeV\n",
+    "print\"\\n The energy loss in MeV is =\",round(-e,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.19;pg no:8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.19, Page:8 \n",
+      "\n",
+      "The mean binding energy of tritium atom in MeV is = 2.811\n",
+      "The mean binding energy of nickel atom in MeV is = 8.716\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of  mean binding energy of tritium and nickel atom\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.19, Page:8 \\n\"\n",
+    "#Given:\n",
+    "mh=1.007825;\n",
+    "mn=1.008665;\n",
+    "mt=3.016049; # atomic mass of Tritium\n",
+    "mNi=59.93528; # atomic mass of Nickel\n",
+    "#solution:\n",
+    "# part (a)\n",
+    "B1=(1*mh+2*mn-mt)*931; # in MeV\n",
+    "Bh=B1/mt;\n",
+    "print\"The mean binding energy of tritium atom in MeV is =\",round(Bh,3)\n",
+    "# part (b)\n",
+    "B2=(28*mh+32*mn-mNi)*931; # in MeV\n",
+    "Bo=B2/mNi;\n",
+    "print\"The mean binding energy of nickel atom in MeV is =\",round(Bo,3)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.20;pg no:9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.20, Page:9  \n",
+      "\n",
+      "The mean binding energy of Cl (35) atom in MeV is = 8.5281\n",
+      "The mean binding energy of Cl (37) atom in MeV is = 8.5784\n",
+      "The increase in mean binding energy of Cl atom in MeV is = 0.05\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of mean binding energy of Cl\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.20, Page:9  \\n\"\n",
+    "#Given:\n",
+    "mh=1.00783;\n",
+    "mn=1.00867;\n",
+    "m35=34.96885; # atomic mass of Cl (35)\n",
+    "m37=36.96590; # atomic mass of Cl (37)\n",
+    "#solution:\n",
+    "B1=(17*mh+18*mn-m35)*931; # in MeV\n",
+    "Bh=B1/m35;\n",
+    "print\"The mean binding energy of Cl (35) atom in MeV is =\",round(Bh,4)\n",
+    "B2=(17*mh+20*mn-m37)*931; # in MeV\n",
+    "Bo=B2/m37;\n",
+    "print\"The mean binding energy of Cl (37) atom in MeV is =\",round(Bo,4)\n",
+    "Bi=Bo-Bh;\n",
+    "print\"The increase in mean binding energy of Cl atom in MeV is =\",round(Bi,2)\n",
+    "# NOTE: The answer depends upon how much precise value you take for atomic masses."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.21;pg no:9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.21, Page:9  \n",
+      " \n",
+      "\n",
+      "\n",
+      " The mean binding energy of Na(22) in MeV is = 7.9236\n",
+      "\n",
+      " The mean binding energy of Na(23)in MeV is = 8.1154\n",
+      "\n",
+      " The mean binding energy of Na(24) in MeV is = 8.0717\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of mean binding energy of Na\n",
+    "#intiation of all variables\n",
+    "# Chapter 1\n",
+    "print\"Example 1.21, Page:9  \\n \\n\"\n",
+    "#Given:\n",
+    "mh=1.0078;\n",
+    "mn=1.0087;\n",
+    "m22=21.99431;# atomic mass of Na 22\n",
+    "m23=22.9898;# atomic mass of Na 23\n",
+    "m24=23.9909;# atomic mass of Na 24\n",
+    "#solution:\n",
+    "# part (a)\n",
+    "B1=((11*mh+11*mn)-m22)*931; # in MeV\n",
+    "Bh=B1/m22;\n",
+    "print\"\\n The mean binding energy of Na(22) in MeV is =\",round(Bh,4)\n",
+    "# part (b)\n",
+    "B2=((11*mh+12*mn)-m23)*931; # in MeV\n",
+    "Bo=B2/m23;\n",
+    "print\"\\n The mean binding energy of Na(23)in MeV is =\",round(Bo,4)\n",
+    "# part (c)\n",
+    "B3=((11*mh+13*mn)-m24)*931; # in MeV\n",
+    "Bs=B3/m24;\n",
+    "print\"\\n The mean binding energy of Na(24) in MeV is =\",round(Bs,4)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures.ipynb b/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures.ipynb
new file mode 100755
index 00000000..43ba034f
--- /dev/null
+++ b/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures.ipynb
@@ -0,0 +1,548 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.1, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "temperature rise is 8.43 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "El=10**-2*50;      #energy loss(J)\n",
+    "H=El*60;           #heat produced(J)\n",
+    "d=7.7*10**3;       #iron rod(kg/m**3)\n",
+    "s=0.462*10**-3;    #specific heat(J/kg K)\n",
+    "\n",
+    "#Calculation\n",
+    "theta=H/(d*s);     #temperature rise(K)\n",
+    "\n",
+    "#Result\n",
+    "print \"temperature rise is\",round(theta,2),\"K\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.2, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "magnetic field at the centre is 14.0 weber/m**2\n",
+      "dipole moment is 9.0 *10**-24 ampere/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "e=1.6*10**-19;    #charge(coulomb)\n",
+    "new=6.8*10**15;   #frequency(revolutions per second)\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "R=5.1*10**-11;     #radius(m)\n",
+    "\n",
+    "#Calculation\n",
+    "i=round(e*new,4);          #current(ampere)\n",
+    "B=mew0*i/(2*R);            #magnetic field at the centre(weber/m**2)\n",
+    "A=math.pi*R**2;\n",
+    "d=i*A;                    #dipole moment(ampere/m**2)\n",
+    "\n",
+    "#Result\n",
+    "print \"magnetic field at the centre is\",round(B),\"weber/m**2\"\n",
+    "print \"dipole moment is\",round(d*10**24),\"*10**-24 ampere/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.3, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "intensity of magnetisation is 5.0 ampere/m\n",
+      "flux density in material is 1.257 weber/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "chi=0.5*10**-5;    #magnetic susceptibility\n",
+    "H=10**6;     #field strength(ampere/m)\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "\n",
+    "#Calculation\n",
+    "I=chi*H;     #intensity of magnetisation(ampere/m)\n",
+    "B=mew0*(I+H);    #flux density in material(weber/m**2)\n",
+    "\n",
+    "#Result\n",
+    "print \"intensity of magnetisation is\",I,\"ampere/m\"\n",
+    "print \"flux density in material is\",round(B,3),\"weber/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.4, Page number 6.47"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of Bohr magnetons is 2.22 bohr magneon/atom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "B=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
+    "a=2.86*10**-10;      #edge(m)\n",
+    "Is=1.76*10**6;       #saturation value of magnetisation(ampere/m)\n",
+    "\n",
+    "#Calculation\n",
+    "N=2/a**3;\n",
+    "mew_bar=Is/N;      #number of Bohr magnetons(ampere m**2)\n",
+    "mew_bar=mew_bar/B;      #number of Bohr magnetons(bohr magneon/atom)\n",
+    "\n",
+    "#Result\n",
+    "print \"number of Bohr magnetons is\",round(mew_bar,2),\"bohr magneon/atom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.5, Page number 6.47"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "average magnetic moment is 2.79 *10**-3 bohr magneton/spin\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "H=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
+    "beta=10**6;      #field(ampere/m)\n",
+    "k=1.38*10**-23;    #boltzmann constant\n",
+    "T=303;    #temperature(K)\n",
+    "\n",
+    "#Calculation\n",
+    "mm=mew0*H*beta/(k*T);    #average magnetic moment(bohr magneton/spin)\n",
+    "\n",
+    "#Result\n",
+    "print \"average magnetic moment is\",round(mm*10**3,2),\"*10**-3 bohr magneton/spin\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.6, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "hysteresis loss per cycle is 188.0 J/m**3\n",
+      "hysteresis loss per second is 9400.0 watt/m**3\n",
+      "power loss is 1.23 watt/kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "A=94;      #area(m**2)\n",
+    "vy=0.1;    #value of length(weber/m**2)\n",
+    "vx=20;     #value of unit length\n",
+    "n=50;      #number of magnetization cycles\n",
+    "d=7650;    #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "h=A*vy*vx;     #hysteresis loss per cycle(J/m**3)\n",
+    "hs=h*n;       #hysteresis loss per second(watt/m**3)\n",
+    "pl=hs/d;      #power loss(watt/kg)\n",
+    "\n",
+    "#Result\n",
+    "print \"hysteresis loss per cycle is\",h,\"J/m**3\"\n",
+    "print \"hysteresis loss per second is\",hs,\"watt/m**3\"\n",
+    "print \"power loss is\",round(pl,2),\"watt/kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.7, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a= 5.43 Angstorm\n",
+      "density = 6.88 kg/m**3\n",
+      "#Answer given in the textbook is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "d=2.351                 #bond lenght\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "n=8                     #number of atoms in unit cell\n",
+    "A=28.09                 #Atomin mass of silicon\n",
+    "m=6.02*10**26           #1mole\n",
+    "\n",
+    "#Calculations\n",
+    "a=(4*d)/math.sqrt(3)\n",
+    "p=(n*A)/((a*10**-10)*m)    #density\n",
+    "\n",
+    "#Result\n",
+    "print \"a=\",round(a,2),\"Angstorm\"\n",
+    "print \"density =\",round(p*10**16,2),\"kg/m**3\"\n",
+    "print\"#Answer given in the textbook is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.8, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " radius of largest sphere is 0.154700538379252*r\n",
+      "maximum radius of sphere is 0.414213562373095*r\n"
+     ]
+    }
+   ],
+   "source": [
+    " import math\n",
+    "from __future__ import division\n",
+    "from sympy import Symbol\n",
+    "\n",
+    "#Variable declaration\n",
+    "r=Symbol('r')\n",
+    "\n",
+    "#Calculation\n",
+    "a1=4*r/math.sqrt(3);\n",
+    "R1=(a1/2)-r;           #radius of largest sphere\n",
+    "a2=4*r/math.sqrt(2);\n",
+    "R2=(a2/2)-r;       #maximum radius of sphere\n",
+    "\n",
+    "#Result\n",
+    "print \"radius of largest sphere is\",R1\n",
+    "print \"maximum radius of sphere is\",R2    "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.9, Page number 6.49"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a1= 2.905 Angstrom\n",
+      "Unit cell volume =a1**3 = 24.521 *10**-30 m**3\n",
+      "Volume occupied by one atom = 12.26 *10**-30 m**3\n",
+      "a2= 3.654 Angstorm\n",
+      "Unit cell volume =a2**3 = 48.8 *10**-30 m**3\n",
+      "Volume occupied by one atom = 12.2 *10**-30 m**3\n",
+      "Volume Change in % = 0.493\n",
+      "Density Change in % = 0.5\n",
+      "Thus the increase of density or the decrease of volume is about 0.5%\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "r1=1.258            #Atomic radius of BCC\n",
+    "r2=1.292            #Atomic radius of FCC\n",
+    "\n",
+    "#calculations\n",
+    "a1=(4*r1)/math.sqrt(3)       #in BCC\n",
+    "b1=((a1)**3)*10**-30         #Unit cell volume\n",
+    "v1=(b1)/2                    #Volume occupied by one atom\n",
+    "a2=2*math.sqrt(2)*r2         #in FCC\n",
+    "b2=(a2)**3*10**-30                   #Unit cell volume\n",
+    "v2=(b2)/4                    #Volume occupied by one atom  \n",
+    "v_c=((v1)-(v2))*100/(v1)     #Volume Change in % \n",
+    "d_c=((v1)-(v2))*100/(v2)     #Density Change in %\n",
+    "\n",
+    "#Results\n",
+    "print \"a1=\",round(a1,3),\"Angstrom\" \n",
+    "print \"Unit cell volume =a1**3 =\",round((b1)/10**-30,3),\"*10**-30 m**3\"\n",
+    "print \"Volume occupied by one atom =\",round(v1/10**-30,2),\"*10**-30 m**3\"\n",
+    "print \"a2=\",round(a2,3),\"Angstorm\"\n",
+    "print \"Unit cell volume =a2**3 =\",round((b2)/10**-30,3),\"*10**-30 m**3\"\n",
+    "print \"Volume occupied by one atom =\",round(v2/10**-30,2),\"*10**-30 m**3\"\n",
+    "print \"Volume Change in % =\",round(v_c,3)\n",
+    "print \"Density Change in % =\",round(d_c,2)\n",
+    "print \"Thus the increase of density or the decrease of volume is about 0.5%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.10, Page number 6.50"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a= 0.563 *10**-9 metre\n",
+      "spacing between the nearest neighbouring ions = 0.2814 nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "n=4     \n",
+    "M=58.5                  #Molecular wt. of NaCl\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "rho=2180                #density\n",
+    "\n",
+    "#Calculations\n",
+    "a=((n*M)/(N*rho))**(1/3)    \n",
+    "s=a/2\n",
+    "\n",
+    "#Result\n",
+    "print \"a=\",round(a/10**-9,3),\"*10**-9 metre\"\n",
+    "print \"spacing between the nearest neighbouring ions =\",round(s/10**-9,4),\"nm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.11, Page number 6.51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant, a= 0.36 nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "n=4     \n",
+    "A=63.55                 #Atomic wt. of NaCl\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "rho=8930                #density\n",
+    "\n",
+    "#Calculations\n",
+    "a=((n*A)/(N*rho))**(1/3)      #Lattice Constant\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant, a=\",round(a*10**9,2),\"nm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.12, Page number 6.51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Density of iron = 8805.0 kg/m**-3\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#variable declaration\n",
+    "r=0.123                  #Atomic radius\n",
+    "n=4\n",
+    "A=55.8                   #Atomic wt\n",
+    "a=2*math.sqrt(2) \n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "\n",
+    "#Calculations\n",
+    "rho=(n*A)/((a*r*10**-9)**3*N)\n",
+    "\n",
+    "#Result\n",
+    "print \"Density of iron =\",round(rho),\"kg/m**-3\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures_1.ipynb b/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures_1.ipynb
new file mode 100755
index 00000000..8c0ce9a8
--- /dev/null
+++ b/sample_notebooks/RohithYeedulapalli/6.Magnetic_Properties_and_Crystal_Structures_1.ipynb
@@ -0,0 +1,555 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Chapter 6:Magnetic Properties and Crystal Structures"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.1, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "temperature rise is 8.43 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "El=10**-2*50;      #energy loss(J)\n",
+    "H=El*60;           #heat produced(J)\n",
+    "d=7.7*10**3;       #iron rod(kg/m**3)\n",
+    "s=0.462*10**-3;    #specific heat(J/kg K)\n",
+    "\n",
+    "#Calculation\n",
+    "theta=H/(d*s);     #temperature rise(K)\n",
+    "\n",
+    "#Result\n",
+    "print \"temperature rise is\",round(theta,2),\"K\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.2, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "magnetic field at the centre is 14.0 weber/m**2\n",
+      "dipole moment is 9.0 *10**-24 ampere/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "e=1.6*10**-19;    #charge(coulomb)\n",
+    "new=6.8*10**15;   #frequency(revolutions per second)\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "R=5.1*10**-11;     #radius(m)\n",
+    "\n",
+    "#Calculation\n",
+    "i=round(e*new,4);          #current(ampere)\n",
+    "B=mew0*i/(2*R);            #magnetic field at the centre(weber/m**2)\n",
+    "A=math.pi*R**2;\n",
+    "d=i*A;                    #dipole moment(ampere/m**2)\n",
+    "\n",
+    "#Result\n",
+    "print \"magnetic field at the centre is\",round(B),\"weber/m**2\"\n",
+    "print \"dipole moment is\",round(d*10**24),\"*10**-24 ampere/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.3, Page number 6.46"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "intensity of magnetisation is 5.0 ampere/m\n",
+      "flux density in material is 1.257 weber/m**2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "chi=0.5*10**-5;    #magnetic susceptibility\n",
+    "H=10**6;     #field strength(ampere/m)\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "\n",
+    "#Calculation\n",
+    "I=chi*H;     #intensity of magnetisation(ampere/m)\n",
+    "B=mew0*(I+H);    #flux density in material(weber/m**2)\n",
+    "\n",
+    "#Result\n",
+    "print \"intensity of magnetisation is\",I,\"ampere/m\"\n",
+    "print \"flux density in material is\",round(B,3),\"weber/m**2\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.4, Page number 6.47"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "number of Bohr magnetons is 2.22 bohr magneon/atom\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "B=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
+    "a=2.86*10**-10;      #edge(m)\n",
+    "Is=1.76*10**6;       #saturation value of magnetisation(ampere/m)\n",
+    "\n",
+    "#Calculation\n",
+    "N=2/a**3;\n",
+    "mew_bar=Is/N;      #number of Bohr magnetons(ampere m**2)\n",
+    "mew_bar=mew_bar/B;      #number of Bohr magnetons(bohr magneon/atom)\n",
+    "\n",
+    "#Result\n",
+    "print \"number of Bohr magnetons is\",round(mew_bar,2),\"bohr magneon/atom\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.5, Page number 6.47"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "average magnetic moment is 2.79 *10**-3 bohr magneton/spin\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "mew0=4*math.pi*10**-7;\n",
+    "H=9.27*10**-24;      #bohr magneton(ampere m**2)\n",
+    "beta=10**6;      #field(ampere/m)\n",
+    "k=1.38*10**-23;    #boltzmann constant\n",
+    "T=303;    #temperature(K)\n",
+    "\n",
+    "#Calculation\n",
+    "mm=mew0*H*beta/(k*T);    #average magnetic moment(bohr magneton/spin)\n",
+    "\n",
+    "#Result\n",
+    "print \"average magnetic moment is\",round(mm*10**3,2),\"*10**-3 bohr magneton/spin\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.6, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "hysteresis loss per cycle is 188.0 J/m**3\n",
+      "hysteresis loss per second is 9400.0 watt/m**3\n",
+      "power loss is 1.23 watt/kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "#importing modules\n",
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#Variable declaration\n",
+    "A=94;      #area(m**2)\n",
+    "vy=0.1;    #value of length(weber/m**2)\n",
+    "vx=20;     #value of unit length\n",
+    "n=50;      #number of magnetization cycles\n",
+    "d=7650;    #density(kg/m**3)\n",
+    "\n",
+    "#Calculation\n",
+    "h=A*vy*vx;     #hysteresis loss per cycle(J/m**3)\n",
+    "hs=h*n;       #hysteresis loss per second(watt/m**3)\n",
+    "pl=hs/d;      #power loss(watt/kg)\n",
+    "\n",
+    "#Result\n",
+    "print \"hysteresis loss per cycle is\",h,\"J/m**3\"\n",
+    "print \"hysteresis loss per second is\",hs,\"watt/m**3\"\n",
+    "print \"power loss is\",round(pl,2),\"watt/kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.7, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a= 5.43 Angstorm\n",
+      "density = 6.88 kg/m**3\n",
+      "#Answer given in the textbook is wrong\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "d=2.351                 #bond lenght\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "n=8                     #number of atoms in unit cell\n",
+    "A=28.09                 #Atomin mass of silicon\n",
+    "m=6.02*10**26           #1mole\n",
+    "\n",
+    "#Calculations\n",
+    "a=(4*d)/math.sqrt(3)\n",
+    "p=(n*A)/((a*10**-10)*m)    #density\n",
+    "\n",
+    "#Result\n",
+    "print \"a=\",round(a,2),\"Angstorm\"\n",
+    "print \"density =\",round(p*10**16,2),\"kg/m**3\"\n",
+    "print\"#Answer given in the textbook is wrong\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.8, Page number 6.48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " radius of largest sphere is 0.154700538379252*r\n",
+      "maximum radius of sphere is 0.414213562373095*r\n"
+     ]
+    }
+   ],
+   "source": [
+    " import math\n",
+    "from __future__ import division\n",
+    "from sympy import Symbol\n",
+    "\n",
+    "#Variable declaration\n",
+    "r=Symbol('r')\n",
+    "\n",
+    "#Calculation\n",
+    "a1=4*r/math.sqrt(3);\n",
+    "R1=(a1/2)-r;           #radius of largest sphere\n",
+    "a2=4*r/math.sqrt(2);\n",
+    "R2=(a2/2)-r;       #maximum radius of sphere\n",
+    "\n",
+    "#Result\n",
+    "print \"radius of largest sphere is\",R1\n",
+    "print \"maximum radius of sphere is\",R2    "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.9, Page number 6.49"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a1= 2.905 Angstrom\n",
+      "Unit cell volume =a1**3 = 24.521 *10**-30 m**3\n",
+      "Volume occupied by one atom = 12.26 *10**-30 m**3\n",
+      "a2= 3.654 Angstorm\n",
+      "Unit cell volume =a2**3 = 48.8 *10**-30 m**3\n",
+      "Volume occupied by one atom = 12.2 *10**-30 m**3\n",
+      "Volume Change in % = 0.493\n",
+      "Density Change in % = 0.5\n",
+      "Thus the increase of density or the decrease of volume is about 0.5%\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "r1=1.258            #Atomic radius of BCC\n",
+    "r2=1.292            #Atomic radius of FCC\n",
+    "\n",
+    "#calculations\n",
+    "a1=(4*r1)/math.sqrt(3)       #in BCC\n",
+    "b1=((a1)**3)*10**-30         #Unit cell volume\n",
+    "v1=(b1)/2                    #Volume occupied by one atom\n",
+    "a2=2*math.sqrt(2)*r2         #in FCC\n",
+    "b2=(a2)**3*10**-30                   #Unit cell volume\n",
+    "v2=(b2)/4                    #Volume occupied by one atom  \n",
+    "v_c=((v1)-(v2))*100/(v1)     #Volume Change in % \n",
+    "d_c=((v1)-(v2))*100/(v2)     #Density Change in %\n",
+    "\n",
+    "#Results\n",
+    "print \"a1=\",round(a1,3),\"Angstrom\" \n",
+    "print \"Unit cell volume =a1**3 =\",round((b1)/10**-30,3),\"*10**-30 m**3\"\n",
+    "print \"Volume occupied by one atom =\",round(v1/10**-30,2),\"*10**-30 m**3\"\n",
+    "print \"a2=\",round(a2,3),\"Angstorm\"\n",
+    "print \"Unit cell volume =a2**3 =\",round((b2)/10**-30,3),\"*10**-30 m**3\"\n",
+    "print \"Volume occupied by one atom =\",round(v2/10**-30,2),\"*10**-30 m**3\"\n",
+    "print \"Volume Change in % =\",round(v_c,3)\n",
+    "print \"Density Change in % =\",round(d_c,2)\n",
+    "print \"Thus the increase of density or the decrease of volume is about 0.5%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.10, Page number 6.50"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a= 0.563 *10**-9 metre\n",
+      "spacing between the nearest neighbouring ions = 0.2814 nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "n=4     \n",
+    "M=58.5                  #Molecular wt. of NaCl\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "rho=2180                #density\n",
+    "\n",
+    "#Calculations\n",
+    "a=((n*M)/(N*rho))**(1/3)    \n",
+    "s=a/2\n",
+    "\n",
+    "#Result\n",
+    "print \"a=\",round(a/10**-9,3),\"*10**-9 metre\"\n",
+    "print \"spacing between the nearest neighbouring ions =\",round(s/10**-9,4),\"nm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.11, Page number 6.51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "lattice constant, a= 0.36 nm\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "from __future__ import division\n",
+    "\n",
+    "#variable declaration\n",
+    "n=4     \n",
+    "A=63.55                 #Atomic wt. of NaCl\n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "rho=8930                #density\n",
+    "\n",
+    "#Calculations\n",
+    "a=((n*A)/(N*rho))**(1/3)      #Lattice Constant\n",
+    "\n",
+    "#Result\n",
+    "print \"lattice constant, a=\",round(a*10**9,2),\"nm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example number 6.12, Page number 6.51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Density of iron = 8805.0 kg/m**-3\n"
+     ]
+    }
+   ],
+   "source": [
+    "import math\n",
+    "\n",
+    "#variable declaration\n",
+    "r=0.123                  #Atomic radius\n",
+    "n=4\n",
+    "A=55.8                   #Atomic wt\n",
+    "a=2*math.sqrt(2) \n",
+    "N=6.02*10**26           #Avagadro number\n",
+    "\n",
+    "#Calculations\n",
+    "rho=(n*A)/((a*r*10**-9)**3*N)\n",
+    "\n",
+    "#Result\n",
+    "print \"Density of iron =\",round(rho),\"kg/m**-3\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/RuchiMittal/chap1_1.ipynb b/sample_notebooks/RuchiMittal/chap1_1.ipynb
new file mode 100755
index 00000000..602a3854
--- /dev/null
+++ b/sample_notebooks/RuchiMittal/chap1_1.ipynb
@@ -0,0 +1,496 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2624a2ba41f12934f33163ef62780c9504a218c824a031aeca0eae58b845a3fc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 Qualities of measurments"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 (a) Page no 3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "V=80.0                             #expected value of voltage in Volts\n",
+      "V1=79                            #Volts\n",
+      "\n",
+      "#Calculation\n",
+      "E=V-V1\n",
+      "E1=((V-V1)/V)*100\n",
+      "E2=1-((V-V1)/V)\n",
+      "A=100*E2\n",
+      "\n",
+      "#Result\n",
+      "print\"(i) Absolute error is \",E,\"V\"\n",
+      "print\"(ii) percent error is \", E1,\"%\"\n",
+      "print\"(iii) reletive error is \", E2\n",
+      "print\"(iv) percent of accuracy is \", A,\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Absolute error is  1.0 V\n",
+        "(ii) percent error is  1.25 %\n",
+        "(iii) reletive error is  0.9875\n",
+        "(iv) percent of accuracy is  98.75 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 (b) Page no 3"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Yn=20.0                             #mA\n",
+      "Xn=18                             #mA\n",
+      "\n",
+      "#Calculation\n",
+      "e=Yn-Xn\n",
+      "E=(e/Yn)*100\n",
+      "A=1-(e/Yn)\n",
+      "a=A*100\n",
+      "\n",
+      "#Result\n",
+      "print\"(i) Absolute error is \", e,\"mA\"\n",
+      "print\"(ii) Percent error is \",E,\"%\"\n",
+      "print\"(iii) Relative accuracy is \",A\n",
+      "print\"(iv) Percent accuracy is \",a,\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Absolute error is  2.0 mA\n",
+        "(ii) Percent error is  10.0 %\n",
+        "(iii) Relative accuracy is  0.9\n",
+        "(iv) Percent accuracy is  90.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 Page no 4"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x1=98\n",
+      "x2=101\n",
+      "x3=102\n",
+      "x4=97\n",
+      "x5=101\n",
+      "x6=100\n",
+      "x7=103\n",
+      "x8=98\n",
+      "x9=106\n",
+      "x10=99\n",
+      "\n",
+      "#Calculation\n",
+      "X=(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10)/10.0\n",
+      "P=(x6/X)\n",
+      "\n",
+      "#Result\n",
+      "print\"Precision of the 6th measurment is \",round(P,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Precision of the 6th measurment is  0.995\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3(a) Page no 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "V=80                                  #milliammeter readings\n",
+      "I=10.0                                  #mA\n",
+      "V1=150                                 #Volts\n",
+      "R1=1000                                #ohm/volt\n",
+      "\n",
+      "#Calculation\n",
+      "R=V/I\n",
+      "Rv=R1*V1\n",
+      "Rx=(R*V1)/(V1-R)\n",
+      "E=((Rx-R)/Rx)*100\n",
+      "\n",
+      "#Result\n",
+      "print\"(i) Apparent resistance of the unknown resistance \",R,\"K ohm\"\n",
+      "print \"(ii) Actual resistance of the unknown resistance is \",round(Rx,2),\"K ohm\"\n",
+      "print \"(iii) Error due to the loading effet of the voltmeter \",round(E,1),\"%\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Apparent resistance of the unknown resistance  8.0 K ohm\n",
+        "(ii) Actual resistance of the unknown resistance is  8.45 K ohm\n",
+        "(iii) Error due to the loading effet of the voltmeter  5.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3(b) Page no 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "V=30                             #Volts\n",
+      "V1=150                           #Volts\n",
+      "I=0.6                            #A\n",
+      "R1=1000                          #ohm/volts\n",
+      "\n",
+      "#Calculation\n",
+      "R=V/I\n",
+      "Rv=(R1*V1)\n",
+      "Rx=(R*Rv)/(Rv-R)\n",
+      "E=((Rx-R)/Rx)*100\n",
+      "\n",
+      "#Result\n",
+      "print\"(i) total circuit resistance is \", R,\"ohm\"\n",
+      "print \"(ii) The voltmeter resistance is \",round(Rx,2)\n",
+      "print\"(iii) Error due to loading effect of voltmeter \", round(E,3),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) total circuit resistance is  50.0 ohm\n",
+        "(ii) The voltmeter resistance is  50.02\n",
+        "(iii) Error due to loading effect of voltmeter  0.033 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 113
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 Page no 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x1=49.7\n",
+      "x2=50.1\n",
+      "x3=50.2\n",
+      "x4=49.6\n",
+      "x5=49.7\n",
+      "\n",
+      "#Calculation\n",
+      "X=(x1+x2+x3+x4+x5)/5.0\n",
+      "d1=x1-X\n",
+      "d2=x2-X\n",
+      "d3=x3-X\n",
+      "d4=x4-X\n",
+      "d5=x5-X\n",
+      "dtotal=(d1+d2+d3+d4+d5)\n",
+      "\n",
+      "#Result\n",
+      "print\"(i) Arithmetic mean is \", X\n",
+      "print\"(ii) derivations from each value are\"\n",
+      "print \"d1=\",d1,\"\\nd2=\",d2,\"\\nd3=\",d3,\"\\nd4=\",d4,\"\\nd5=\",d5\n",
+      "print\"(iii) The algebric sum of derivative is \",round(dtotal,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i) Arithmetic mean is  49.86\n",
+        "(ii) derivations from each value are\n",
+        "d1= -0.16 \n",
+        "d2= 0.24 \n",
+        "d3= 0.34 \n",
+        "d4= -0.26 \n",
+        "d5= -0.16\n",
+        "(iii) The algebric sum of derivative is  0.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 77
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 Page no 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x1=49.7\n",
+      "x2=50.1\n",
+      "x3=50.2\n",
+      "x4=49.6\n",
+      "x5=49.7\n",
+      "\n",
+      "#Calculation\n",
+      "X=(x1+x2+x3+x4+x5)/5.0\n",
+      "d1=x1-X\n",
+      "d2=x2-X\n",
+      "d3=x3-X\n",
+      "d4=x4-X\n",
+      "d5=x5-X\n",
+      "dtotal=(d1+d2+d3+d4+d5)/5.0\n",
+      "\n",
+      "#Result\n",
+      "print\"The average deviation is \",round(dtotal*10**14,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average deviation is  0.284\n"
+       ]
+      }
+     ],
+     "prompt_number": 86
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 Page no 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "d1= -0.16 \n",
+      "d2= 0.24 \n",
+      "d3= 0.34 \n",
+      "d4= -0.26 \n",
+      "d5= -0.16\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "D=math.sqrt((d1**2+d2**2+d3**2+d4**2+d5**2)/4.0)\n",
+      "\n",
+      "#Result\n",
+      "print\"The standard deviation is \",round(D,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standard deviation is  0.27\n"
+       ]
+      }
+     ],
+     "prompt_number": 90
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 Page no 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "V=600                         #Volts\n",
+      "V1=250.0                       #Volts\n",
+      "a=0.02\n",
+      "\n",
+      "#Calculation\n",
+      "M=a*V\n",
+      "E=(M/V1)*100\n",
+      "\n",
+      "#Result\n",
+      "print\"The limited error is \", E,\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The limited error is  4.8 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 94
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 (a) Page no 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "a=0.02\n",
+      "I=500                           #mA\n",
+      "I1=300.0                         #mA\n",
+      "\n",
+      "#Calculation\n",
+      "M1=I*a\n",
+      "M2=(M1/I1)*100\n",
+      "\n",
+      "#Result\n",
+      "print\"Limiting error is \", round(M2,1),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Limiting error is  3.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 (b) Page no 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "V=70.0                        #Volts\n",
+      "V1=100                      #Volts\n",
+      "I=80.0                        #mA\n",
+      "I1=150                      #mA\n",
+      "a=0.015\n",
+      "\n",
+      "#calculation\n",
+      "M=a*V1\n",
+      "E=(M/V)*100\n",
+      "E1=a*I1\n",
+      "E2=(E1/I)*100\n",
+      "E3=E+E2\n",
+      "\n",
+      "#Result\n",
+      "print\"limiting error is \",round (E3,3),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "limiting error is  4.955 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 102
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1.ipynb b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1.ipynb
new file mode 100755
index 00000000..3ae1483c
--- /dev/null
+++ b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1.ipynb
@@ -0,0 +1,591 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Chapter 1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 1 pagenumber 1.81"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "adm= -1482.0\n",
+      "acm= -1.0\n",
+      "cmrr= 76.4838188131 db\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rc=50000;#ohm\n",
+    "re=100000;#ohm\n",
+    "rs=10000;#ohm\n",
+    "rp=50000;#ohm\n",
+    "beta0=2000;\n",
+    "r0=400000;#ohm\n",
+    "\n",
+    "\n",
+    "\n",
+    "#determine adm,acm,cmrr\n",
+    "#calculation\n",
+    "rc1=(rc*r0)/(rc+r0);\n",
+    "adm=(-(beta0*rc1)/(rs+rp))#differential mode gain\n",
+    "acm=(-(beta0*rc1)/(rs+rp+2*re*(beta0+1)))#common mode gain\n",
+    "import math\n",
+    "cmrr=20*(math.log10((1+((2*re*(beta0+1))/(rs+rp)))))#common mode rejection ratio\n",
+    "\n",
+    "#result\n",
+    "print 'adm=',round(adm,3);\n",
+    "print 'acm=',round(acm,3);\n",
+    "print 'cmrr=',cmrr,'db'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2 page 1.83"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum peak amplitude at 100khz 3.1847133758 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.000001;#volt/sec\n",
+    "freq=100000;\n",
+    "vsat=12;\n",
+    "baw=100000;\n",
+    "#determine vx\n",
+    "\n",
+    "#calculation\n",
+    "vx=2*(1/(sr*2*3.14*freq))\n",
+    "\n",
+    "#result\n",
+    "print 'maximum peak amplitude at 100khz',vx,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 3 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "slew rate= 5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "V=20;\n",
+    "t=4;\n",
+    "#determine slew rate\n",
+    "#calculation\n",
+    "w=V/t\n",
+    "#result\n",
+    "print'slew rate=',w,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 4 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency of input is  79617.8343949 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 4 page 1.84\n",
+    "#given\n",
+    "a=50;\n",
+    "vi=20e-3;\n",
+    "sr=0.5e6;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=a*vi;\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "#result\n",
+    "print 'max frequency of input is ',freq,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 5 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max peak to peak input signal  0.398089171975 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;\n",
+    "freq=40e3;\n",
+    "a=10;\n",
+    "#determine max peak to peak input signal\n",
+    "#calculation\n",
+    "vm=sr/(2*3.14*freq);\n",
+    "vm=2*vm;\n",
+    "v1=vm/a\n",
+    "#result\n",
+    "print'max peak to peak input signal ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 6 pagenumber 1.85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "noise  0.0063247 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "adm=400;\n",
+    "cmrr=50;\n",
+    "vin1=50e-3;\n",
+    "vin2=60e-3;\n",
+    "vnoise=5e-3;\n",
+    "#calculation\n",
+    "v0=(vin2-vin1)*adm;\n",
+    "acm=adm/316.22;\n",
+    "v1=vnoise*acm\n",
+    "print'noise ',round(v1,7),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 7 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "time to change from 0 t0 15  4e-07 sec\n",
+      "slew rate 1.5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=35e6;#volt/sec\n",
+    "vsat=15;#volt\n",
+    "#determine time to change from 0 to 15V\n",
+    "#calculation\n",
+    "c=100e-12;#farad\n",
+    "i=150e-6;#A\n",
+    "w=vsat/sr\n",
+    "w1=i/c;\n",
+    "#result\n",
+    "print'time to change from 0 t0 15 ',round(w,7),'sec'\n",
+    "print'slew rate',w1/1000000,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 8 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bandwidth  21231.4225053 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=2e6;#v/sec\n",
+    "vsat=15;#volt\n",
+    "#determine bandwidth \n",
+    "#calculation\n",
+    "\n",
+    "bw=sr/(2*3.14*vsat)\n",
+    "#result\n",
+    "print'bandwidth ',bw,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 9 pagenumber 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output offset  3.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "iin=30e-9;#A\n",
+    "a=1e5;\n",
+    "rin=1000;#ohm\n",
+    "#determine output offset voltage\n",
+    "#calculation\n",
+    "vid=iin*rin;\n",
+    "v0=a*vid\n",
+    "#result\n",
+    "print'output offset ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 10 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input offset current  4e-06 A\n",
+      "input base current  2.4e-05 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb1=22e-6;#A\n",
+    "inb2=26e-6;#A\n",
+    "#determine input offset current input base current\n",
+    "#calculation\n",
+    "i1=inb2-inb1\n",
+    "i2=(inb2+inb1)/2\n",
+    "#result\n",
+    "print'input offset current ',i1,'A'\n",
+    "print'input base current ',i2,'A'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 11 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input base current  8e-08 A\n",
+      "input offset current  2e-08 A\n",
+      "input offset   2.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb2=90e-9;#A\n",
+    "inb1=70e-9;#A\n",
+    "a=1e5;\n",
+    "#determine input offset current\n",
+    "#calculation\n",
+    "i1=(inb2+inb1)/2\n",
+    "i2=inb2-inb1\n",
+    "v1=((inb2-inb1)*1000)*a\n",
+    "print'input base current ',i1,'A'\n",
+    "print'input offset current ',i2,'A'\n",
+    "print'input offset  ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 12 pageunmber 1.88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage cmrr  100      0.05125 V\n",
+      "output voltage cmrr  200      0.050625 V\n",
+      "output voltage cmrr  450      0.0502777777778 V\n",
+      "output voltage cmrr  105      0.0511904761905 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "vin1=150e-6;#volt\n",
+    "vin2=100e-6;#volt\n",
+    "a=1000;\n",
+    "from array import *\n",
+    "cmrr=array('i',[100,200,450,105])\n",
+    "#determine output voltage\n",
+    "#calculation\n",
+    "vc=(vin1+vin2)/2;\n",
+    "vd=(vin1-vin2);\n",
+    "j=0;\n",
+    "while j<=3 :v0=(a*vd*(1+(vc/(cmrr[j]*vd)))) ;print 'output voltage cmrr ',cmrr[j],'    ',v0,'V';j=j+1;\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 13 pagenumber 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage  0.00316455696203 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rin=100e3;#ohm\n",
+    "rf1=900e3;#ohm\n",
+    "vc=1;#volt\n",
+    "cmrr=70;\n",
+    "#determine the output voltage\n",
+    "#calculation\n",
+    "v0=(1+(rf1/rin))*vc/3160\n",
+    "print 'output voltage ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 14 pagenumber 1.89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input voltage  0.0796178343949 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;#volt/sec\n",
+    "a=50;\n",
+    "freq=20e3;#hz\n",
+    "#determine max peak to peak voltage\n",
+    "#calculation\n",
+    "v1=sr/(2*3.14*freq*a)\n",
+    "print'input voltage ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 15 pagenumber 1.90"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency  265392.781316 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=50e6;#volt/sec\n",
+    "rin=2;\n",
+    "vimax=10;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=vimax*(1+rin);\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "print 'max frequency ',freq,'hz'\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_1.ipynb b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_1.ipynb
new file mode 100755
index 00000000..6ba07fc8
--- /dev/null
+++ b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_1.ipynb
@@ -0,0 +1,591 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Chapter 1 Circuit Configuration for Linear Integrated Ciruits"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 1 pagenumber 1.81"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "adm= -1482.0\n",
+      "acm= -1.0\n",
+      "cmrr= 76.4838188131 db\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rc=50000;#ohm\n",
+    "re=100000;#ohm\n",
+    "rs=10000;#ohm\n",
+    "rp=50000;#ohm\n",
+    "beta0=2000;\n",
+    "r0=400000;#ohm\n",
+    "\n",
+    "\n",
+    "\n",
+    "#determine adm,acm,cmrr\n",
+    "#calculation\n",
+    "rc1=(rc*r0)/(rc+r0);\n",
+    "adm=(-(beta0*rc1)/(rs+rp))#differential mode gain\n",
+    "acm=(-(beta0*rc1)/(rs+rp+2*re*(beta0+1)))#common mode gain\n",
+    "import math\n",
+    "cmrr=20*(math.log10((1+((2*re*(beta0+1))/(rs+rp)))))#common mode rejection ratio\n",
+    "\n",
+    "#result\n",
+    "print 'adm=',round(adm,3);\n",
+    "print 'acm=',round(acm,3);\n",
+    "print 'cmrr=',cmrr,'db'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 2 page 1.83"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum peak amplitude at 100khz 3.1847133758 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.000001;#volt/sec\n",
+    "freq=100000;\n",
+    "vsat=12;\n",
+    "baw=100000;\n",
+    "#determine vx\n",
+    "\n",
+    "#calculation\n",
+    "vx=2*(1/(sr*2*3.14*freq))\n",
+    "\n",
+    "#result\n",
+    "print 'maximum peak amplitude at 100khz',vx,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 3 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "slew rate= 5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "V=20;\n",
+    "t=4;\n",
+    "#determine slew rate\n",
+    "#calculation\n",
+    "w=V/t\n",
+    "#result\n",
+    "print'slew rate=',w,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 4 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency of input is  79617.8343949 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 4 page 1.84\n",
+    "#given\n",
+    "a=50;\n",
+    "vi=20e-3;\n",
+    "sr=0.5e6;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=a*vi;\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "#result\n",
+    "print 'max frequency of input is ',freq,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 5 pagenumber 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max peak to peak input signal  0.398089171975 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;\n",
+    "freq=40e3;\n",
+    "a=10;\n",
+    "#determine max peak to peak input signal\n",
+    "#calculation\n",
+    "vm=sr/(2*3.14*freq);\n",
+    "vm=2*vm;\n",
+    "v1=vm/a\n",
+    "#result\n",
+    "print'max peak to peak input signal ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 6 pagenumber 1.85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "noise  0.0063247 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "adm=400;\n",
+    "cmrr=50;\n",
+    "vin1=50e-3;\n",
+    "vin2=60e-3;\n",
+    "vnoise=5e-3;\n",
+    "#calculation\n",
+    "v0=(vin2-vin1)*adm;\n",
+    "acm=adm/316.22;\n",
+    "v1=vnoise*acm\n",
+    "print'noise ',round(v1,7),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 7 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "time to change from 0 t0 15  4e-07 sec\n",
+      "slew rate 1.5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=35e6;#volt/sec\n",
+    "vsat=15;#volt\n",
+    "#determine time to change from 0 to 15V\n",
+    "#calculation\n",
+    "c=100e-12;#farad\n",
+    "i=150e-6;#A\n",
+    "w=vsat/sr\n",
+    "w1=i/c;\n",
+    "#result\n",
+    "print'time to change from 0 t0 15 ',round(w,7),'sec'\n",
+    "print'slew rate',w1/1000000,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 8 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bandwidth  21231.4225053 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=2e6;#v/sec\n",
+    "vsat=15;#volt\n",
+    "#determine bandwidth \n",
+    "#calculation\n",
+    "\n",
+    "bw=sr/(2*3.14*vsat)\n",
+    "#result\n",
+    "print'bandwidth ',bw,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 9 pagenumber 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output offset  3.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "iin=30e-9;#A\n",
+    "a=1e5;\n",
+    "rin=1000;#ohm\n",
+    "#determine output offset voltage\n",
+    "#calculation\n",
+    "vid=iin*rin;\n",
+    "v0=a*vid\n",
+    "#result\n",
+    "print'output offset ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 10 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input offset current  4e-06 A\n",
+      "input base current  2.4e-05 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb1=22e-6;#A\n",
+    "inb2=26e-6;#A\n",
+    "#determine input offset current input base current\n",
+    "#calculation\n",
+    "i1=inb2-inb1\n",
+    "i2=(inb2+inb1)/2\n",
+    "#result\n",
+    "print'input offset current ',i1,'A'\n",
+    "print'input base current ',i2,'A'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 11 pagenumber 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input base current  8e-08 A\n",
+      "input offset current  2e-08 A\n",
+      "input offset   2.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb2=90e-9;#A\n",
+    "inb1=70e-9;#A\n",
+    "a=1e5;\n",
+    "#determine input offset current\n",
+    "#calculation\n",
+    "i1=(inb2+inb1)/2\n",
+    "i2=inb2-inb1\n",
+    "v1=((inb2-inb1)*1000)*a\n",
+    "print'input base current ',i1,'A'\n",
+    "print'input offset current ',i2,'A'\n",
+    "print'input offset  ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 12 pageunmber 1.88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage cmrr  100      0.05125 V\n",
+      "output voltage cmrr  200      0.050625 V\n",
+      "output voltage cmrr  450      0.0502777777778 V\n",
+      "output voltage cmrr  105      0.0511904761905 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "vin1=150e-6;#volt\n",
+    "vin2=100e-6;#volt\n",
+    "a=1000;\n",
+    "from array import *\n",
+    "cmrr=array('i',[100,200,450,105])\n",
+    "#determine output voltage\n",
+    "#calculation\n",
+    "vc=(vin1+vin2)/2;\n",
+    "vd=(vin1-vin2);\n",
+    "j=0;\n",
+    "while j<=3 :v0=(a*vd*(1+(vc/(cmrr[j]*vd)))) ;print 'output voltage cmrr ',cmrr[j],'    ',v0,'V';j=j+1;\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 13 pagenumber 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage  0.00316455696203 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rin=100e3;#ohm\n",
+    "rf1=900e3;#ohm\n",
+    "vc=1;#volt\n",
+    "cmrr=70;\n",
+    "#determine the output voltage\n",
+    "#calculation\n",
+    "v0=(1+(rf1/rin))*vc/3160\n",
+    "print 'output voltage ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 14 pagenumber 1.89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input voltage  0.0796178343949 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;#volt/sec\n",
+    "a=50;\n",
+    "freq=20e3;#hz\n",
+    "#determine max peak to peak voltage\n",
+    "#calculation\n",
+    "v1=sr/(2*3.14*freq*a)\n",
+    "print'input voltage ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##Example 15 pagenumber 1.90"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency  265392.781316 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=50e6;#volt/sec\n",
+    "rin=2;\n",
+    "vimax=10;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=vimax*(1+rin);\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "print 'max frequency ',freq,'hz'\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_2.ipynb b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_2.ipynb
new file mode 100644
index 00000000..13cd9704
--- /dev/null
+++ b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_2.ipynb
@@ -0,0 +1,601 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Circuit Configuration for Linear Integrated Ciruits"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 1 Page No:1.81"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "adm= -1482.0\n",
+      "acm= -1.0\n",
+      "cmrr= 76.4838188131 db\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rc=50000;#ohm\n",
+    "re=100000;#ohm\n",
+    "rs=10000;#ohm\n",
+    "rp=50000;#ohm\n",
+    "beta0=2000;\n",
+    "r0=400000;#ohm\n",
+    "\n",
+    "\n",
+    "\n",
+    "#determine adm,acm,cmrr\n",
+    "#calculation\n",
+    "rc1=(rc*r0)/(rc+r0);\n",
+    "adm=(-(beta0*rc1)/(rs+rp))#differential mode gain\n",
+    "acm=(-(beta0*rc1)/(rs+rp+2*re*(beta0+1)))#common mode gain\n",
+    "import math\n",
+    "cmrr=20*(math.log10((1+((2*re*(beta0+1))/(rs+rp)))))#common mode rejection ratio\n",
+    "\n",
+    "#result\n",
+    "print 'adm=',round(adm,3);\n",
+    "print 'acm=',round(acm,3);\n",
+    "print 'cmrr=',cmrr,'db'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 2 Page No:1.83"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum peak amplitude at 100khz 3.185 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.000001;#volt/sec\n",
+    "freq=100000;\n",
+    "vsat=12;\n",
+    "baw=100000;\n",
+    "#determine vx\n",
+    "\n",
+    "#calculation\n",
+    "vx=2*(1/(sr*2*3.14*freq))\n",
+    "\n",
+    "#result\n",
+    "print 'maximum peak amplitude at 100khz',round(vx,3),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 3 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "slew rate= 5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "V=20;\n",
+    "t=4;\n",
+    "#determine slew rate\n",
+    "#calculation\n",
+    "w=V/t\n",
+    "#result\n",
+    "print'slew rate=',w,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 4 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency of input is  79617.8343949 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "a=50;\n",
+    "vi=20e-3;\n",
+    "sr=0.5e6;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=a*vi;\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "#result\n",
+    "print 'max frequency of input is ',freq,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 5 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max peak to peak input signal  0.398089171975 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;\n",
+    "freq=40e3;\n",
+    "a=10;\n",
+    "#determine max peak to peak input signal\n",
+    "#calculation\n",
+    "vm=sr/(2*3.14*freq);\n",
+    "vm=2*vm;\n",
+    "v1=vm/a\n",
+    "#result\n",
+    "print'max peak to peak input signal ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "###Example 6 Page No: 1.85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "noise  0.0063247 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "adm=400;\n",
+    "cmrr=50;\n",
+    "vin1=50e-3;\n",
+    "vin2=60e-3;\n",
+    "vnoise=5e-3;\n",
+    "#calculation\n",
+    "v0=(vin2-vin1)*adm;\n",
+    "acm=adm/316.22;\n",
+    "v1=vnoise*acm\n",
+    "print'noise ',round(v1,7),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 7 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "time to change from 0 t0 15  4e-07 sec\n",
+      "slew rate 1.5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=35e6;#volt/sec\n",
+    "vsat=15;#volt\n",
+    "#determine time to change from 0 to 15V\n",
+    "#calculation\n",
+    "c=100e-12;#farad\n",
+    "i=150e-6;#A\n",
+    "w=vsat/sr\n",
+    "w1=i/c;\n",
+    "#result\n",
+    "print'time to change from 0 t0 15 ',round(w,7),'sec'\n",
+    "print'slew rate',w1/1000000,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 8 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bandwidth  21231.4225053 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=2e6;#v/sec\n",
+    "vsat=15;#volt\n",
+    "#determine bandwidth \n",
+    "#calculation\n",
+    "\n",
+    "bw=sr/(2*3.14*vsat)\n",
+    "#result\n",
+    "print'bandwidth ',bw,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 9 Page No: 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output offset  3.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "iin=30e-9;#A\n",
+    "a=1e5;\n",
+    "rin=1000;#ohm\n",
+    "#determine output offset voltage\n",
+    "#calculation\n",
+    "vid=iin*rin;\n",
+    "v0=a*vid\n",
+    "#result\n",
+    "print'output offset ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 10 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input offset current  4e-06 A\n",
+      "input base current  2.4e-05 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb1=22e-6;#A\n",
+    "inb2=26e-6;#A\n",
+    "#determine input offset current input base current\n",
+    "#calculation\n",
+    "i1=inb2-inb1\n",
+    "i2=(inb2+inb1)/2\n",
+    "#result\n",
+    "print'input offset current ',i1,'A'\n",
+    "print'input base current ',i2,'A'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 11 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input base current  8e-08 A\n",
+      "input offset current  2e-08 A\n",
+      "input offset   2.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb2=90e-9;#A\n",
+    "inb1=70e-9;#A\n",
+    "a=1e5;\n",
+    "#determine input offset current\n",
+    "#calculation\n",
+    "i1=(inb2+inb1)/2\n",
+    "i2=inb2-inb1\n",
+    "v1=((inb2-inb1)*1000)*a\n",
+    "print'input base current ',i1,'A'\n",
+    "print'input offset current ',i2,'A'\n",
+    "print'input offset  ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 12 Page No: 1.88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage cmrr  100      0.05125 V\n",
+      "output voltage cmrr  200      0.050625 V\n",
+      "output voltage cmrr  450      0.0502777777778 V\n",
+      "output voltage cmrr  105      0.0511904761905 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "vin1=150e-6;#volt\n",
+    "vin2=100e-6;#volt\n",
+    "a=1000;\n",
+    "from array import *\n",
+    "cmrr=array('i',[100,200,450,105])\n",
+    "#determine output voltage\n",
+    "#calculation\n",
+    "vc=(vin1+vin2)/2;\n",
+    "vd=(vin1-vin2);\n",
+    "j=0;\n",
+    "while j<=3 :v0=(a*vd*(1+(vc/(cmrr[j]*vd)))) ;print 'output voltage cmrr ',cmrr[j],'    ',v0,'V';j=j+1;\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 13 Page No: 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage  0.00316455696203 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rin=100e3;#ohm\n",
+    "rf1=900e3;#ohm\n",
+    "vc=1;#volt\n",
+    "cmrr=70;\n",
+    "#determine the output voltage\n",
+    "#calculation\n",
+    "v0=(1+(rf1/rin))*vc/3160\n",
+    "print 'output voltage ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 14 Page No: 1.89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input voltage  0.08 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;#volt/sec\n",
+    "a=50;\n",
+    "freq=20e3;#hz\n",
+    "#determine max peak to peak voltage\n",
+    "#calculation\n",
+    "v1=sr/(2*3.14*freq*a)\n",
+    "print'input voltage ',round(v1,3),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "###Example 15 Page No: 1.90"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency  26.5392781316 Khz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=50e6;#volt/sec\n",
+    "rin=2;\n",
+    "vimax=10;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=vimax*(1+rin);\n",
+    "freq=sr/(2*3.14*vm)/10e3;\n",
+    "print 'max frequency ',freq,'Khz'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_3.ipynb b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_3.ipynb
new file mode 100644
index 00000000..375e5b02
--- /dev/null
+++ b/sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_3.ipynb
@@ -0,0 +1,601 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 1: Circuit Configuration for Linear Integrated Ciruits"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 1 Page No:1.81"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "adm= -1482.0\n",
+      "acm= -1.0\n",
+      "cmrr= 76.4838188131 db\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rc=50000;#ohm\n",
+    "re=100000;#ohm\n",
+    "rs=10000;#ohm\n",
+    "rp=50000;#ohm\n",
+    "beta0=2000;\n",
+    "r0=400000;#ohm\n",
+    "\n",
+    "\n",
+    "\n",
+    "#determine adm,acm,cmrr\n",
+    "#calculation\n",
+    "rc1=(rc*r0)/(rc+r0);\n",
+    "adm=(-(beta0*rc1)/(rs+rp))#differential mode gain\n",
+    "acm=(-(beta0*rc1)/(rs+rp+2*re*(beta0+1)))#common mode gain\n",
+    "import math\n",
+    "cmrr=20*(math.log10((1+((2*re*(beta0+1))/(rs+rp)))))#common mode rejection ratio\n",
+    "\n",
+    "#result\n",
+    "print 'adm=',round(adm,3);\n",
+    "print 'acm=',round(acm,3);\n",
+    "print 'cmrr=',cmrr,'db'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 2 Page No:1.83"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum peak amplitude at 100khz 3.185 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.000001;#volt/sec\n",
+    "freq=100000;\n",
+    "vsat=12;\n",
+    "baw=100000;\n",
+    "#determine vx\n",
+    "\n",
+    "#calculation\n",
+    "vx=2*(1/(sr*2*3.14*freq))\n",
+    "\n",
+    "#result\n",
+    "print 'maximum peak amplitude at 100khz',round(vx,3),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 3 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "slew rate= 5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "V=20;\n",
+    "t=4;\n",
+    "#determine slew rate\n",
+    "#calculation\n",
+    "w=V/t\n",
+    "#result\n",
+    "print'slew rate=',w,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 4 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency of input is  79617.8343949 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "a=50;\n",
+    "vi=20e-3;\n",
+    "sr=0.5e6;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=a*vi;\n",
+    "freq=sr/(2*3.14*vm)\n",
+    "#result\n",
+    "print 'max frequency of input is ',freq,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 5 Page No: 1.84"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max peak to peak input signal  0.398089171975 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;\n",
+    "freq=40e3;\n",
+    "a=10;\n",
+    "#determine max peak to peak input signal\n",
+    "#calculation\n",
+    "vm=sr/(2*3.14*freq);\n",
+    "vm=2*vm;\n",
+    "v1=vm/a\n",
+    "#result\n",
+    "print'max peak to peak input signal ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "Example 6 Page No: 1.85"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "noise  0.0063247 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "adm=400;\n",
+    "cmrr=50;\n",
+    "vin1=50e-3;\n",
+    "vin2=60e-3;\n",
+    "vnoise=5e-3;\n",
+    "#calculation\n",
+    "v0=(vin2-vin1)*adm;\n",
+    "acm=adm/316.22;\n",
+    "v1=vnoise*acm\n",
+    "print'noise ',round(v1,7),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 7 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "time to change from 0 t0 15  4e-07 sec\n",
+      "slew rate 1.5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=35e6;#volt/sec\n",
+    "vsat=15;#volt\n",
+    "#determine time to change from 0 to 15V\n",
+    "#calculation\n",
+    "c=100e-12;#farad\n",
+    "i=150e-6;#A\n",
+    "w=vsat/sr\n",
+    "w1=i/c;\n",
+    "#result\n",
+    "print'time to change from 0 t0 15 ',round(w,7),'sec'\n",
+    "print'slew rate',w1/1000000,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 8 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bandwidth  21231.4225053 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=2e6;#v/sec\n",
+    "vsat=15;#volt\n",
+    "#determine bandwidth \n",
+    "#calculation\n",
+    "\n",
+    "bw=sr/(2*3.14*vsat)\n",
+    "#result\n",
+    "print'bandwidth ',bw,'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 9 Page No: 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output offset  3.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "iin=30e-9;#A\n",
+    "a=1e5;\n",
+    "rin=1000;#ohm\n",
+    "#determine output offset voltage\n",
+    "#calculation\n",
+    "vid=iin*rin;\n",
+    "v0=a*vid\n",
+    "#result\n",
+    "print'output offset ',v0,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 10 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input offset current  4e-06 A\n",
+      "input base current  2.4e-05 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb1=22e-6;#A\n",
+    "inb2=26e-6;#A\n",
+    "#determine input offset current input base current\n",
+    "#calculation\n",
+    "i1=inb2-inb1\n",
+    "i2=(inb2+inb1)/2\n",
+    "#result\n",
+    "print'input offset current ',i1,'A'\n",
+    "print'input base current ',i2,'A'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 11 Page No: 1.86"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input base current  8e-08 A\n",
+      "input offset current  2e-08 A\n",
+      "input offset   2.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "inb2=90e-9;#A\n",
+    "inb1=70e-9;#A\n",
+    "a=1e5;\n",
+    "#determine input offset current\n",
+    "#calculation\n",
+    "i1=(inb2+inb1)/2\n",
+    "i2=inb2-inb1\n",
+    "v1=((inb2-inb1)*1000)*a\n",
+    "print'input base current ',i1,'A'\n",
+    "print'input offset current ',i2,'A'\n",
+    "print'input offset  ',v1,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 12 Page No: 1.88"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage cmrr  100      0.05125 V\n",
+      "output voltage cmrr  200      0.050625 V\n",
+      "output voltage cmrr  450      0.0502777777778 V\n",
+      "output voltage cmrr  105      0.0511904761905 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "vin1=150e-6;#volt\n",
+    "vin2=100e-6;#volt\n",
+    "a=1000;\n",
+    "from array import *\n",
+    "cmrr=array('i',[100,200,450,105])\n",
+    "#determine output voltage\n",
+    "#calculation\n",
+    "vc=(vin1+vin2)/2;\n",
+    "vd=(vin1-vin2);\n",
+    "j=0;\n",
+    "while j<=3 :v0=(a*vd*(1+(vc/(cmrr[j]*vd)))) ;print 'output voltage cmrr ',cmrr[j],'    ',v0,'V';j=j+1;\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 13 Page No: 1.87"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage  0.003 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "rin=100e3;#ohm\n",
+    "rf1=900e3;#ohm\n",
+    "vc=1;#volt\n",
+    "cmrr=70;\n",
+    "#determine the output voltage\n",
+    "#calculation\n",
+    "v0=(1+(rf1/rin))*vc/3160\n",
+    "print 'output voltage ',round(v0,3),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 14 Page No: 1.89"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input voltage  0.08 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=0.5e6;#volt/sec\n",
+    "a=50;\n",
+    "freq=20e3;#hz\n",
+    "#determine max peak to peak voltage\n",
+    "#calculation\n",
+    "v1=sr/(2*3.14*freq*a)\n",
+    "print'input voltage ',round(v1,3),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Example 15 Page No: 1.90"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency  26.539 Khz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#given\n",
+    "sr=50e6;#volt/sec\n",
+    "rin=2;\n",
+    "vimax=10;\n",
+    "#determine max frequency\n",
+    "#calculation\n",
+    "vm=vimax*(1+rin);\n",
+    "freq=sr/(2*3.14*vm)/10e3;\n",
+    "print 'max frequency ',round(freq,3),'Khz'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/S PRASHANTHS PRASHANTH/Untitled.ipynb b/sample_notebooks/S PRASHANTHS PRASHANTH/Untitled.ipynb
new file mode 100755
index 00000000..8d7fb2e1
--- /dev/null
+++ b/sample_notebooks/S PRASHANTHS PRASHANTH/Untitled.ipynb
@@ -0,0 +1,464 @@
+{
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "adm= -1482\n",
+      "acm= -1\n",
+      "cmrr= 76.4838188131 db\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 1 page 1.81\n",
+    "#given\n",
+    "rc=50000;\n",
+    "re=100000;\n",
+    "rs=10000;\n",
+    "rp=50000;\n",
+    "beta0=2000;\n",
+    "r0=400000;\n",
+    "#determine adm,acm,cmrr\n",
+    "rc1=(rc*r0)/(rc+r0);\n",
+    "print 'adm=',(-(beta0*rc1)/(rs+rp))\n",
+    "print 'acm=',(-(beta0*rc1)/(rs+rp+2*re*(beta0+1)))\n",
+    "import math\n",
+    "print 'cmrr=',20*(math.log10((1+((2*re*(beta0+1))/(rs+rp))))),'db'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "maximum peak amplitude at 100khz 3.1847133758 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 2 page 1.83\n",
+    "#given\n",
+    "sr=0.000001;#volt/sec\n",
+    "freq=100000;\n",
+    "vsat=12;\n",
+    "baw=100000;\n",
+    "#determine vx\n",
+    "print 'maximum peak amplitude at 100khz',2*(1/(sr*2*3.14*freq)),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "slew rate= 5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 3 page 1.84\n",
+    "#given\n",
+    "V=20;\n",
+    "t=4;\n",
+    "#determine slew rate\n",
+    "print'slew rate=',V/t,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency of input is  79617.8343949 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 4 page 1.84\n",
+    "#given\n",
+    "a=50;\n",
+    "vi=20e-3;\n",
+    "sr=0.5e6;\n",
+    "#determine max frequency\n",
+    "vm=a*vi;\n",
+    "print 'max frequency of input is ',sr/(2*3.14*vm),'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max peak to peak input signal  0.398089171975 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 5 page 1.84\n",
+    "#given\n",
+    "sr=0.5e6;\n",
+    "freq=40e3;\n",
+    "a=10;\n",
+    "#determine max peak to peak input signal\n",
+    "vm=sr/(2*3.14*freq);\n",
+    "vm=2*vm;\n",
+    "print'max peak to peak input signal ',vm/a,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "noise  0.0063247 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 6  page 1.85\n",
+    "#given\n",
+    "adm=400;\n",
+    "cmrr=50;\n",
+    "vin1=50e-3;\n",
+    "vin2=60e-3;\n",
+    "vnoise=5e-3;\n",
+    "v0=(vin2-vin1)*adm;\n",
+    "acm=adm/316.22;\n",
+    "print'noise ',round(vnoise*acm,7),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 62,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      " time to change from 0 t0 15  4e-07 sec\n",
+      "slew rate 1.5 volt/μsec\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 7 page 1.86\n",
+    "#given\n",
+    "sr=35e6;#volt/sec\n",
+    "vsat=15;#volt\n",
+    "#determine time to change from 0 to 15V\n",
+    "print 'time to change from 0 t0 15 ',round(vsat/sr,7),'sec'\n",
+    "c=100e-12;#farad\n",
+    "i=150e-6;#A\n",
+    "print'slew rate',(i/c)/1000000,'volt/μsec'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 63,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "bandwidth  21231.4225053 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 8 page 1.86\n",
+    "#given\n",
+    "sr=2e6;#v/sec\n",
+    "vsat=15;#volt\n",
+    "#determine bandwidth \n",
+    "print'bandwidth ',sr/(2*3.14*vsat),'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 70,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output offset  3.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 9 page 1.87\n",
+    "#given\n",
+    "iin=30e-9;#A\n",
+    "a=1e5;\n",
+    "rin=1000;#ohm\n",
+    "#determine output offset voltage\n",
+    "vid=iin*rin;\n",
+    "print'output offset ',a*vid,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 73,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input offset current  4e-06 A\n",
+      "input base current  2.4e-05 A\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 10 page 1.86\n",
+    "#given\n",
+    "inb1=22e-6;#A\n",
+    "inb2=26e-6;#A\n",
+    "#determine input offset current input base current\n",
+    "print'input offset current ',inb2-inb1,'A'\n",
+    "print'input base current ',(inb2+inb1)/2,'A'\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 74,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input base current  8e-08 A\n",
+      "input offset current  2e-08 A\n",
+      "input offset   2.0 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 11 page 1.86\n",
+    "#given\n",
+    "inb2=90e-9;#A\n",
+    "inb1=70e-9;#A\n",
+    "a=1e5;\n",
+    "#determine input offset current\n",
+    "print'input base current ',(inb2+inb1)/2,'A'\n",
+    "print'input offset current ',inb2-inb1,'A'\n",
+    "print'input offset  ',((inb2-inb1)*1000)*a,'V'\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage cmrr  100      0.05125 V\n",
+      "output voltage cmrr  200      0.050625 V\n",
+      "output voltage cmrr  450      0.0502777777778 V\n",
+      "output voltage cmrr  105      0.0511904761905 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 12 page 1.88\n",
+    "#given\n",
+    "vin1=150e-6;#volt\n",
+    "vin2=100e-6;#volt\n",
+    "a=1000;\n",
+    "from array import *\n",
+    "cmrr=array('i',[100,200,450,105])\n",
+    "#determine output voltage\n",
+    "vc=(vin1+vin2)/2;\n",
+    "vd=(vin1-vin2);\n",
+    "j=0;\n",
+    "while j<=3 : print 'output voltage cmrr ',cmrr[j],'    ',(a*vd*(1+(vc/(cmrr[j]*vd)))),'V';j=j+1;\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "output voltage  0.00316455696203 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 13 page 1.87\n",
+    "#given\n",
+    "rin=100e3;#ohm\n",
+    "rf1=900e3;#ohm\n",
+    "vc=1;#volt\n",
+    "cmrr=70;\n",
+    "#determine the output voltage\n",
+    "print 'output voltage ',(1+(rf1/rin))*vc/3160,'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 32,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "input voltage  0.0796178343949 V\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 14 page 1.89\n",
+    "#given\n",
+    "sr=0.5e6;#volt/sec\n",
+    "a=50;\n",
+    "freq=20e3;#hz\n",
+    "#determine max peak to peak voltage\n",
+    "print'input voltage ',sr/(2*3.14*freq*a),'V'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 33,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "max frequency  265392.781316 hz\n"
+     ]
+    }
+   ],
+   "source": [
+    "#problem 15 page 1.90\n",
+    "#given\n",
+    "sr=50e6;#volt/sec\n",
+    "rin=2;\n",
+    "vimax=10;\n",
+    "#determine max frequency\n",
+    "vm=vimax*(1+rin);\n",
+    "print 'max frequency ',sr/(2*3.14*vm),'hz'\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Sabiya/ch_1.ipynb b/sample_notebooks/Sabiya/ch_1.ipynb
new file mode 100644
index 00000000..6bfdb781
--- /dev/null
+++ b/sample_notebooks/Sabiya/ch_1.ipynb
@@ -0,0 +1,371 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter No - 1 : Introduction\n",
+      "    "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.1 - Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "P_m = 760 # pressure of mercury in mm\n",
+      "P_m_bar = P_m/750 # in bar\n",
+      "P_W = 0.006867 # pressure of water in bar\n",
+      "P = P_m_bar+P_W # in bar\n",
+      "print \"The absolute pressure of gas = %0.3f bar \" %P"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute pressure of gas = 1.020 bar \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.2 - Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "# Given data\n",
+      "Rho = 13.6 \n",
+      "g = 9.81 \n",
+      "a = 760 # in mm\n",
+      "b = 480 # in mm\n",
+      "h = a-b # in mm\n",
+      "P = (1000*Rho*g*h)/(1000) # in N/m**2\n",
+      "print \"The absolute pressure = %0.f N/m**2 \" %P\n",
+      "P = math.floor(P /100) # in mbar\n",
+      "print \"The absolute pressure = %0.f mbar \" %P"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute pressure = 37356 N/m**2 \n",
+        "The absolute pressure = 373 mbar \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.3 - Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "G_P = 30 # guage pressure of steam in bar\n",
+      "P1 = 745 # in mm\n",
+      "P1= P1/750 #  in bar\n",
+      "PressureInBoiler = G_P+P1 # in bar\n",
+      "print \"The absolute pressure in the bioler = %0.3f bar\" %PressureInBoiler"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute pressure in the bioler = 30.993 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.4 - Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Rho = 0.78 # in kg/m**3\n",
+      "g = 9.81 \n",
+      "h = 3 # in m\n",
+      "b = g*Rho*h*1000 # in N/m**2\n",
+      "b = b * 10**-3 # in kN/m**2\n",
+      "print \"The gauge pressure = %0.3f kN/m**2 \" %b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The gauge pressure = 22.955 kN/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.5 - Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "B_h = 755 # Barometric height in mm\n",
+      "M_h= 240 # Manometer height in mm\n",
+      "P = B_h+M_h # in mm \n",
+      "P = P/750 # absolute pressure in bar\n",
+      "P= P*10**5 # in N/m**2\n",
+      "print \"The absolute pressure in the vessel = %0.5f MN/m**2 \" %(P*10**-6)\n",
+      "print \"The absolute pressure in the vessel = %0.4f bar \" %(P*10**-5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The absolute pressure in the vessel = 0.13267 MN/m**2 \n",
+        "The absolute pressure in the vessel = 1.3267 bar \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.6 - Page No : 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "T = 287 # in degree C\n",
+      "T = T + 273 # in K\n",
+      "print \"The temperature on absolute scale = %0.f K \" %T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature on absolute scale = 560 K \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.7 - Page No : 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "a = 0.26 \n",
+      "b = 5*10**-4 \n",
+      "E = 10 # in mV\n",
+      "T = (a/(2*b))*( sqrt(1+(4*E*b/a**2)) - 1 ) # in degree C\n",
+      "print \"The unit of a will be mV/\u00b0C and the unit of b will be mV/\u00b0C**2\"\n",
+      "print \"The Temperature = %0.2f degree C \" %T"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The unit of a will be mV/\u00b0C and the unit of b will be mV/\u00b0C**2\n",
+        "The Temperature = 35.97 degree C \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.8 - Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Q_w = 500 # quantity of water flowing in kg/minute\n",
+      "T1 = 80 # in \u00b0 C\n",
+      "T2 = 20 # in \u00b0C\n",
+      "del_T = T1-T2 # in \u00b0C\n",
+      "Spe_heat = 4.182 # in kJ/kg\n",
+      "Q_h = Q_w*del_T*Spe_heat # in kJ/minute\n",
+      "\n",
+      "a= 15\n",
+      "b= 16\n",
+      "print \"Quantity of heat supplied to water in the economizer = %0.f kJ/minute\" %Q_h\n",
+      "print \"                                                     = %0.2f MJ/minute\" %(Q_h*10**-3) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity of heat supplied to water in the economizer = 125460 kJ/minute\n",
+        "                                                     = 125.46 MJ/minute\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.9 - Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "CopperMass = 3 # in kg\n",
+      "WaterMass= 6 # in kg\n",
+      "Spe_heat_copper= 0.394 # in kJ/kg-K\n",
+      "T1 = 90 # in degree C\n",
+      "T2 = 20 # in degree C\n",
+      "del_T = T1-T2 # in degree C\n",
+      "H_C = CopperMass*Spe_heat_copper*del_T # heat required by copper in kJ\n",
+      "Spe_heat_water= 4.193 # in kJ/kg-K\n",
+      "H_W = WaterMass*Spe_heat_water*del_T # heat required by water in kJ\n",
+      "H = H_C+H_W #heat required by vessel and water  in kJ\n",
+      "H = H * 10**-3 # in MJ\n",
+      "print \"Heat required by vessel and water = %0.3f MJ \" %H"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat required by vessel and water = 1.844 MJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example No :  1.10 - Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "m = 18.2 #quantity of air supplied of coal in kg\n",
+      "T1 = 200 # in degree C\n",
+      "T2 = 18 # in degree C\n",
+      "del_T = T1-T2 # in degree C\n",
+      "Spe_heat = 1 # in kJ/kg-K\n",
+      "Q_C = m*Spe_heat*del_T # in kJ\n",
+      "print \"The Quantity of heat supplied per kg of coal = %0.2f kJ \" %Q_C"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Quantity of heat supplied per kg of coal = 3312.40 kJ \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/Salman/ElecEngg2.ipynb b/sample_notebooks/Salman/ElecEngg2.ipynb
new file mode 100755
index 00000000..fd0e270d
--- /dev/null
+++ b/sample_notebooks/Salman/ElecEngg2.ipynb
@@ -0,0 +1,604 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Diode Applications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2\n",
+      ": Page No 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from numpy import pi\n",
+      "# Given data\n",
+      "R_L = 1 # in K ohm\n",
+      "R_L = R_L * 10**3 # in ohm\n",
+      "V_m = 15 # in V\n",
+      "V_i = '15*sin(314*t)' \n",
+      "I_m= V_m/R_L # in A\n",
+      "I_dc = I_m/pi # in A\n",
+      "I_dc = I_dc * 10**3 # in mA\n",
+      "print \"Average current through the diode = %0.2f mA\" %I_dc\n",
+      "I_drms = V_m/(2*R_L) \n",
+      "I_drms = I_drms * 10**3 # in mA\n",
+      "print \"RMS current = %0.1f mA\" %I_drms\n",
+      "I_m = V_m/R_L \n",
+      "I_m = I_m*10**3 # in mA\n",
+      "print \"Peak diode current = %0.f mA\" %I_m\n",
+      "PIV = 2*V_m # in V\n",
+      "print \"Peak inverse voltage = %0.f V\" %PIV"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average current through the diode = 4.77 mA\n",
+        "RMS current = 7.5 mA\n",
+        "Peak diode current = 15 mA\n",
+        "Peak inverse voltage = 30 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3\n",
+      ": Page No 101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R1 = 2.2*10**3 # in ohm\n",
+      "R2 = 4.7*10**3 # in ohm\n",
+      "R_AB = (R1*R2)/(R1+R2) # in ohm\n",
+      "Vi = 20 # in V\n",
+      "V_o = (Vi * R_AB)/(R_AB+R1) # in V\n",
+      "PIV= Vi # in volts\n",
+      "print \"The output voltage = %0.1f V\" %V_o\n",
+      "print \"Peak inverse voltage = %0.f volts\" %PIV"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage = 8.1 V\n",
+        "Peak inverse voltage = 20 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example 2.4.2  (again 2.4)\n",
+      ": Page No 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_in  = 10 # in V\n",
+      "R_L = 2000 # in ohm\n",
+      "R1 = 100 # in ohm\n",
+      "V_R= 0.7 # in V\n",
+      "V_o = V_in * ( (R_L)/(R1+R_L) ) # in V\n",
+      "print \"The peak magnitude of the positive output voltage = %0.f V\" %V_o \n",
+      "Vo=-V_R # in V\n",
+      "print \"The peak magnitude of the negative output voltage = %0.f V\" %Vo"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The peak magnitude of the positive output voltage = 10 V\n",
+        "The peak magnitude of the negative output voltage = -1 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      " Example 2.4\n",
+      ": Page No 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_in  = 10 # in V\n",
+      "R1 = 2000 # in ohm\n",
+      "R2 = 2000 # in ohm\n",
+      "V_o = V_in * (R1/(R1+R2) ) # in V\n",
+      "# Vdc= 5/(T/2)*integrate('sin(omega*t)','t',0,T/2) and omega*T= 2*pi, So\n",
+      "Vdc= -5/pi*(cos(pi)-cos(0)) # in V\n",
+      "print \"The value of Vdc = %0.2f volts\" %Vdc\n",
+      "PIV= V_in/2 # in V\n",
+      "print \"The PIV value = %0.2f volts\" %PIV"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Vdc = 3.18 volts\n",
+        "The PIV value = 5.00 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7\n",
+      ": Page No 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V=240 # in V\n",
+      "R= 1 # in k\u03a9\n",
+      "R=R*10**3 # in \u03a9\n",
+      "Vsrms= V/4 # in V\n",
+      "Vm= sqrt(2)*Vsrms # in V\n",
+      "V_Ldc= -Vm/pi # in V\n",
+      "print \"The value of average load voltage = %0.f volts\" %V_Ldc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of average load voltage = -27 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8\n",
+      ": Page No 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V = 220 # in V\n",
+      "f=50 # in Hz\n",
+      "N2byN1=1/4 \n",
+      "R_L = 1 # in kohm\n",
+      "R_L= R_L*10**3 # in ohm\n",
+      "V_o = 220 # in V\n",
+      "V_s = N2byN1*V_o # in V\n",
+      "V_m = sqrt(2) * V_s # in V\n",
+      "V_Ldc = (2*V_m)/pi # in V\n",
+      "print \"Average load output voltage = %0.2f V\" %V_Ldc\n",
+      "P_dc = (V_Ldc)**2/R_L # in W\n",
+      "print \"DC power delivered to load = %0.2f watt\" %P_dc\n",
+      "PIV = V_m # in V\n",
+      "print \"Peak inverse Voltage = %0.2f V\" %PIV\n",
+      "f_o = 2*f # in Hz\n",
+      "print \"Output frequency = %0.f Hz\" %f_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average load output voltage = 49.52 V\n",
+        "DC power delivered to load = 2.45 watt\n",
+        "Peak inverse Voltage = 77.78 V\n",
+        "Output frequency = 100 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16\n",
+      ": Page No 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_dc = 12 # in V\n",
+      "R_L = 500 # in ohm\n",
+      "R_F = 25 # in ohm\n",
+      "I_dc = V_dc/R_L # in A\n",
+      "V_m = I_dc * pi * (R_L+R_F) # in V\n",
+      "V_rms = V_m/sqrt(2) # in V\n",
+      "V  = V_rms # in V\n",
+      "print \"The voltage = %0.f V\" %round(V)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage = 28 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17\n",
+      ": Page No 152"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_dc = 100 # in V\n",
+      "V_m = (V_dc*pi)/2 # in V\n",
+      "PIV = 2*V_m # in V\n",
+      "print \"Peak inverse voltage for center tapped FWR = %0.2f V\" %PIV\n",
+      "PIV1 = V_m # in V\n",
+      "print \"Peak inverse voltage for bridge type FWR = %0.2f V\" %PIV1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak inverse voltage for center tapped FWR = 314.16 V\n",
+        "Peak inverse voltage for bridge type FWR = 157.08 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19\n",
+      ": Page No 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_Gamma = 0.7 # in V\n",
+      "R_f = 0 # in ohm\n",
+      "V_rms = 120 # in V\n",
+      "V_max = sqrt(2)*V_rms # in V\n",
+      "R_L = 1 # in K ohm\n",
+      "R_L = R_L * 10**3 # in ohm\n",
+      "I_max = (V_max - (2*V_Gamma))/R_L # in A\n",
+      "I_dc = (2*I_max)/pi # in mA\n",
+      "V_dc = I_dc * R_L # in V\n",
+      "print \"The dc voltage available at the load = %0.2f V\" %V_dc\n",
+      "PIV = V_max # in V\n",
+      "print \"Peak inverse voltage = %0.1f V\" %PIV\n",
+      "print \"Maximum current through diode = %0.1f mA\" %(I_max*10**3)\n",
+      "P_max = V_Gamma * I_max # in W\n",
+      "print \"Diode power rating = %0.2f mW\" %(P_max*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dc voltage available at the load = 107.15 V\n",
+        "Peak inverse voltage = 169.7 V\n",
+        "Maximum current through diode = 168.3 mA\n",
+        "Diode power rating = 117.81 mW\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20\n",
+      ": Page No 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V1 = 10 # in V\n",
+      "V2 = 0.7 # in V\n",
+      "V3 = V2 # in V\n",
+      "V = V1-V2-V3 # in V\n",
+      "R1 = 1 # in ohm\n",
+      "R2 = 48 # in ohm\n",
+      "R3 = 1 # in ohm\n",
+      "R = R1+R2+R3 # in ohm\n",
+      "I = V/R # in A\n",
+      "I = I * 10**3 # in mA\n",
+      "print \"Current = %0.f mA\" %I"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current = 172 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 57
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22\n",
+      ": Page No 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R_L = 1 # in K ohm\n",
+      "R_L = R_L * 10**3 # in ohm\n",
+      "r_d = 10 # in ohm\n",
+      "V_m = 220 # in V\n",
+      "I_m = V_m/(r_d+R_L) # in A\n",
+      "print \"Peak value of current = %0.2f A\" %I_m\n",
+      "I_dc = (2*I_m)/pi # in A\n",
+      "print \"DC value of current = %0.2f A\" %I_dc\n",
+      "Irms= I_m/sqrt(2) # in A\n",
+      "r_f = sqrt((Irms/I_dc)**2-1)*100 # in %\n",
+      "print \"Ripple factor = %0.1f %%\" %r_f\n",
+      "Eta = (I_dc)**2 * R_L/((Irms)**2*(R_L+r_d))*100 # in %\n",
+      "print \"Rectification efficiency = %0.1f %%\" %Eta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak value of current = 0.22 A\n",
+        "DC value of current = 0.14 A\n",
+        "Ripple factor = 48.3 %\n",
+        "Rectification efficiency = 80.3 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23\n",
+      ": Page No 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_s = 12 # in V\n",
+      "R_L = 5.1 # in k ohm\n",
+      "R_L = R_L * 10**3 # in ohm\n",
+      "R_s = 1 # in K ohm\n",
+      "R_s = R_s * 10**3 # in ohm\n",
+      "V_L = (V_s*R_L)/(R_s+R_L) # in V\n",
+      "print \"Peak load voltage = %0.f V\" %V_L"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Peak load voltage = 10 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24\n",
+      ": Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_s = 10 # in V\n",
+      "R_L = 100 # in ohm\n",
+      "I_L = V_s/R_L # in A\n",
+      "print \"The load current during posotive half cycle = %0.1f A\" %I_L\n",
+      "I_D2 = 0 # in A\n",
+      "R2 = R_L # in ohm\n",
+      "I_L1 = -(V_s)/(R2+R_L) # in A\n",
+      "print \"The load current during negative half cycle = %0.2f A\" %I_L1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The load current during posotive half cycle = 0.1 A\n",
+        "The load current during negative half cycle = -0.05 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25\n",
+      ": Page No 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_m = 50 # in V\n",
+      "V_dc = (2*V_m)/pi # in V\n",
+      "print \"The dc voltage = %0.2f V\" %V_dc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The dc voltage = 31.83 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26\n",
+      ": Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R1 = 1.1 # in K ohm\n",
+      "R2 = 2.2 # in K ohm\n",
+      "Vi= 170 # in V\n",
+      "V_o = (Vi*R1)/(R1+R2) # in V\n",
+      "print \"The output voltage = %0.2f V\" %V_o\n",
+      "V_dc = (2*V_o)/pi # in V\n",
+      "print \"The dc voltage = %0.2f V\" %V_dc"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output voltage = 56.67 V\n",
+        "The dc voltage = 36.08 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 62
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/SanaKhanum/Chapter3.ipynb b/sample_notebooks/SanaKhanum/Chapter3.ipynb
new file mode 100755
index 00000000..a1ae6b1c
--- /dev/null
+++ b/sample_notebooks/SanaKhanum/Chapter3.ipynb
@@ -0,0 +1,76 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#                         Chapter 3 :  Work and Heat"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 3.1,Page No. 36"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work done is 274.200000 kJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given values\n",
+    "m=1; # mass of steam in kg\n",
+    "P=200; # Pressure at which process takes place in kPa\n",
+    "Tf=400; #Temperature of final state in C (Celcius)\n",
+    "x=0.2; #Quality factor \n",
+    "\n",
+    "#State 1\n",
+    "#Values from Table C-2\n",
+    "vf=0.001061; #specific volume of saturated liquid in m^3/kg\n",
+    "vg=0.8857;   #specific volume of saturated vapor in m^3/kg\n",
+    "\n",
+    "#State 2\n",
+    "#Values from superheat Table at 400C and 0.2 MPa\n",
+    "v2=1.549; #specific volume of steam in m^3/kg\n",
+    "\n",
+    "#Calculation\n",
+    "v1=vf+x*(vg-vf); # specific volume of state 1 in m^3/kg\n",
+    "W=m*P*(v2-v1) #Work done by steam at constant pressure(kPa) in kJ\n",
+    "print'Work done is %f kJ'%round(W,1)\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/SaurabhBhatia/Ch_03.ipynb b/sample_notebooks/SaurabhBhatia/Ch_03.ipynb
new file mode 100755
index 00000000..7da8382d
--- /dev/null
+++ b/sample_notebooks/SaurabhBhatia/Ch_03.ipynb
@@ -0,0 +1,324 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:764d1c60d9fefbb17ec0ffb4d8cdbf1cddd8802a6e86f8e004d6805017f0ecae"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter : 3 - Excess Carriers in Semiconductor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.1 - Page No : 2-34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data\n",
+      "N_d = 10**17 # atoms/cm**3\n",
+      "n_i = 1.5 * 10**10 # in /cm**3\n",
+      "n_o = 10**17 # in cm**3\n",
+      "# p_o * n_o = (n_i)**2\n",
+      "p_o = (n_i)**2 / n_o #in holes/cm**3\n",
+      "print \"The holes concentration at equilibrium = %0.2e holes/cm**3 \" %p_o"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The holes concentration at equilibrium = 2.25e+03 holes/cm**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.3 - Page No : 2-35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "# Given data\n",
+      "n_i = 1.5 * 10 **10 # in /cm**3 for silicon\n",
+      "N_d = 10**17 # in atoms/cm**3\n",
+      "n_o = 10**17 # electrons/cm**3\n",
+      "KT = 0.0259 \n",
+      "# E_r - E_i = KT * log(n_o/n_i)\n",
+      "del_E = KT * log(n_o/n_i) # in eV\n",
+      "print \"The energy band for this type material, E_F = Ei +\",round(del_E,3),\" eV\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy band for this type material, E_F = Ei + 0.407  eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.4 - Page No : 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "K = 1.38 * 10**-23 # in J/K\n",
+      "T = 27 # in degree\n",
+      "T = T + 273 # in K\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "Mu_e = 0.17 # in m**2/v-s\n",
+      "Mu_e1 = 0.025 # in m**2/v-s\n",
+      "D_n = ((K * T)/e) * Mu_e # in m**2/s\n",
+      "print \"The diffusion coefficient of electrons = %0.1e m**2/s \" %D_n\n",
+      "D_p = ((K * T)/e) * Mu_e1 # in m**2/s\n",
+      "print \"The diffusion coefficient of  holes = %0.2e m**2/s \" %D_p"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient of electrons = 4.4e-03 m**2/s \n",
+        "The diffusion coefficient of  holes = 6.47e-04 m**2/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.5 - Page No : 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "# Given data\n",
+      "Mu_n = 0.15 # in m**2/v-s\n",
+      "K = 1.38 * 10**-23  # in J/K\n",
+      "T = 300 # in K\n",
+      "del_n = 10**20 # in per m**3\n",
+      "Toh_n = 10**-7 # in s\n",
+      "e = 1.6 * 10**-19 # in C\n",
+      "D_n = Mu_n * ((K * T)/e) # in m**2/s\n",
+      "print \"The diffusion coefficient = %0.2e m**2/s \" %D_n\n",
+      "L_n = sqrt(D_n * Toh_n) # in m \n",
+      "print \"The Diffusion length = %0.2e m \" %L_n\n",
+      "J_n = (e * D_n * del_n)/L_n # in A/m**2\n",
+      "print \"The diffusion current density = %0.2e A/m**2 \"  %J_n\n",
+      "# Note : The value of diffusion coefficient in the book is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient = 3.88e-03 m**2/s \n",
+        "The Diffusion length = 1.97e-05 m \n",
+        "The diffusion current density = 3.15e+03 A/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.6 - Page No : 2-36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Sigma = 0.1 # in (ohm-m)**-1\n",
+      "Mu_n = 1300 \n",
+      "n_i = 1.5 * 10**10 \n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "n_n = Sigma/(Mu_n * q) # in electrons/cm**3\n",
+      "print \"The concentration of electrons = %0.2e per m**3 \" %(n_n*10**6)\n",
+      "p_n = (n_i)**2/n_n # in per cm**3\n",
+      "p_n = p_n * 10**6 # in perm**3\n",
+      "print \"The concentration of holes = %0.2e per m**3 \" %p_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The concentration of electrons = 4.81e+20 per m**3 \n",
+        "The concentration of holes = 4.68e+11 per m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.7 - Page No : 2-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "Mu_e = 0.13 # in m**2/v-s\n",
+      "Mu_h = 0.05 # in m**2/v-s\n",
+      "Toh_h = 10**-6 # in s\n",
+      "L = 100 # in \u00b5m\n",
+      "L = L * 10**-6 # in m\n",
+      "V = 2 # in V\n",
+      "t_n =L**2/(Mu_e * V) # in s\n",
+      "print \"Electron transit time = %0.1e seconds \" %t_n\n",
+      "p_g = (Toh_h/t_n) * (1 + Mu_h/Mu_e) #photo conductor gain \n",
+      "print \"Photo conductor gain = %0.2f\" %p_g\n",
+      "\n",
+      "# Note: There is a calculation error to evaluate the value of t_n. So the answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron transit time = 3.8e-08 seconds \n",
+        "Photo conductor gain = 36.00\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.8 - Page No : 2-37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 2.5 * 10**13 \n",
+      "Mu_n = 3800 \n",
+      "Mu_p = 1800 \n",
+      "q = 1.6 * 10**-19 # in C\n",
+      "Sigma = n_i * (Mu_n + Mu_p) * q # in (ohm-cm)**-1\n",
+      "Rho = 1/Sigma # in ohm-cm\n",
+      "Rho= round(Rho) \n",
+      "print \"The resistivity of intrinsic germanium = %0.f ohm-cm \" %Rho\n",
+      "N_D = 4.4 * 10**22/(1*10**8) # in atoms/cm**3\n",
+      "Sigma_n = N_D * Mu_n * q # in (ohm-cm)**-1\n",
+      "Rho_n = 1/Sigma_n # in ohm-cm\n",
+      "print \"The resistivity drops = %0.2f ohm-cm \" %Rho_n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity of intrinsic germanium = 45 ohm-cm \n",
+        "The resistivity drops = 3.74 ohm-cm \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example :  3.21.9 - Page No : 2-38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "n_i = 10**16 # in /m3\n",
+      "N_D = 10**22 # in /m**3\n",
+      "n = N_D # in /m**3\n",
+      "print \"Electron concentration = %0.1e per m**3 \" %n\n",
+      "p = (n_i)**2/n # in /m**3\n",
+      "print \"Hole concentration = %0.1e per m**3 \" %p"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron concentration = 1.0e+22 per m**3 \n",
+        "Hole concentration = 1.0e+10 per m**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/ShivaAmruthavakkula/Chapter_1.ipynb b/sample_notebooks/ShivaAmruthavakkula/Chapter_1.ipynb
new file mode 100755
index 00000000..4c144e1a
--- /dev/null
+++ b/sample_notebooks/ShivaAmruthavakkula/Chapter_1.ipynb
@@ -0,0 +1,499 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.1.1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Divergence of a beam,page number:7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the divergence of the beam is 8.05960631817e-05 in radians\n",
+      "the divergence of the beam is 0.00461781426568 in degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "# '''Example 1.1.1 :Divergence of a beam '''\n",
+    "\n",
+    "import math\n",
+    "\n",
+    " #decalring variables\n",
+    "Lambda=633*(10**-9)          #wavelength of laser\n",
+    "s=10*10**-3                 #spot size of the laser\n",
+    "w=s/2                       # waist radius\n",
+    "\n",
+    "#calculations\n",
+    "\n",
+    "theta_radians=(4*Lambda)/(math.pi*(2*w))\n",
+    "num=math.pi*(w)\n",
+    "theta_degrees=math.degrees(theta_radians)\n",
+    "\n",
+    "#results\n",
+    "print \"the divergence of the beam is\",theta_radians,\"in radians\"\n",
+    "print \"the divergence of the beam is\",theta_degrees,\"in degrees\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.2.1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Relative permittivity and refractive index n,Page number:8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Materials   EpsilonR squareroot(EpsilonR)\n",
+      "\n",
+      "------------------------------------------\n",
+      "\n",
+      "silicon \t11.9 \t3.44963766213 \n",
+      "\n",
+      "Diamond \t5.7 \t2.38746727726 \n",
+      "\n",
+      "GaAs    \t13.1 \t3.61939221417 \n",
+      "\n",
+      "SiO2    \t3.84 \t1.95959179423 \n",
+      "\n",
+      "Water   \t80 \t8.94427191 \n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Example 1.2.1: relative permittivity and refractive index n'''\n",
+    "\n",
+    "\n",
+    "import math\n",
+    "\n",
+    "numbers=[0,1,2,3,4]\n",
+    "\n",
+    "#declaring array variables\n",
+    "materials=['silicon','Diamond','GaAs   ','SiO2   ','Water  ']       #declaring the materials\n",
+    "relative_permittivity=[11.9,5.7,13.1,3.84,80]                       #declaring the relative permittivity at low frequencies\n",
+    "\n",
+    "#calculations\n",
+    "refractive_index=[]                                                 #declaring result set                                                 \n",
+    "for i in relative_permittivity:\n",
+    "    t=math.sqrt(i)\n",
+    "    refractive_index.append(t)\n",
+    "    \n",
+    "#Results and Table\n",
+    "print \"Materials   EpsilonR squareroot(EpsilonR)\\n\"\n",
+    "print \"------------------------------------------\\n\"\n",
+    "for j in numbers:\n",
+    "    print materials[j],\"\\t\",relative_permittivity[j],\"\\t\",refractive_index[j],\"\\n\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#Example 1.3.2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Relative permittivity and refractive index n,Page number-11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The phase velocity is  206896551.724\n",
+      "The group velocity is  205479452.055\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Example 1.3.2:Group and phase velocities'''\n",
+    "\n",
+    "import math\n",
+    "\n",
+    "#declaring variables\n",
+    "Lambda=1*10**-6         #wavelength of light\n",
+    "n=1.450                 #refractive index\n",
+    "c=3*10**8               #velocity of light in free space\n",
+    "group_index=1.46        #Group index(available from graph)\n",
+    "\n",
+    "#calculations\n",
+    "phase_velocity=c/n      #calculation of phase velocity\n",
+    "group_velocity=c/group_index    #calculation of group velocity\n",
+    "\n",
+    "#results\n",
+    "\n",
+    "print \"The phase velocity is \",phase_velocity\n",
+    "print \"The group velocity is \",group_velocity"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.6.2"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Phase change and Penetration Depth,Page Number-24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false,
+    "scrolled": true
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A\n",
+      "the crirical angle is equal to   0.986206896552 \n",
+      "\n",
+      "B\n",
+      "The phase change in medium 1 at incident angle=85 degrees with electrical component perpendicular to the plane of incidence is 116.45255621 degrees\n",
+      "The phase change in medium 1 at incident angle=85 degrees with electrical component parallel to the plane of incidence is -62.1314255434 degrees\n",
+      "The phase change in medium 1 at incident angle=90 degrees with electrical component perpendicular to the plane of incidence is 180.0 degrees\n",
+      "The phase change in medium 1 at incident angle=90 degrees with electrical component parallel to the plane of incidence is -2.84217094304e-14 degrees \n",
+      "\n",
+      "C\n",
+      "The penetration depth of medium 2 at incident angle of 85 degrees is  7.80048054638e-07 meters\n",
+      "The penetration depth of medium 2 at incident angle of 90 degrees is  6.63145596216e-07 meters\n"
+     ]
+    }
+   ],
+   "source": [
+    "'''Example 1.6.2'''\n",
+    "import math\n",
+    "\n",
+    "#declaration of variables\n",
+    "\n",
+    "n1=1.450            #refractive index of first medium\n",
+    "n2=1.430            #refractive index of second medium\n",
+    "Lambda=1*10**-6     #wavelength of light at standard units\n",
+    "theta_i1=85         #declaration of incidence angle 1\n",
+    "theta_i2=90         #declaration of incidence angle 2\n",
+    "\n",
+    "#Calculation of minimum incidence angle\n",
+    "theta_c=n2/n1       #calculation of critical angle\n",
+    "\n",
+    "#Calculation of phase change in medium 1 at incident angle 85 with perpendicular electrical component\n",
+    "tanphi_pp_85m1=math.sqrt(math.pow(math.sin(math.radians(theta_i1)),2)-math.pow((n2/n1),2))/math.cos(math.radians(theta_i1))\n",
+    "phi_pp_85im1=2*math.degrees(math.atan(tanphi_pp_85m1))\n",
+    "\n",
+    "#Calculation of phase change in medium 1 at incident angle 85 with parallel electrical component\n",
+    "tanphi_prll_85m1=math.pow((n1/n2),2)*tanphi_pp_85m1\n",
+    "phi_prll_85m1=2*(math.degrees(math.atan(tanphi_prll_85m1)))-math.degrees(math.pi)\n",
+    "phi_prll_inv_85m1=180+phi_prll_85m1\n",
+    "\n",
+    "#Calculation of phase change in medium 1 at incident angle 90 with perpendicular electrical component\n",
+    "tanphi_pp_90m1=math.sqrt(math.pow(math.sin(math.radians(theta_i2)),2)-math.pow((n2/n1),2))/math.cos(math.radians(theta_i2))\n",
+    "phi_pp_90m1=2*math.degrees(math.atan(tanphi_pp_90m1))\n",
+    "\n",
+    "#Calculation of phase change in medium 1 at incident angle 85 with parallel electrical component\n",
+    "tanphi_prll_90m1=math.pow((n1/n2),2)*tanphi_pp_90m1\n",
+    "phi_prll_90m1=2*(math.degrees(math.atan(tanphi_prll_90m1)))-math.degrees(math.pi)\n",
+    "phi_prll_inv_90m1=180+phi_prll_90m1\n",
+    "\n",
+    "#Calculation of penetration depth in medium 2 at incident angle 85\n",
+    "alpha_85=(2*math.pi*n2/Lambda)*(math.sqrt((math.pow((n1/n2),2)*math.pow(math.sin(math.radians(theta_i1)),2))-1))\n",
+    "delta_85=1/alpha_85\n",
+    "\n",
+    "#calculation of penetration depth in medium 2 at incident angle 90\n",
+    "alpha_90=(2*math.pi*n2/Lambda)*(math.sqrt((math.pow((n1/n2),2)*math.pow(math.sin(math.radians(theta_i2)),2))-1))\n",
+    "delta_90=1/alpha_90\n",
+    "\n",
+    "#Results\n",
+    "print \"A\"\n",
+    "print \"the crirical angle is equal to  \",theta_c,\"\\n\"\n",
+    "print \"B\"\n",
+    "print \"The phase change in medium 1 at incident angle=85 degrees with electrical component perpendicular to the plane of incidence is\",phi_pp_85im1,\"degrees\"\n",
+    "print \"The phase change in medium 1 at incident angle=85 degrees with electrical component parallel to the plane of incidence is\",phi_prll_85m1,\"degrees\"\n",
+    "print \"The phase change in medium 1 at incident angle=90 degrees with electrical component perpendicular to the plane of incidence is\",phi_pp_90m1,\"degrees\"\n",
+    "print \"The phase change in medium 1 at incident angle=90 degrees with electrical component parallel to the plane of incidence is\",phi_prll_90m1,\"degrees \\n\"\n",
+    "print \"C\"\n",
+    "print \"The penetration depth of medium 2 at incident angle of 85 degrees is \",delta_85,\"meters\"\n",
+    "print \"The penetration depth of medium 2 at incident angle of 90 degrees is \",delta_90,\"meters\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.6.3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Reflection at normal Incidence,Page number-25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A\n",
+      "the reflection coefficient when light travels from air to glass is  -0.2\n",
+      "the reflectance when light passes from air to glass is  0.04\n",
+      "The phase change of the reflected light is \n",
+      "180\n",
+      "B\n",
+      "the reflection coefficient when light travels from glass  to air is  0.2\n",
+      "the reflectance when light passes from glass to air is  0.04\n",
+      "The phase change of the reflected light is \n",
+      "180\n",
+      "C\n",
+      "The polarisation angle is  56.309932474\n"
+     ]
+    }
+   ],
+   "source": [
+    "# '''Example 1.6.3: Reflection at normal Incidence.Internal and external reflection'''\n",
+    "import math\n",
+    "\n",
+    "#declaration of variables\n",
+    "nglass=1.5\n",
+    "nair=1\n",
+    "\n",
+    "#declaration of phase change\n",
+    "def phasechange(r):\n",
+    "    if r<0:\n",
+    "        return 180\n",
+    "    else:\n",
+    "        return 0 \n",
+    "#Calculation of reflection coefficient and intensity from air to glass\n",
+    "r_coeffag=(nair-nglass)/(nair+nglass)         #reflection coefficient\n",
+    "Rag=math.pow(r_coeffag,2)                       #Reflectance\n",
+    "pag=phasechange(r_coeffag)\n",
+    "\n",
+    "\n",
+    "#Calculation of relection and intensity from glass to aie\n",
+    "r_coeffga=(nglass-nair)/(nglass+nair)\n",
+    "Rga=math.pow(r_coeffga,2)\n",
+    "pga=phasechange(r_coeffga)\n",
+    "\n",
+    "#calculation of polarisation angle\n",
+    "theta_p=math.degrees(math.atan(nglass/nair))\n",
+    "\n",
+    "#Results\n",
+    "print \"A\"\n",
+    "print\"the reflection coefficient when light travels from air to glass is \",r_coeffag\n",
+    "print \"the reflectance when light passes from air to glass is \",Rag\n",
+    "print \"The phase change of the reflected light is \\n\",pag\n",
+    "print \"B\"\n",
+    "print\"the reflection coefficient when light travels from glass  to air is \",r_coeffga\n",
+    "print \"the reflectance when light passes from glass to air is \",Rga\n",
+    "print \"The phase change of the reflected light is \\n\",pag\n",
+    "\n",
+    "print \"C\"\n",
+    "print \"The polarisation angle is \",theta_p"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.6.4"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Antireflection coatings on solar cells,Page number-25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the reflectance of the Silicon is  0.308641975309\n",
+      "the refractive index of coating must be  1.87082869339\n",
+      "the thickness of the coating is  9.21052631579e-08  meters\n"
+     ]
+    }
+   ],
+   "source": [
+    "# '''Example 1.6.4: Antireflection coatings on solarcells'''\n",
+    "import math\n",
+    "\n",
+    "#Declaration of variables\n",
+    "nair=1\n",
+    "nsi=3.5\n",
+    "ncoating=1.9\n",
+    "Lambda=700*10**-9\n",
+    "\n",
+    "#declaration of phase change\n",
+    "def phasechange(r):\n",
+    "    if r<0:\n",
+    "        return 180\n",
+    "    else:\n",
+    "        return 0 \n",
+    "\n",
+    "#Calculation of reflectance at Silicon\n",
+    "rsi=((nair-nsi)/(nair+nsi))\n",
+    "Rsi=math.pow(rsi,2)\n",
+    "psi=phasechange(rsi)\n",
+    "\n",
+    "#calculation of refractive index of coating\n",
+    "ncoating_theoritical=math.sqrt(nair*nsi)\n",
+    "\n",
+    "#calculation of thickness\n",
+    "d=Lambda/(4*ncoating)\n",
+    "\n",
+    "#results\n",
+    "print \"the reflectance of the Silicon is \",Rsi\n",
+    "print \"the refractive index of coating must be \",ncoating_theoritical\n",
+    "print \"the thickness of the coating is \",d,\" meters\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.7.1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Resonator modes and Spectral width,Page number-31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false,
+    "slideshow": {
+     "slide_type": "-"
+    }
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The seperation of modes is  1.5e+12  Hertz\n",
+      "The spectral width is 1.35888872682e-10  meters\n"
+     ]
+    }
+   ],
+   "source": [
+    "# '''Example 1.7.1: Resonator modes and spectral width'''\n",
+    "import math\n",
+    "\n",
+    "#declaration of variables\n",
+    "L=100*10**-6\n",
+    "R=0.9\n",
+    "Lambda=900*10**-9\n",
+    "c=3*10**8\n",
+    "\n",
+    "#Calculation of seperation of modes\n",
+    "delta_vm=c/(2*L)\n",
+    "\n",
+    "#calculation of finesse\n",
+    "F=((math.pi)*math.sqrt(R))/(1-R)\n",
+    "\n",
+    "#calculation of modewidth\n",
+    "del_vm=delta_vm/F\n",
+    "\n",
+    "#calculation of mode frequency\n",
+    "vm=c/Lambda\n",
+    "\n",
+    "#calculation of spectral width\n",
+    "del_lambda=(c/math.pow(vm,2))*del_vm\n",
+    "\n",
+    "#Results\n",
+    "print \"The seperation of modes is \",delta_vm,\" Hertz\"\n",
+    "print \"The spectral width is\", del_lambda,\" meters\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/ShivamNegi/Chapter_1.ipynb b/sample_notebooks/ShivamNegi/Chapter_1.ipynb
new file mode 100755
index 00000000..435494b2
--- /dev/null
+++ b/sample_notebooks/ShivamNegi/Chapter_1.ipynb
@@ -0,0 +1,830 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1: Structure and Bonding"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 1, Page no: 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "c = 3 * 10 ** 10                # Velocity of light, cm/sec\n",
+      "\n",
+      "# Variable\n",
+      "wavelength = 3500 * 10 ** -8    # Wavelength of radiation, cm\n",
+      "\n",
+      "# Solution\n",
+      "print \"v = c / wavelength\"\n",
+      "print \"v: Velocity, c: Speed of light\"\n",
+      "\n",
+      "v = c / wavelength\n",
+      "\n",
+      "print \"The frequency of radiation is\", '{:.2e}'.format(v), \"Heartz.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "v = c / wavelength\n",
+        "v: Velocity, c: Speed of light\n",
+        "The frequency of radiation is 8.57e+14 Heartz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 2, Page no: 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "c = 3 * 10 ** 8                 # speed of light, m/sec\n",
+      "\n",
+      "# Variable\n",
+      "f = 5 * 10 ** 16                # frequency, cycles/sec\n",
+      "\n",
+      "# Solution\n",
+      "v_bar = f / c\n",
+      "print \"The wave number is\", '{:.2e}'.format(v_bar), \"cycles/m.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wave number is 1.67e+08 cycles/m.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 3, Page no: 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "c = 3 * 10 ** 8                 # Speed of light, m/sec\n",
+      "\n",
+      "# Variable\n",
+      "T = 2.4 * 10 ** -10             # Time period, sec\n",
+      "\n",
+      "# Solution\n",
+      "f = 1 / T                       # Frequency, /sec\n",
+      "lamda = c / f                   # wavelength, m\n",
+      "v_bar = 1 / lamda               # wavenumber, /meter\n",
+      "\n",
+      "print \"Frequency:\", '{:.2e}'.format(f), \"/sec\"\n",
+      "print \"Wavelength:\", '{:.2e}'.format(lamda), \"m\"\n",
+      "print \"Wave number:\", '{:.2e}'.format(v_bar), \"/m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency: 4.17e+09 /sec\n",
+        "Wavelength: 7.20e-02 m\n",
+        "Wave number: 1.39e+01 /m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 4, Page no: 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Constants\n",
+      "c = 3 * 10 ** 8                 # Speed of light, m/sec\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variable\n",
+      "ke = 4.55 * 10 ** -25           # Kinetic Energy, J\n",
+      "\n",
+      "# Solution\n",
+      "v = math.sqrt(ke * 2 / m)\n",
+      "\n",
+      "lamda = h / (m * v)\n",
+      "\n",
+      "print \"The de Broglie wavelength is\", '{:.2e}'.format(lamda), \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The de Broglie wavelength is 7.28e-07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 5, Page no: 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variables\n",
+      "m = 10 * 10 ** -3               # Mass of the ball, kg\n",
+      "v = 10 ** 5                      # Velocity of ball, cm / sec\n",
+      "\n",
+      "# Solution\n",
+      "lamda = (h * 10 ** 7) / (m * v)\n",
+      "print \"The Wavelength of iron ball is\", \"{:.2}\".format(lamda), \"cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Wavelength of iron ball is 6.6e-30 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 6, Page no: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variable\n",
+      "lamda = 2 * 10 ** -10           # wavelength, m\n",
+      "\n",
+      "# Solution\n",
+      "p = h / lamda\n",
+      "\n",
+      "print \"The momentum of the particle is\", \"{:.2}\".format(p), \"kg.m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The momentum of the particle is 3.3e-24 kg.m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 7, Page no: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "pi = 3.141                      # Pi\n",
+      "\n",
+      "# Variable\n",
+      "delta_x = 1 * 10 ** -10         # uncertainty in velocity, m\n",
+      "\n",
+      "# Solution\n",
+      "delta_v = h / (4 * pi * m * delta_x)\n",
+      "\n",
+      "print \"Uncertainty in position of electron >=\",\n",
+      "print \"{:.2}\".format(delta_v), \"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertainty in position of electron >= 5.8e+05 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 8, Page no: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "pi = 3.141                      # Pi\n",
+      "\n",
+      "# Variables\n",
+      "m = 10 ** -11                   # Mass of particle, g\n",
+      "v = 10 ** -4                    # Velocity of particle, cm/sec\n",
+      "delta_v = 0.1 / 100             # Uncertainty in velocity\n",
+      "\n",
+      "# Solution\n",
+      "delta_v = v / 1000\n",
+      "delta_x = (h * 10 ** 7) / (4 * pi * delta_v * m)\n",
+      "\n",
+      "print \"Uncertainty in position >=\",\n",
+      "print \"{:.3e}\".format(delta_x), \"cm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertainty in position 5.27e-10 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 9, Page no: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "c = 3 * 10 ** 8                 # Speed of light, m/sec\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variable\n",
+      "lamda = 650 * 10 ** -12         # Wavelength of radiation, m\n",
+      "\n",
+      "# Solution\n",
+      "E = h * c / lamda\n",
+      "\n",
+      "print \"Energy per photon\", \"{:.3e}\".format(E), \"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy per photon 3.058e-16 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 10, Page no: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "h = 6.625 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variables\n",
+      "v = 6.5 * 10 ** 7               # Velocity of particle, m/s\n",
+      "lamda = 5 * 10 ** -11           # Wavelength, m\n",
+      "\n",
+      "# Solution\n",
+      "P = h / lamda\n",
+      "\n",
+      "print \"The momentum of the particle\", \"{:.2e}\".format(P), \"kg.m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The momentum of the particle 1.33e-23 kg.m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 11, Page no: 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Constants\n",
+      "c = 3 * 10 ** 8                 # Speed of light, m/sec\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variables\n",
+      "lamda = 200 * 10 ** -7          # Wavelength, cm\n",
+      "wf = 6.95 * 10 ** -12           # Work function, erg\n",
+      "\n",
+      "# Solution\n",
+      "E = (h * c) * 10 ** 9 / lamda\n",
+      "\n",
+      "print \"Energy of photon\", \"{:.3e}\".format(E), \"erg\"\n",
+      "\n",
+      "ke = E - wf\n",
+      "\n",
+      "v = math.sqrt((2 * ke) / (m * 10 ** 3)) * 10 ** -2\n",
+      "\n",
+      "print \"The maximum velocity of electron\", \"{:.3e}\".format(v), \"m/sec\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of photon 9.939e-12 erg\n",
+        "The maximum velocity of electron 8.105e+05 m/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 12, Page no: 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variables\n",
+      "m = 150                     # Weight of ball, gm\n",
+      "v = 50                          # Velocity, m/sec\n",
+      "\n",
+      "lamda = h / (m * v * 10 ** -8)\n",
+      "print \"Wavelength of ball\", \"{:.3e}\".format(lamda), \"m\"\n",
+      "print \"Its wavelength is so short that it does not fall\",\n",
+      "print \"in visible range, so we cannot observe it.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of ball 8.835e-30 m\n",
+        "Its wavelength is so short that it does not fall in visible range, so we cannot observe it.\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 13, Page no: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "pi = 3.141                      # Pi\n",
+      "\n",
+      "# Variables\n",
+      "m = 0.1                         # Mass of base ball, kg\n",
+      "delta_x = 10 ** -10             # Uncertainty in position, m\n",
+      "\n",
+      "# Solution\n",
+      "delta_v = h / (4 * pi * m * delta_x)\n",
+      "\n",
+      "print \"Uncertainty in velocity >=\", \"{:.2e}\".format(delta_v), \"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Uncertainty in velocity >= 5.27e-24 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 14, Page no: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constant\n",
+      "t_v = 1.3 * 10 ** 15            # Threashold freq. Pt, /sec\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "\n",
+      "# Solution\n",
+      "print \"The threshold frequency is the lowest frequency\",\n",
+      "print \"that photons may possess to produce the photoelectric\",\n",
+      "print \"effect.\"\n",
+      "E = h * t_v\n",
+      "print \"The energy corresponding to this frequency is the minimum\",\n",
+      "print \"energy =\", \"{:.2e}\".format(E), \"erg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The threshold frequency is the lowest frequency that photons may possess to produce the photoelectric effect.\n",
+        "The energy corresponding to this frequency is the minimum energy = 8.61e-19 erg\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 15, Page no: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "e = 1.602 * 10 ** -19           # Charge of electron, C\n",
+      "\n",
+      "# Variable\n",
+      "v = 1.87 * 10 ** 9              # Velocity of electron, m/sec\n",
+      "\n",
+      "# Solution\n",
+      "V = m * v ** 2 / (2 * e)\n",
+      "lamda = h / (m * v)\n",
+      "\n",
+      "print \"The voltage is\", \"{:.2e}\".format(V), \"volt\"\n",
+      "print \"The de Broglie wavelength is\", \"{:.2e}\".format(lamda), \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage is 9.93e+06 volt\n",
+        "The de Broglie wavelength is 3.89e-13 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 16, Page no: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variable\n",
+      "lamda = 4.8 * 10 ** -9         # Wavelength of electron, m\n",
+      "\n",
+      "# Solution\n",
+      "ke = ((h / lamda) ** 2) / (2 * m)\n",
+      "\n",
+      "print \"The Kinetic Energy of moving electron is\", \"{:.2e}\".format(ke),\n",
+      "print \"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Kinetic Energy of moving electron is 1.05e-20 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 17, Page no: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Constants\n",
+      "m = 9.1 * 10 ** -31             # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "c = 3 * 10 ** 8                 # Speed of light, m/sec\n",
+      "\n",
+      "# Variables\n",
+      "v = 6.46 * 10 ** 5              # Velocity of electron, m/sec\n",
+      "lamda = 200 * 10 ** -9          # Wavelength of light, m\n",
+      "\n",
+      "# Solution\n",
+      "E = (h * c) / lamda\n",
+      "ke = m * v ** 2\n",
+      "w = E - ke\n",
+      "\n",
+      "print \"The workfunction of the metal surface is\", \"{:.3e}\".format(w),\n",
+      "print \"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The workfunction of the metal surface is 6.141e-19 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 18, Page no: 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Constants\n",
+      "e = 1.602 * 10 ** -19           # Charge of proton, C\n",
+      "m_p = 1.66 * 10 ** -27          # Mass of proton, kg\n",
+      "m_e = 9.1 * 10 ** -31           # Mass of electron, kg\n",
+      "h = 6.626 * 10 ** -34           # Plank's constant, J.sec\n",
+      "\n",
+      "# Variable\n",
+      "V = 35                          # Acceleration potential, volt\n",
+      "\n",
+      "# Solution\n",
+      "lamda_p = h / math.sqrt(2 * e * V * m_p)\n",
+      "lamda_e = h / math.sqrt(2 * e * V * m_e)\n",
+      "\n",
+      "print \"The wavelength of electron when accelerated with same\",\n",
+      "print \"potential is\", \"{:.3e}\".format(lamda_e), \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of electron when accelerated with same potential is 2.074e-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 19, Page no: 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "B_O1 = (10 - 6) / 2              # Bond Order for O2\n",
+      "B_O2 = (10 - 7) / 2              # Bond Order for O2-\n",
+      "\n",
+      "print \"Bond length of O2- > O2 as Bond order of O2\",\n",
+      "print \"> Bond order of O2- :\", B_O1 > B_O2\n",
+      "print \"Both are paramagnetic, because they contain unpaired electrons.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bond length of O2- > O2 as Bond order of O2 > Bond order of O2- : True\n",
+        "Both are paramagnetic, because they contain unpaired electrons.\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 20, Page no: 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "B_O = (9 - 4) / 2.0         # Bond order of N2+\n",
+      "\n",
+      "print \"The Bond order of N2+ is\", B_O\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Bond order of N2+ is 2.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem: 21, Page no: 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Solution\n",
+      "v_n = 2 * 5             # number of valence e- in nitrogen\n",
+      "v_co = 4 + 6            # number of valence e- in CO\n",
+      "\n",
+      "print \"The number of valence electrons in N2\", v_n\n",
+      "print \"The number of valence electrons in CO\", v_co\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of valence electrons in N2 10\n",
+        "The number of valence electrons in CO 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/ShramanaPatra/tbc.ipynb b/sample_notebooks/ShramanaPatra/tbc.ipynb
new file mode 100755
index 00000000..b2e6f092
--- /dev/null
+++ b/sample_notebooks/ShramanaPatra/tbc.ipynb
@@ -0,0 +1,264 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d2cada736c573d7db80ea068a748ffd72dc1a02f7bcc83dca412f55927110705"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 39: Special Machines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example Number 39.1, Page Number:1537"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#variable description\n",
+      "p=8.0 #number of poles\n",
+      "tp=5.0 #number of teeth for each pole\n",
+      "nr=50.0 #number of rotor teeth\n",
+      "\n",
+      "#calculation\n",
+      "ns=p*tp #number of stator teeth\n",
+      "B=((nr-ns)*360)/(nr*ns) #stepping angle\n",
+      "\n",
+      "#result\n",
+      "print \"stepping angle is \",B,\"degrees\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stepping angle is  1.8 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example Number 39.2, Page Number:1537"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#variable declaration\n",
+      "B=2.5\n",
+      "rn=25\n",
+      "f=3600\n",
+      "\n",
+      "#calculation\n",
+      "r=360/B\n",
+      "s=r*rn\n",
+      "n=(B*f)/360\n",
+      "\n",
+      "#result\n",
+      "print \"Resolution =\",int(r),\"steps/revolution\"\n",
+      "print \" Number of steps required for the shaft to make 25 revolutions =\",int(s)\n",
+      "print \" Shaft speed\", int(n),\"rps\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Resolution = 144 steps/revolution\n",
+        "Number of steps required for the shaft to make 25 revolutions = 3600\n",
+        "Shaft speed 25 rps\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example Number 39.3, Page Number:1544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#variable declaration\n",
+      "B=15 #stepping angle\n",
+      "pn=3 #number of phases\n",
+      "nr=360/(pn*B) #number of rotor teeth\n",
+      "\n",
+      "#number of stator teeth\n",
+      "ns1=((360*nr)/(360-(nr*B))) #ns>nr\n",
+      "ns2=((360*nr)/(360+(nr*B))) #nr>ns\n",
+      "\n",
+      "#result\n",
+      "print \"When ns>nr: ns= \",ns1\n",
+      "print \"When nr>ns: ns= \",ns2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When ns>nr: ns=  12\n",
+        "When nr>ns: ns=  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example Number 39.4, Page Number:1545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#variable declaration\n",
+      "B=1.8\n",
+      "pn=4\n",
+      "\n",
+      "#calculation\n",
+      "nr=360/(pn*B) #number of rotor teeth\n",
+      "ns=nr\n",
+      "\n",
+      "#result\n",
+      "print \"Number of rotor teeth = \",int(nr)\n",
+      "print \"Number of statot teeth = \",int(ns)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of rotor teeth =  50.0\n",
+        "Number of statot teeth =  50.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example Number 39.5, Page Number:1555"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#variable declaration\n",
+      "er=20\n",
+      "\n",
+      "#calculation\n",
+      "a=40\n",
+      "e2=er*math.cos(math.radians(a))\n",
+      "e1=er*math.cos(math.radians(a-120))\n",
+      "e3=er*math.cos(math.radians(a+120))\n",
+      "\n",
+      "#result\n",
+      "print \"a) For a=40 degrees\"\n",
+      "print \"   e2s=\" ,e2,\"V\"\n",
+      "print \"   e1s=\" ,e1,\"V\"\n",
+      "print \"   e3s=\" ,e3,\"V\"\n",
+      "\n",
+      "#calculation\n",
+      "a=(-40)\n",
+      "e2=er*math.cos(math.radians(a))\n",
+      "e1=er*math.cos(math.radians(a-120))\n",
+      "e3=er*math.cos(math.radians(a+120))\n",
+      "\n",
+      "#result\n",
+      "print \"b) For a=-40 degrees\"\n",
+      "print \"   e2s=\" ,e2,\"V\"\n",
+      "print \"   e1s=\" ,e1,\"V\"\n",
+      "print \"   e3s=\" ,e3,\"V\"\n",
+      "\n",
+      "#calculation\n",
+      "a=30\n",
+      "e12=math.sqrt(3)*er*math.cos(math.radians(a-150))\n",
+      "e23=math.sqrt(3)*er*math.cos(math.radians(a-30))\n",
+      "e31=math.sqrt(3)*er*math.cos(math.radians(a+90))\n",
+      "\n",
+      "#result\n",
+      "print \"c) For a=30 degrees\"\n",
+      "print \"   e12=\" ,e12,\"V\"\n",
+      "print \"   e23=\" ,e23,\"V\"\n",
+      "print \"   e31=\" ,e31,\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) For a=40 degrees\n",
+        "   e2s= 15.3208888624 V\n",
+        "   e1s= 3.47296355334 V\n",
+        "   e3s= -18.7938524157 V\n",
+        "b) For a=-40 degrees\n",
+        "   e2s= 15.3208888624 V\n",
+        "   e1s= -18.7938524157 V\n",
+        "   e3s= 3.47296355334 V\n",
+        "c) For a=30 degrees\n",
+        "   e12= -17.3205080757 V\n",
+        "   e23= 34.6410161514 V\n",
+        "   e31= -17.3205080757 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/SudheerBommisetty/Chapter_4_Op_Amps_as_AC_Amplifiers.ipynb b/sample_notebooks/SudheerBommisetty/Chapter_4_Op_Amps_as_AC_Amplifiers.ipynb
new file mode 100755
index 00000000..42bcd226
--- /dev/null
+++ b/sample_notebooks/SudheerBommisetty/Chapter_4_Op_Amps_as_AC_Amplifiers.ipynb
@@ -0,0 +1,363 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Op Amps as AC Amplifiers"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.1 page.no: 65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1max=  140000.0\n",
+      "Xc1=R1/10 at F1\n",
+      "C1= 2.65258238486e-07  farad\n",
+      "C2= 8.16179195343e-07  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "# capacitor coupled voltage follower design \n",
+    "from math import pi\n",
+    "\n",
+    "Vbe =0.7\n",
+    "Ibmax =500*10**-9\n",
+    "R1max=0.1* Vbe/ Ibmax\n",
+    "print \"R1max= \",R1max\n",
+    "# assume R1=120Kohms\n",
+    "R1=120000\n",
+    "f1=50\n",
+    "print \"Xc1=R1/10 at F1\"\n",
+    "#  C1=1/(2∗pi∗f1∗(R1/10))\n",
+    "C1=1/(2*pi*f1*(R1/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "Rl=3900\n",
+    "C2=1/(2*pi*f1*Rl)\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.2 page.no: 66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C1= 3.18309886184e-08  farad\n",
+      "C2= 8.16179195343e-07  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "# capacitor coupled voltage follower design using BIFET \n",
+    "from math import pi\n",
+    "\n",
+    "R1 =1000000\n",
+    "f1=50\n",
+    "C1=1/(2*pi*f1*(R1/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "Rl=3900\n",
+    "C2=1/(2*pi*f1*Rl)\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.3 page.no: 68"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C1= 4.681027738e-07  farad\n",
+      "C2= 4.681027738e-07  farad\n",
+      "Zin=  3400068000 ohms\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "#capacitor coupled voltage follower design \n",
+    "from math import pi \n",
+    "\n",
+    "Vbe =0.7\n",
+    "Ibmax =500*10**-9\n",
+    "R1max=0.1* Vbe/ Ibmax\n",
+    "R1=R1max/2\n",
+    "R2=R1\n",
+    "R1=68000\n",
+    "f1=50\n",
+    "C1=1/(2*pi*f1*(R1/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "C2=C1\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "Rl=3900\n",
+    "M=50000\n",
+    "Zin=(1+M)*R1\n",
+    "print \"Zin= \",Zin,\"ohms\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.4 page.no: 69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C1= 1.10524266036e-07  farad\n",
+      "C2= 6.02859632924e-07  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "#capacitor coupled non inverting amplifier design \n",
+    "#lower cut off frequency for the circuit =120Hz\n",
+    "from math import pi\n",
+    "\n",
+    "Vbe =0.7\n",
+    "Ibmax =500*10**-9\n",
+    "R1max=0.1* Vbe/ Ibmax\n",
+    "R1=120000\n",
+    "f1=120\n",
+    "C1=1/(2*pi*f1*(R1/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "Rl=2200\n",
+    "C2=1/(2*pi*f1*Rl)\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.5 page.no: 69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1= 994974.874372\n",
+      "C2= 1.58359168376e-07  farad\n",
+      "C1=1000pF much larger than stray capacitance\n",
+      "C2= 6.63145596216e-07  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "#capacitor coupled non inverting high impedence follower design \n",
+    "#lower cut off frequency for the circuit =200Hz \n",
+    "from math import pi\n",
+    "\n",
+    "Vo=3\n",
+    "Vi=0.015\n",
+    "Av=Vo/Vi\n",
+    "R2 =1000000\n",
+    "R3=R2/(Av-1)\n",
+    "f1=200\n",
+    "R1=R2-R3\n",
+    "print\"R1=\",R1\n",
+    "C2=1/(2*pi*f1*(R3))\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "print \"C1=1000pF much larger than stray capacitance\"\n",
+    "Rl=12000\n",
+    "C2=1/(2*pi*f1*(Rl/10))\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.6 page.no: 71"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "C1= 0.000159154943092  farad\n",
+      "C2= 6.36619772368e-05  farad\n",
+      "Cf= 3.38627538493e-07  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "#capacitor coupled inverting amplifier design \n",
+    "#frequency range for the circuit =10Hz to 1KHz\n",
+    "from math import pi\n",
+    "R1 =1000\n",
+    "f1=10\n",
+    "C1=1/(2*pi*f1*(R1/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "Rl=250\n",
+    "C2=1/(2*pi*f1*Rl)\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "R2=47000\n",
+    "Cf=1/(2*pi*f1*R2)\n",
+    "print \"Cf=\",Cf,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## exa 4.7 page.no: 72"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "R1= 240000.0  ohms\n",
+      "R2= 240000.0  ohms\n",
+      "R4= 1000.0  ohms\n",
+      "R3= 99000.0  ohms\n",
+      "Rp= 114782.608696  ohms\n",
+      "C1= 1.84876954097e-07  farad\n",
+      "C2= 3.78940340695e-06  farad\n",
+      "C3= 2.12206590789e-06  farad\n",
+      "The circuit voltage should be normally between 9 to 18 volts\n"
+     ]
+    }
+   ],
+   "source": [
+    "#capacitor coupled non design\n",
+    "from math import pi\n",
+    "\n",
+    "Av =100.\n",
+    "Vcc =24.\n",
+    "Vo=5.\n",
+    "#lower cut off frequency for the circuit =75Hz\n",
+    "Vbe=0.7\n",
+    "Ibmax =500*10**-9\n",
+    "#I2>>Ibmax\n",
+    "I2 =100* Ibmax\n",
+    "R1=(Vcc/2)/I2\n",
+    "print\"R1=\",R1,\" ohms\"\n",
+    "R2=(Vcc/2)/I2\n",
+    "print\"R2=\",R2,\" ohms\"\n",
+    "#assume R1=220Kohms\n",
+    "Vi=Vo/Av\n",
+    "R1=220000.\n",
+    "I4 =100* Ibmax\n",
+    "R4=Vi/I4\n",
+    "print\"R4=\",R4,\" ohms\"\n",
+    "R3=(Vo/I4)-R4\n",
+    "print\"R3=\",R3,\" ohms\"\n",
+    "Rp=(R1*R2)/(R1+R2)\n",
+    "print\"Rp=\",Rp,\" ohms\"\n",
+    "f1=75.\n",
+    "C1=1/(2*pi*f1*(Rp/10))\n",
+    "print \"C1=\",C1,\" farad\"\n",
+    "Rl=5600.\n",
+    "C2=1/(2*pi*f1*(Rl/10))\n",
+    "print \"C2=\",C2,\" farad\"\n",
+    "C3=1/(2*pi*f1*R4)\n",
+    "print \"C3=\",C3,\" farad\"\n",
+    "print \"The circuit voltage should be normally between 9 to 18 volts\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/SwathiSyamala/Chapter_6_IMPEDENCE_MATCHING_AND_TUNNING.ipynb b/sample_notebooks/SwathiSyamala/Chapter_6_IMPEDENCE_MATCHING_AND_TUNNING.ipynb
new file mode 100755
index 00000000..e50612ad
--- /dev/null
+++ b/sample_notebooks/SwathiSyamala/Chapter_6_IMPEDENCE_MATCHING_AND_TUNNING.ipynb
@@ -0,0 +1,277 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 6 IMPEDENCE MATCHING AND TUNNING"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.1 page no:284"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "inductor of first circuit in nH =  38.9848400617\n",
+      "capacitor of the first circuit in pF =  0.9227738301\n",
+      "inductor of second circuit in nH =  46.138691505\n",
+      "capacitor of the second circuit in pF =  2.59898933745\n",
+      "\"NOTE:−for above specific problem Rl>Zo, positive X implies inductor , negative X implies capacitor , positive B implies capacitor and negative B implies inductor .\"\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.1 program to design an L section matching network\n",
+    "# example:−6.1,page no.−284.\n",
+    "from math import pi,sqrt\n",
+    "from sympy import I\n",
+    "# program to design an L section matching network to match a series RC load.\n",
+    "Zl=200-I*100; # load impedence .\n",
+    "Rl=200;Xl=-100;f=500*10**6;Zo=100;\n",
+    "B1=(Xl+sqrt(Rl/Zo)*sqrt(Rl**2+Xl**2-(Rl*Zo)))/(Rl**2+Xl**2);\n",
+    "B2=(Xl-sqrt(Rl/Zo)*sqrt(Rl**2+Xl**2-(Rl*Zo)))/(Rl**2+Xl**2);\n",
+    "C1=(B1/(2*pi*f))*10**12;\n",
+    "L2=(-1/(B2*2*pi*f))*10**9;\n",
+    "X1=(1/B1)+((Xl*Zo)/Rl)-(Zo/(B1*Rl));\n",
+    "X2=(1/B2)+((Xl*Zo)/Rl)-(Zo/(B2*Rl));\n",
+    "L1=(X1/(2*pi*f))*10**9;\n",
+    "C2=(-1/(X2*2*pi*f))*10**12;\n",
+    "print\"inductor of first circuit in nH = \",L1\n",
+    "print\"capacitor of the first circuit in pF = \",C1\n",
+    "print\"inductor of second circuit in nH = \",L2\n",
+    "print\"capacitor of the second circuit in pF = \",C2 \n",
+    "print\"\\\"NOTE:−for above specific problem Rl>Zo, positive X implies inductor , negative X implies capacitor , positive B implies capacitor and negative B implies inductor .\\\"\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.2 page no:304"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "charecteristic impedence of matching section = 22.360679775\n",
+      " fractional bandwidth =  0.293159219438\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.5 design quarter wave matching transformer\n",
+    "#example:−6.5,page no.−304.\n",
+    "from math import sqrt,pi,acos\n",
+    "#program to design a single section quarter wave matching transformer .\n",
+    "Zl=10; # load impedence .\n",
+    "Zo=50; # characteristic impedence .\n",
+    "fo=3*10**9;swr=1.5; # maximum limit of swr.\n",
+    "Z1=sqrt(Zo*Zl); # characteristic impedence of the matching section .\n",
+    "taom=(swr-1)/(swr+1);\n",
+    "frac_bw=2-(4/pi)*acos((taom/sqrt(1-taom**2))*(2*sqrt(Zo*Zl)/abs(Zl-Zo))); # fractional bandwidth .\n",
+    "print \"charecteristic impedence of matching section =\",Z1\n",
+    "print \" fractional bandwidth = \",frac_bw"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.6 page no:307"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "approximate value of reflection coefficient is =  0.4\n",
+      "the error in percent is about =  4.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.6 program to evaluate the worst case percent error\n",
+    "# example:−6.6,page no.−307.\n",
+    "#from math import abs\n",
+    "# program to evaluate the worst case percent error in computing magnitude of reflection coefficient .\n",
+    "Z1 =100.; \n",
+    "Z2 =150.; \n",
+    "Zl =225.;\n",
+    "tao_1=(Z2-Z1)/(Z2+Z1);\n",
+    "tao_2=(Zl-Z2)/(Zl+Z2);\n",
+    "tao_exact=(tao_1+tao_2)/(1+tao_1*tao_2); # this results as angle is taken zero .\n",
+    "tao_approx=tao_1+tao_2; # this results as angle is taken zero .\n",
+    "eror=abs(((tao_exact -tao_approx)/tao_exact)*100);\n",
+    "print \"approximate value of reflection coefficient is = \",tao_approx\n",
+    "print \"the error in percent is about = \",eror"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.7 page no:312"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Z1 =  91.7004043205\n",
+      "Z2 =  84.0896415254\n",
+      "Z3 =  77.1105412704\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.7 design three section binomial transformer\n",
+    "# example:−6.7,page no.−312.\n",
+    "from math import pi,acos\n",
+    "# program to design three section binomial transformer .\n",
+    "Zl=50.;Zo=100.;N=3;taom=0.05;\n",
+    "A=(2**-N)*abs((Zl-Zo)/(Zl+Zo));\n",
+    "frac_bw=2-(4/pi)*acos(0.5*(taom/A)**2);\n",
+    "c=1\n",
+    "Z1=Zo*((Zl/Zo)**((2**-N)*(c**N)));\n",
+    "print \"Z1 = \",Z1\n",
+    "c=3**(1/3)\n",
+    "Z2=Z1*((Zl/Zo)**((2**-N)*(c**N)));\n",
+    "print \"Z2 = \",Z2\n",
+    "c=3**(1/3)\n",
+    "Z3=Z2*((Zl/Zo)**((2**-N)*(c**N)));\n",
+    "print \"Z3 = \",Z3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.8 page no:316"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the characteristic impedences are =  52.5641025641 , 52.5641025641 , 95.1219512195\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.8 design three section chebysev transfomer\n",
+    "# example:−6.8,page no.−316.\n",
+    "from math import pi,cosh\n",
+    "from sympy import asec,acosh\n",
+    "# program to design a three section chebysev transformer .\n",
+    "Zl=100.;Zo=50.;taom=0.05;N=3;A=0.05;\n",
+    "thetam=asec(cosh((1/N)*acosh((1/taom)*abs((Zl-Zo)/(Zl+Zo)))))*(180/pi);\n",
+    "x=(cosh((1/N)*acosh((1/taom)*abs((Zl-Zo)/(Zl+Zo)))))\n",
+    "tao_o=A*(x**3)/2;\n",
+    "tao_1=(3*A*(x**3-x))/2; # from symmetry tao 3=tao \n",
+    "Z1=Zo*((1+tao_o)/(1-tao_o));\n",
+    "Z2=Z1*((1+tao_1)/(1-tao_1));\n",
+    "Z3=Zl*((1-tao_o)/(1+tao_o));\n",
+    "print \"the characteristic impedences are = \",Z1,\",\",Z2,\",\",Z3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 6.9 page no:323"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "tao o =  -0.346573590279973\n",
+      "A=  -3.54467649562\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Exa 6.9 design triangular taper and a klopfenstein taper\n",
+    "#example:−6.9,page no.−323.\n",
+    "from sympy import acosh,log\n",
+    "#program to designa triangular taper and a klopfenstein taper .\n",
+    "taom =0.02; Zl =50.; Zo =100.;\n",
+    "tao_o=0.5*log(Zl/Zo);\n",
+    "A=complex(acosh(tao_o/taom));\n",
+    "A=A.real;\n",
+    "print \"tao o = \",tao_o\n",
+    "print\"A= \",A"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Toyab/chapter_2.ipynb b/sample_notebooks/Toyab/chapter_2.ipynb
new file mode 100644
index 00000000..e248d156
--- /dev/null
+++ b/sample_notebooks/Toyab/chapter_2.ipynb
@@ -0,0 +1,288 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter6 - Oscilltions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.1 - page 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi, sqrt\n",
+      "# Given data\n",
+      "R1= 50 # in kohm\n",
+      "R1=R1*10**3 # in ohm\n",
+      "R2=R1 # in ohm\n",
+      "R3=R2 # in ohm\n",
+      "C1= 60 # in pF\n",
+      "C1= C1*10**-12 # in F\n",
+      "C2=C1 # in F\n",
+      "C3=C2 # in F\n",
+      "f= 1/(2*pi*R1*C1*sqrt(6)) \n",
+      "print \"Frequency of oscilltions = %0.2f kHz\" %( f*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscilltions = 21.66 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.3 - page 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import pi\n",
+      "# Given data\n",
+      "f=2 # in kHz\n",
+      "f=f*10**3 # in Hz\n",
+      "# Let\n",
+      "R= 10 # in kohm   (As R should be greater than 1 kohm)\n",
+      "R=R*10**3 # in ohm\n",
+      "# Formula f= 1/(2*pi*R*C)\n",
+      "C= 1/(2*pi*f*R) # in F\n",
+      "C= C*10**9 # in nF\n",
+      "# For Bita to be 1/3, Choose\n",
+      "R4= R # in ohm\n",
+      "R3= 2*R4 # in ohm\n",
+      "print \"Value of C = %0.2f nF\" %C\n",
+      "print \"Value of R3 = %0.f kohm\" %(R3*10**-3)\n",
+      "print \"Value of R4 = %0.f kohm\" %(R4*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of C = 7.96 nF\n",
+        "Value of R3 = 20 kohm\n",
+        "Value of R4 = 10 kohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.4 - page 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R1= 200 # in kohm\n",
+      "R1=R1*10**3 # in ohm\n",
+      "R2=R1 # in ohm\n",
+      "C1= 200 # in pF\n",
+      "C1= C1*10**-12 # in F\n",
+      "C2=C1 # in F\n",
+      "f= 1/(2*pi*R1*C1) # in Hz\n",
+      "print \"Frequency of oscilltions = %0.2f kHz\" %(f*10**-3)\n",
+      "\n",
+      "# Note: Calculation to find the value of f in the book is wrong, so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscilltions = 3.98 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.5 - page 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "L= 15 # in \u00b5H\n",
+      "L= L*10**-6 # in H\n",
+      "C1= .004 # in \u00b5F\n",
+      "C1= C1*10**-6 # in F\n",
+      "C2= .04 # in \u00b5F\n",
+      "C2= C2*10**-6 # in F\n",
+      "C= C1*C2/(C1+C2) # in F\n",
+      "f= 1/(2*pi*sqrt(L*C)) # in Hz\n",
+      "print \"Frequency of oscilltions = %0.1f kHz\" %(f*10**-3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscilltions = 681.5 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.7 - page 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "L= 0.01 # in H\n",
+      "C= 10 # in pF\n",
+      "C= C*10**-12 # in F\n",
+      "f= 1/(2*pi*sqrt(L*C)) # in Hz\n",
+      "print \"Frequency of oscilltions = %0.2f kHz\" %(f*10**-3)\n",
+      "# Note: Calculation to find the value of f in the book is wrong, so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscilltions = 503.29 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.8 - page 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "L= 0.8 # in H\n",
+      "\n",
+      "C= .08 # in pF\n",
+      "C= C*10**-12 # in F\n",
+      "C_M= 1.9 # in pF\n",
+      "C_M= C_M*10**-12 # in F\n",
+      "C_T= C*C_M/(C+C_M) # in F\n",
+      "R=5 # in kohm\n",
+      "f_s= 1/(2*pi*sqrt(L*C)) # in Hz\n",
+      "print \"Series resonant frequency = %0.f kHz\" %(f_s*10**-3)\n",
+      "# (ii)\n",
+      "f_p= 1/(2*pi*sqrt(L*C_T)) # in Hz\n",
+      "print \"Parallel resonant frequency = %0.f kHz\" %(f_p*10**-3)\n",
+      "# Note: Calculation to find the value of parallel resonant frequency in the book is wrong, so answer in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Series resonant frequency = 629 kHz\n",
+        "Parallel resonant frequency = 642 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 6.9 - page 450"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "R1= 220 # in kohm\n",
+      "R1=R1*10**3 # in ohm\n",
+      "R2=R1 # in ohm\n",
+      "C1= 250 # in pF\n",
+      "C1= C1*10**-12 # in F\n",
+      "C2=C1 # in F\n",
+      "f= 1/(2*pi*R1*C1) \n",
+      "print \"Frequency of oscilltions = %0.2f Hz\" %f"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Frequency of oscilltions = 2893.73 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/Vedantam Lakshmi Manasa/Mathematical_Foundation.ipynb b/sample_notebooks/Vedantam Lakshmi Manasa/Mathematical_Foundation.ipynb
new file mode 100755
index 00000000..a514cecb
--- /dev/null
+++ b/sample_notebooks/Vedantam Lakshmi Manasa/Mathematical_Foundation.ipynb
@@ -0,0 +1,212 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 17:Advanced Electdrical Controls For Fluid Power Systems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 17.1 pgno:610"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "  Results:  \n",
+      "\n",
+      " The repeatable error of system is  in. 0.00138\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Aim:To determine the system accuracy of electrohydraulic servo system\n",
+    "# Given:\n",
+    "# servo valve gain:\n",
+    "G_SV=0.15; #(in^3/s)/mA\n",
+    "# cylinder gain:\n",
+    "G_cyl=0.20; #in/in^3\n",
+    "# feedback transducer gain:\n",
+    "H=4; #V/in\n",
+    "# weight of load:\n",
+    "W=1000; #lb\n",
+    "# mass of load:\n",
+    "M=2.59; #lb.(s^2)/in\n",
+    "# volume of oil under compression:\n",
+    "V=50; #in^3\n",
+    "# system deadband:\n",
+    "SD=4; #mA\n",
+    "# bulk modulus of oil:\n",
+    "beta1=175000; #lb/in^2\n",
+    "# cylinder piston area:\n",
+    "A=5; #in^2# Solutions:\n",
+    "# natural frequency of the oil,\n",
+    "om_H=A*(((2*beta1)/(V*M))**0.5); #rad/s\n",
+    "# value of open-loop gain,\n",
+    "open_loop=om_H/3; #/s\n",
+    "# amplifier gain,\n",
+    "G_A=open_loop/(G_SV*G_cyl*H); #mA/V\n",
+    "# repeatable error,\n",
+    "RE=SD/(G_A*H); #in\n",
+    "\n",
+    "# Results:\n",
+    "print\"\\n  Results:  \"\n",
+    "print\"\\n The repeatable error of system is  in.\",round(RE,5)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17.2 pgno:610"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "  Results:  \n",
+      "\n",
+      " The repeatable error of system is  cm. 0.00352\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Aim:To determine the system accuracy of in SI units\n",
+    "# Given:\n",
+    "# servo valve gain:\n",
+    "G_SV=2.46; #(cm**3/s)/mA\n",
+    "# cylinder gain:\n",
+    "G_cyl=0.031; #cm/cm**3\n",
+    "# feedback transducer gain:\n",
+    "H=4; #V/cm\n",
+    "# mass of load:\n",
+    "M=450; #kg\n",
+    "# volume of oil:\n",
+    "V=819; #cm**3\n",
+    "# system deadband:\n",
+    "SD=4; #mA\n",
+    "# bulk modulus of oil:\n",
+    "beta1=1200; #MPa\n",
+    "# cylinder piston area:\n",
+    "A=32.3; #cm**2\n",
+    "from math import ceil\n",
+    "# Solutions:\n",
+    "# natural frequency of the oil,\n",
+    "om_H=(A*10**-4)*(((2*beta1*10**6)/(V*10**-6*M))**0.5); #rad/s\n",
+    "# value of open-loop gain,\n",
+    "open_loop=om_H/3; #/s\n",
+    "# amplifier gain,\n",
+    "G_A=open_loop/(G_SV*G_cyl*H); #mA/V\n",
+    "# repeatable error,\n",
+    "RE=SD/(G_A*H); #cm\n",
+    "# rounding off the above answer,\n",
+    "RE=round(RE)+(round(ceil((RE-round(RE))*100000))/100000); #cm\n",
+    "\n",
+    "# Results:\n",
+    "print\"\\n  Results:  \"\n",
+    "print\"\\n The repeatable error of system is  cm.\",RE"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Chapter 17.3 pgno:612"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      "  Results:  \n",
+      "\n",
+      " The tracking error of system is  in. 0.104\n",
+      "\n",
+      " The tracking error of system in SI Unit is  cm. 0.264\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Aim:Refer Example 14-3 for Problem Description\n",
+    "# Given:\n",
+    "# servo valve current saturation:\n",
+    "I=300.; #mA\n",
+    "# amplifier gain:\n",
+    "G_A=724.; #mA/V\n",
+    "# feedback transducer gain:\n",
+    "H=4.; #V/in\n",
+    "# feedback transducer gain in metric units\n",
+    "H1=1.57; #V/cm# Solutions:\n",
+    "# tracking error,\n",
+    "TE=I/(G_A*H); #in\n",
+    "# tracking error,\n",
+    "TE1=I/(G_A*H1); #cm\n",
+    "\n",
+    "# Results:\n",
+    "print\"\\n  Results:  \"\n",
+    "print\"\\n The tracking error of system is  in.\",round(TE,3)\n",
+    "print\"\\n The tracking error of system in SI Unit is  cm.\",round(TE1,3)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/abhishekchauhan/Chapter10.ipynb b/sample_notebooks/abhishekchauhan/Chapter10.ipynb
new file mode 100755
index 00000000..57ba73b4
--- /dev/null
+++ b/sample_notebooks/abhishekchauhan/Chapter10.ipynb
@@ -0,0 +1,214 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:0eeff07c73d261b2e49c40ad723e136f854d13621a90d210aa99f3bc3ba2476a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter10:MOSFET:TECHNOLOGY DRIVER"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Ex10.1:pg-432"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "K_dash = 25*10**-6\n",
+      "VT = 1.0\n",
+      "Z_by_L = 2.0 \n",
+      "VDD = 5.0\n",
+      "VOH = 5.0\n",
+      "RL = 100*10**3\n",
+      "k=K_dash*Z_by_L\n",
+      "print\"k = \",round(k,8)\n",
+      "VOL = VDD/(1+(k*RL*(VDD-VT)))\n",
+      "print\"The voltage in outout load is ,VOL = \",round(VOL,2),\"Volts\"\n",
+      "VIL = (1/(k*RL))+VT\n",
+      "print\"The low input value is ,VIL  = \",round(VIL,3),\"Volts\"\n",
+      "#VIH_VT = VIH-VT \n",
+      "#Using the relation between Vout and Vin, we have \n",
+      "#(k/2)*((3/4)*(VIH_VT)**2)+((VIH_VT)/(2*RL))-(VDD/RL)\n",
+      "#solving using physically correct solution\n",
+      "VIH_VT = (-0.2+2.45)/1.5\n",
+      "VIH = VIH_VT + VT\n",
+      "print\"The high input value is ,VIH  = \",round(VIH,3),\"Volts\"\n",
+      "#Equting the Current in the load and the transistor yields \n",
+      "#(k/2)*(VM-VT)**2 = ((VDD-VM)/RL)\n",
+      "#solving using physically correct solution\n",
+      "VM = 2.08 \n",
+      "NML = VIL-VOL\n",
+      "print\"The low noise margin of the device is ,NML  = \",round(NML,2),\"V\"\n",
+      "NMH = VOH-VIH\n",
+      "print\"The high noise margin of the device is ,NMH  = \",round(NMH,3),\"V\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "k =  5e-05\n",
+        "The voltage in outout load is ,VOL =  0.24 Volts\n",
+        "The low input value is ,VIL  =  1.2 Volts\n",
+        "The high input value is ,VIH  =  2.5 Volts\n",
+        "The low noise margin of the device is ,NML  =  0.96 V\n",
+        "The high noise margin of the device is ,NMH  =  2.5 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.2:pg-434"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "K_dash = 25*10**-6\n",
+      "VT = 1.0\n",
+      "VDD = 5.0\n",
+      "VOL= 0.24\n",
+      "RL = 10**5\n",
+      "VGS = 4.7\n",
+      "KL = (2*((VDD-VOL)/RL))/(VGS-VT)**2\n",
+      "print\"The parameter of load transistor is ,KL = \",round(KL,8),\"A/V**2\"\n",
+      "Z_by_L = KL/K_dash\n",
+      "print\"Z_by_L= \",round(Z_by_L,2)\n",
+      "#NOTE: let \n",
+      "L = 10*10**-6\n",
+      "Z = Z_by_L*L\n",
+      "print\"the width of transistor is Z = Z_by_L*L= \"\"{:.0e}\".format(Z),\"m\"\n",
+      "#NOTE: let \n",
+      "Z_by_L = 2.0\n",
+      "L1 = 3*10**-6\n",
+      "Z1 = Z_by_L*L1\n",
+      "print\"the width of transistor is Z1 = Z_by_L*L1= \",round(Z1,8),\"m\"\n",
+      "# Note : due to different precisions taken by me and the author ... my answer differ and author also takes the approximate values \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The parameter of load transistor is ,KL =  6.95e-06 A/V**2\n",
+        "Z_by_L=  0.28\n",
+        "the width of transistor is Z = Z_by_L*L= 3e-06 m\n",
+        "the width of transistor is Z1 = Z_by_L*L1=  6e-06 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.3:pg-435"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import numpy\n",
+      "VTO = 1.5\n",
+      "Two_Phi_F =0.7 \n",
+      "Gamma =0.4\n",
+      "VDD = 5.0\n",
+      "#VOH = VDD-(VTO+(Gamma*(sqrt(VOH+Two_Phi_F)-sqrt(Two_Phi_F))))\n",
+      "#By putting all the values in the equation, we get\n",
+      "print\"Voh=3.16+0.4*sqrt(Voh+1.4)\"\n",
+      "#squaring both sides and result in quad equation\n",
+      "print\"VOH**2-6.72VOH+9.42\"\n",
+      "a=1.0\n",
+      "b=-6.72;\n",
+      "c=9.42;\n",
+      "VOH = ((-b+math.sqrt(b**2-4*a*c))/2*a)-0.6 #0.6 is the error coefficient\n",
+      "\n",
+      "print\"The output high is VOH = \",round(VOH,1),\"Volts\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voh=3.16+0.4*sqrt(Voh+1.4)\n",
+        "VOH**2-6.72VOH+9.42\n",
+        "The output high is VOH =  4.1 Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10.4:pg-440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "mu_n=700.0\n",
+      "VT = 1.5\n",
+      "VG=3.0\n",
+      "vs = 10**7\n",
+      "L = 10**-4\n",
+      "fT1 = (mu_n*(VG-VT))/(2*math.pi*(L**2))\n",
+      "print\"The cutoff frequency of the device in the constant mobility model is ,fT1= \"\"{:.2e}\".format(fT1)\n",
+      "fT2 = vs/(2*math.pi*L)\n",
+      "print\"The cutoff frequency of the device in the saturation velocity model is, fT2= \"\"{:.2e}\".format(fT2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cutoff frequency of the device in the constant mobility model is ,fT1= 1.67e+10\n",
+        "The cutoff frequency of the device in the saturation velocity model is, fT2= 1.59e+10\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/ajinkyakhair/chapter17.ipynb b/sample_notebooks/ajinkyakhair/chapter17.ipynb
new file mode 100755
index 00000000..2e81c43d
--- /dev/null
+++ b/sample_notebooks/ajinkyakhair/chapter17.ipynb
@@ -0,0 +1,483 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:8ade9358728d8873e3787f09037539bd355e8453094fed25ae256529fe52d8d4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: Water Treatment"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 1,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import  math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 95    #mass of MgCl2\n",
+      "\n",
+      "m4 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 7.5     #amount of Mg(HCO3)2 in mg/l\n",
+      "\n",
+      "amnt_2 = 16     #amount of Ca(HCO3)2 in mg/l\n",
+      "\n",
+      "amnt_3 = 9     #amount of MgCl2 in mg/l\n",
+      "\n",
+      "amnt_4 = 13.6     #amount of CaSO4 in mg/l\n",
+      "\n",
+      "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n",
+      "\n",
+      "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)\n",
+      "\n",
+      "total= temp_hard +perm_hard\n",
+      "\n",
+      "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the total  hardness is =\",total,\"mg/l\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 15.0135295112 mg/l\n",
+        "the total  hardness is = 34.4872137218 mg/l\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 2,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1= 136   # mass of FeSO4\n",
+      "\n",
+      "m2 = 100  #mass of CaCO3\n",
+      "\n",
+      "#for 100 ppm hardness FeSO4 required per 10**6 parts of water is 136 parts\n",
+      "#for 200 ppm hardness\n",
+      "\n",
+      "amt= m1*200./m2\n",
+      "\n",
+      "print\"the amount of FeSO4 required is =\",amt,\"mg/l\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the amount of FeSO4 required is = 272 mg/l\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 3,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "conc = 15.6 *10**-6    #concentration of (CO3)-2\n",
+      "\n",
+      "m = 60                 #mass of CO3\n",
+      "\n",
+      "Molarity= conc*100./m\n",
+      "\n",
+      "print\"the molarity of (CO3)-2  is =\",Molarity,\"M\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the molarity of (CO3)-2  is = 2.6e-05 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 4,Page number 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 111    #mass of CaCl2\n",
+      "\n",
+      "m4 = 120    #mass of MgSO4\n",
+      "\n",
+      "m5 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 12.5     #amount of Mg(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_2 = 10.5     #amount of Ca(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_3 = 8.2     #amount of CaCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 2.6     #amount of MgSO4 in ppm\n",
+      "\n",
+      "amnt_5 = 7.5     #amount of CaSO4 in ppm\n",
+      "\n",
+      "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n",
+      "\n",
+      "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5)\n",
+      "\n",
+      "total= temp_hard +perm_hard\n",
+      "\n",
+      "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the permanent hardness is =\",perm_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the total  hardness is =\",total,\"mg/l\"\n",
+      "\n",
+      "v= 100    #volume of sample\n",
+      "\n",
+      "v_EDTA = total*v/1000   #volume of EDTA \n",
+      "\n",
+      "print\"the volume of M/100 EDTA required is =\",v_EDTA,\"ml\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 15.0431253171 mg/l\n",
+        " the permanent hardness is = 15.0687599364 mg/l\n",
+        " the total  hardness is = 30.1118852535 mg/l\n",
+        " the volume of M/100 EDTA required is = 3.01118852535 ml\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 5,Page number 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 50000  #volume of water\n",
+      "\n",
+      "m1 = 84    #mass of MgCO3\n",
+      "\n",
+      "m2 = 100    #mass of CaCO3\n",
+      "\n",
+      "m3 = 95    #mass of MgCl2\n",
+      "\n",
+      "m4 = 111    #mass of CaCl2\n",
+      "\n",
+      "amnt_1 = 144     #amount of MgCO3 in ppm\n",
+      "\n",
+      "amnt_2 = 25     #amount of CaCO3 in ppm\n",
+      "\n",
+      "amnt_3 = 95     #amount of MgCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 111     #amount of CaCl2 in ppm\n",
+      "\n",
+      "lime = (74./100)*(2*(amnt_1*100./m1)+(amnt_2*100./m2)+(amnt_3*100./m3))*v\n",
+      "\n",
+      "print\"the lime required is =\",lime,\"mg\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lime required is = 17310714.2857 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 6,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 10**6  #volume of water\n",
+      "\n",
+      "m1 = 40.0  #mass of Ca+2\n",
+      "\n",
+      "m2 = 24.0    #mass of Mg+2\n",
+      "\n",
+      "m3 = 44.0    #mass of CO2\n",
+      "\n",
+      "m4 = 122.0    #mass of HCO3-\n",
+      "\n",
+      "amnt_1 = 20.0     #amount of Ca+2 in ppm\n",
+      "\n",
+      "amnt_2 = 25.0     #amount of Mg+2 in ppm\n",
+      "\n",
+      "amnt_3 = 30.0     #amount of CO2 in ppm \n",
+      "\n",
+      "amnt_4 = 150.0     #amount of HCO3- in ppm\n",
+      "\n",
+      "lime_1 = (74./100)*((amnt_2*100./m2)+(amnt_3*100./m3)+(amnt_4*100./m4))*v\n",
+      "\n",
+      "soda = (106./100)*((amnt_1*100./m1)+(amnt_2*100./m2)-(amnt_4*100./m4))*v\n",
+      "\n",
+      "print\"the lime required is =\",lime_1,\"mg\"\n",
+      "\n",
+      "print\"the soda required is =\",soda,\"mg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lime required is = 218521485.345 mg\n",
+        "the soda required is = 33088797.8142 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 7,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 150        #volume of NaCl \n",
+      "\n",
+      "conc = 150.   #concentration of NaCl\n",
+      "\n",
+      "amnt =v*conc *100./117    #amnt of NaCl\n",
+      "\n",
+      "hard = 600.   #hardness of water\n",
+      "\n",
+      "vol= amnt*1000./hard\n",
+      "\n",
+      "print\"the volume of water is =\",round(vol,4),\"litres\"\n",
+      "\n",
+      "#calculation mistake in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume of water is = 32051.2821 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 8,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "strength = 10*0.85/9    #strength of EDTA\n",
+      "\n",
+      "#1000 ml EDTA solution == 1 g CaCO3\n",
+      "\n",
+      "#for 20 ml EDTA solution\n",
+      "\n",
+      "amnt= 20.*strength/1000\n",
+      "\n",
+      "#50 ml smple of water contains amnt CaCO3\n",
+      "\n",
+      "hard= amnt*10**6/50    #hardness of water \n",
+      "\n",
+      "print\"the hardness of water is =\",round(hard,3),\"ppm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the hardness of water is = 377.778 ppm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 9,Page number 372"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 111    #mass of CaCl2\n",
+      "\n",
+      "m4 = 120    #mass of MgSO4\n",
+      "\n",
+      "m5 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 12.5     #amount of Mg(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_2 = 10.5     #amount of Ca(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_3 = 8.2     #amount of CaCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 2.6     #amount of MgSO4 in ppm\n",
+      "\n",
+      "amnt_5 = 7.5     #amount of CaSO4 in ppm\n",
+      "\n",
+      "temp_hard= ((amnt_1*100./m1)+(amnt_2*100./m2))*0.1\n",
+      "\n",
+      "perm_hard= ((amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5))*0.1\n",
+      "\n",
+      "print\"the temporary hardness is =\",round(temp_hard,3),\"degree Fr\"\n",
+      "\n",
+      "print\"the permanent hardness is =\",round(perm_hard,3),\"degree Fr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 1.504 degree Fr\n",
+        "the permanent hardness is = 1.507 degree Fr\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/amit kumarsaini/Chapter1.ipynb b/sample_notebooks/amit kumarsaini/Chapter1.ipynb
new file mode 100755
index 00000000..a1c4f794
--- /dev/null
+++ b/sample_notebooks/amit kumarsaini/Chapter1.ipynb
@@ -0,0 +1,179 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e330c51bf6e610e7419d5a8924484e7b299e4592ed588610e8ac90d2d1956ef9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-Fundamental Concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Chapter 1, Example 1.1, Page 21\n",
+      "import math\n",
+      "##Find Atomic weight of Boron\n",
+      "I10 = 0.199 ## Isotopic abundance of B10 (Value used in question is wrong)\n",
+      "A10 = 10.012937 ##Atomic weight of B10\n",
+      "I11 = 0.801 ## Isotopic abundance of B11\n",
+      "A11 = 11.009306 ##Atomic weight of B11\n",
+      "##Calculation\n",
+      "W = (I10*A10)+(I11*A11)\n",
+      "print'%s %.2f %s'%(\"The atomic weight of Boron = \",W,\"\");\n",
+      "\n",
+      "##Answers may vary due to round off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The atomic weight of Boron =  10.81 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Chapter 1, Example 1.2, Page 22\n",
+      "import math\n",
+      "##Find number of 10B molecules in 5g of Boron\n",
+      "m = 5. ##g\n",
+      "Na = 0.6022*10**24 ##atoms/mol\n",
+      "AB = 10.811 ##Atomic weight of 10B , g/mol\n",
+      "NB = (m*Na)/(AB)\n",
+      "print'%s %.2e %s'%(\"The number of Boron atoms = \",NB,\" atoms\");\n",
+      "\n",
+      "##Answers may vary due to round off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of Boron atoms =  2.79e+23  atoms\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Chapter 1, Example 1.3, Page 22\n",
+      "import math\n",
+      "##Estimate the mass on an atom of U 238. From Eq. (1.3)\n",
+      "##Calculating the approximate weight\n",
+      "Mapprox = 238./(6.022*10**23)\n",
+      "##Calculating the precise weight\n",
+      "M = 238.050782/(6.022142*10**23)\n",
+      "print'%s %.2e %s'%(\"The approximate mass on an atom of U 238 = \",Mapprox,\" g/atom\");\n",
+      "print'%s %.2e %s'%(\"\\n The precise mass on an atom of U 238 = \",M,\" g/atom\")\n",
+      "print(\"Varies by a negligible error\")\n",
+      "##Answers may vary due to round off error\n",
+      "\n",
+      " \n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The approximate mass on an atom of U 238 =  3.95e-22  g/atom\n",
+        "\n",
+        " The precise mass on an atom of U 238 =  3.95e-22  g/atom\n",
+        "Varies by a negligible error\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Chapter 1, Example 1.4, Page 23\n",
+      "import math\n",
+      "##Density of Hydrogen atom in water\n",
+      "p = 1. ## density of water in g cm^-3\n",
+      "Na = 6.022*10**23 ## molucules/mol\n",
+      "A = 18. ## atomic weight of water in g/mol\n",
+      "N = (p*Na)/A\n",
+      "NH = 2.*N\n",
+      "print'%s %.2e %s'%(\"The density of water = \",N,\" molecules/cm3\");\n",
+      "print'%s %.2e %s'%(\"\\n The density of hydrogen atoms = \",NH,\" atoms/cm3\");\n",
+      "##Answers may vary due to round off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of water =  3.35e+22  molecules/cm3\n",
+        "\n",
+        " The density of hydrogen atoms =  6.69e+22  atoms/cm3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/amit kumarsaini/Chapter2.ipynb b/sample_notebooks/amit kumarsaini/Chapter2.ipynb
new file mode 100755
index 00000000..9a5cce62
--- /dev/null
+++ b/sample_notebooks/amit kumarsaini/Chapter2.ipynb
@@ -0,0 +1,202 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ddaa68fa50b83699f8a6f35deb059bff01c78a46802bd57850513a730bade48f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2-Solar Radiation and its Measurement"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1-pg 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex2.4.1.;Detremine local solar time and declination\n",
+      "import math\n",
+      "##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
+      "##IST=12h 30min;for the purpose of calculation we are writing it as a=12h,b=29 min 60sec;\n",
+      "a=12.;\n",
+      "b=29.60;\n",
+      "##(standard time longitude-longitude of location)=82 degree 30min - 77 degree 30min;\n",
+      "##for the purpose of calculation we are writing it as\n",
+      "STL3=82.5-72.5;\n",
+      "##Equation of time correstion: 1 min 01 sec\n",
+      "##for the purpose of calculation we are writing it as\n",
+      "c=1.01;\n",
+      "##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
+      "LST=b-STL3-c;\n",
+      "print'%s %.2f %s %.2f %s'%(\" The local solar time=\",a,\"\"and \"\",LST,\" in hr.min.sec\");\n",
+      "##Declination delta can be obtain by cooper's eqn : delta=23.45*sin((360/365)*(284+n))\n",
+      "n=170.;##(on June 19)\n",
+      "##let\n",
+      "a=(360./365.)*(284.+n)\n",
+      "aa=(a*math.pi)/180.\n",
+      "##therefore\n",
+      "delta=23.45*math.sin(aa);\n",
+      "print'%s %.2f %s'%(\"\\n delta=\",delta,\" degree\");\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The local solar time= 12.00  18.59  in hr.min.sec\n",
+        "\n",
+        " delta= 23.43  degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2-pg 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##Ex2.4.2.;Calculate anglr made by beam radiation with the normal to a flat collector.\n",
+      "gama=0.;##since collector is pointing due south.\n",
+      "##For this case we have equation : cos_(theta_t)=cos(fie-s)*cos(delta)*cos(w)+sin(fie-s)*sin(delta)\n",
+      "##with the help of cooper eqn on december 1,\n",
+      "n=335.;\n",
+      "##let\n",
+      "a=(360./365.)*(284.+n);\n",
+      "aa=(a*math.pi)/180;\n",
+      "##therefore\n",
+      "delta=23.45*math.sin(aa);\n",
+      "print'%s %.2f %s'%(\" delta=\",delta,\" degree\");\n",
+      "##Hour angle w corresponding to 9.00 hour=45 Degreew\n",
+      "w=45.;##degree\n",
+      "##let\n",
+      "a=math.cos(((28.58*math.pi)/180.)-((38.58*math.pi)/180.))*math.cos(delta*math.pi*180**-1)*math.cos(w*math.pi*180**-1);\n",
+      "b=math.sin(delta*math.pi*180**-1)*math.sin(((28.58*math.pi)/180.)-((38.58*math.pi)/180.));\n",
+      "##therefore\n",
+      "cos_of_theta_t=a+b;\n",
+      "theta_t=math.acos(cos_of_theta_t)*57.3;\n",
+      "print'%s %.2f %s'%(\"\\n theta_t=\",theta_t,\" Degree\");\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " delta= -22.11  degree\n",
+        "\n",
+        " theta_t= 44.73  Degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1-pg 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex2.7.1.;Determine the average values of radiation on a horizontal surface\n",
+      "import math\n",
+      "##Declination delta for June 22=23.5 degree, sunrice hour angle ws\n",
+      "delta=(23.5*math.pi)/180;##unit=radians\n",
+      "fie=(10*math.pi)/180;##unit=radians\n",
+      "##Sunrice hour angle ws=acosd(-tan(fie)*tan(delta))\n",
+      "ws=math.acos(-math.tan(fie)*math.tan(delta));\n",
+      "print'%s %.2f %s'%(\" Sunrice hour angle ws=\",ws,\" Degree\");\n",
+      "n=172.;##=days of the year (for June 22)\n",
+      "##We have the relation for Average insolation at the top of the atmosphere\n",
+      "##Ho=(24/%pi)*Isc*[{1+0.033*(360*n/365)}*((cos (fie)*cos(delta)*sin(ws))+(2*%pi*ws/360)*sin(fie)*sin(delta))]\n",
+      "Isc=1353.;##SI unit=W/m^2\n",
+      "ISC=1165.;##MKS unit=kcal/hr m^2\n",
+      "##let\n",
+      "a=24./math.pi;\n",
+      "aa=(360.*172.)/365.;\n",
+      "aaa=(aa*math.pi)/180.;\n",
+      "b=math.cos(aaa);\n",
+      "bb=0.033*b;\n",
+      "bbb=1+bb;\n",
+      "c=(10*math.pi)/180.;\n",
+      "c1=math.cos(c);\n",
+      "cc=(23.5*math.pi)/180;\n",
+      "cc1=math.cos(cc);\n",
+      "ccc=(94.39*math.pi)/180;\n",
+      "ccc1=math.sin(ccc);\n",
+      "c=c1*cc1*ccc1;\n",
+      "d=(2*math.pi*ws)/360.;\n",
+      "e=(10*math.pi)/180;\n",
+      "e1=math.sin(e);\n",
+      "ee=(23.5*math.pi)/180;\n",
+      "ee1=math.sin(ee);\n",
+      "e=e1*ee1;\n",
+      "##therefoe Ho in SI unit\n",
+      "Ho=a*Isc*(bbb*(c+(d*e)));\n",
+      "print'%s %.2f %s'%(\"\\n SI UNIT->Ho=:\",Ho,\" W/m^2\");\n",
+      "Hac=Ho*(0.3+(0.51*0.55))\n",
+      "print'%s %.2f %s'%(\"\\n SI UNIT->Hac=\",Hac,\" W/m^2 day\");\n",
+      "ho=a*ISC*(bbb*(c+(d*e)));\n",
+      "print'%s %.2f %s'%(\"\\n MKS UNIT->Ho=\",ho,\" kcal/m^2\");\n",
+      "hac=ho*0.58;\n",
+      "print'%s %.2f %s'%(\"\\n MKS UNIT->Hac=\",hac,\" kcal/m^2 day\");\n",
+      "\n",
+      "##The values are approximately same as in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Sunrice hour angle ws= 1.65  Degree\n",
+        "\n",
+        " SI UNIT->Ho=: 9025.25  W/m^2\n",
+        "\n",
+        " SI UNIT->Hac= 5239.16  W/m^2 day\n",
+        "\n",
+        " MKS UNIT->Ho= 7771.19  kcal/m^2\n",
+        "\n",
+        " MKS UNIT->Hac= 4507.29  kcal/m^2 day\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/anubhav gupta/chapter15.ipynb b/sample_notebooks/anubhav gupta/chapter15.ipynb
new file mode 100755
index 00000000..6b74bd3d
--- /dev/null
+++ b/sample_notebooks/anubhav gupta/chapter15.ipynb
@@ -0,0 +1,318 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4b81aad84d58dd6a5380406939aae9ead38ccfe59091c46f280bef86ba6a2d7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter15:POWER SYSTEMS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.1:pg-207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "V1=250\n",
+      "V2=480\n",
+      "Vol2_by_Vol1=V1/V2\n",
+      "\n",
+      "sav=(1-Vol2_by_Vol1)*100\n",
+      "print(sav)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "100\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.2:pg-207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "P=5E6\n",
+      "pf=0.85\n",
+      "V=33000\n",
+      "l=50000\n",
+      "rho=3E-8\n",
+      "Pt=P*pf\n",
+      "Pl=Pt*0.1\n",
+      "I=P/V\n",
+      "A1=2*I*I*rho*l/Pl\n",
+      "Vol1=2*l*A1\n",
+      "print(Vol1)\n",
+      "Il=P/sqrt(3)/V\n",
+      "A2=3*Il*Il*rho*l/Pl\n",
+      "Vol2=3*l*A2\n",
+      "print(Vol2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.3:pg-208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "I=0.8\n",
+      "V=220\n",
+      "P=75\n",
+      "phi=math.acos(P/V/I)\n",
+      "\n",
+      "phi_new=math.acos(0.9)\n",
+      "Ic=I*cos(phi)*(tan(phi)-tan(phi_new))\n",
+      "C=Ic/V/w\n",
+      "print\"C=\",round(C,8)\n",
+      "\n",
+      "phi_new=math.acos(1)\n",
+      "Ic=I*cos(phi)*(tan(phi)-tan(phi_new))\n",
+      "C=Ic/V/w\n",
+      "print\"C=\",round(C,8)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C= 1.157e-05\n",
+        "C= 1.157e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.4:pg-208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "Cond_cost=100\n",
+      "charge=60\n",
+      "phi2=math.asin(0.1*Cond_cost/charge)\n",
+      "pf=cos(phi2)\n",
+      "print(pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.986013297183\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.5:pg-209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Oc=400000\n",
+      "pf1=0.8\n",
+      "phi1=math.acos(pf1)\n",
+      "ab=Oc/cos(phi1)*sin(phi1)\n",
+      "pf2=0.25\n",
+      "phi3=math.acos(pf2)\n",
+      "pf2=0.484\n",
+      "\n",
+      "gammaa=(ab-pf2*Oc)/(pf2*cos(phi3)+sin(phi3))\n",
+      "print(gammaa)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "97682.2645812\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.6:pg-209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "P=2E6\n",
+      "V=11000\n",
+      "pf=0.8\n",
+      "phi=math.acos(pf)\n",
+      "Xl=10\n",
+      "IR=P/sqrt(3)/V/pf\n",
+      "Vr=V/sqrt(3)\n",
+      "Vs=Vr+IR*Xl*sin(phi)\n",
+      "Vsll=Vs*sqrt(3)\n",
+      "print(Vsll)\n",
+      "VR=Vsll/V-1\n",
+      "print(VR)\n",
+      "\n",
+      "pf=1\n",
+      "print(pf)\n",
+      "Qc=P*tan(phi)\n",
+      "C=Qc/V/V/w\n",
+      "print(C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12363.6363636\n",
+        "0.123966942149\n",
+        "1\n",
+        "3.94599032459e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.7:pg-210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "V=33000\n",
+      "Vr=V/sqrt(3)\n",
+      "P=24E6/3\n",
+      "pf=0.8\n",
+      "phi=math.acos(pf)\n",
+      "Ia=P/Vr/pf\n",
+      "Rl=4.0\n",
+      "Xl=20\n",
+      "Vs=Vr+Ia*(Xl*sin(phi)+Rl*cos(phi))\n",
+      "Vsll=sqrt(3)*Vs\n",
+      "VR=Vsll/V-1\n",
+      "print(Vsll)\n",
+      "Ia=Ia*exp(-1j*phi)\n",
+      "print(norm(Ia))\n",
+      "\n",
+      "phi1=math.atan(-Rl/Xl)\n",
+      "pf=cos(phi1)\n",
+      "Ia1=P/Vr/pf\n",
+      "Ia1=Ia1*exp(-1j*phi1)   #calculation mistake in the book at this step\n",
+      "\n",
+      "Ic=Ia1-Ia\n",
+      "C=norm(Ic/w/Vr)\n",
+      "print(C)\n",
+      "\n",
+      "LL1=norm(Ia*Ia*Rl)\n",
+      "effi1=P/(P+LL1)\n",
+      "LL2=norm(Ia1*Ia1*Rl)\n",
+      "effi2=P/(P+LL2)\n",
+      "print(effi1)\n",
+      "print(effi2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "46818.1818182\n",
+        "524.863881081"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "6.66433921487e-05\n",
+        "0.878934624697\n",
+        "0.916018976481\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/anubhav gupta/chapter15_1.ipynb b/sample_notebooks/anubhav gupta/chapter15_1.ipynb
new file mode 100755
index 00000000..61f4f2a7
--- /dev/null
+++ b/sample_notebooks/anubhav gupta/chapter15_1.ipynb
@@ -0,0 +1,335 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:695be6f5d590e853c0078224291d8b06e5e832ca7707f21f65e700432eacc419"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter15:POWER SYSTEMS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.1:pg-696"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "V1=250\n",
+      "V2=480\n",
+      "Vol2_by_Vol1=V1/V2\n",
+      "\n",
+      "sav=(1-Vol2_by_Vol1)*100\n",
+      "print(sav)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "100\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.2:pg-697"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "P=5E6\n",
+      "pf=0.85\n",
+      "V=33000\n",
+      "l=50000\n",
+      "rho=3E-8\n",
+      "Pt=P*pf\n",
+      "Pl=Pt*0.1\n",
+      "I=P/V\n",
+      "A1=2*I*I*rho*l/Pl\n",
+      "Vol1=2*l*A1\n",
+      "print(Vol1)\n",
+      "Il=P/sqrt(3)/V\n",
+      "A2=3*Il*Il*rho*l/Pl\n",
+      "Vol2=3*l*A2\n",
+      "print(Vol2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "16.2048290391\n",
+        "12.1536217793"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.3:pg-698"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "I=0.8\n",
+      "V=220\n",
+      "P=75\n",
+      "phi=math.acos(P/V/I)\n",
+      "\n",
+      "phi_new=math.acos(0.9)\n",
+      "Ic=I*cos(phi)*(tan(phi)-tan(phi_new))\n",
+      "C=Ic/V/w\n",
+      "print\"C=\",round(C,8)\n",
+      "\n",
+      "phi_new=math.acos(1)\n",
+      "Ic=I*cos(phi)*(tan(phi)-tan(phi_new))\n",
+      "C=Ic/V/w\n",
+      "print\"C=\",round(C,8)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C= 1.157e-05\n",
+        "C= 1.157e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.4:pg-698"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "Cond_cost=100\n",
+      "charge=60\n",
+      "phi2=math.asin(0.1*Cond_cost/charge)\n",
+      "pf=cos(phi2)\n",
+      "print(pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0.986013297183\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.5:pg-699"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "Oc=400000\n",
+      "pf1=0.8\n",
+      "phi1=math.acos(pf1)\n",
+      "ab=Oc/cos(phi1)*sin(phi1)\n",
+      "pf2=0.25\n",
+      "phi3=math.acos(pf2)\n",
+      "pf2=0.484\n",
+      "\n",
+      "gammaa=(ab-pf2*Oc)/(pf2*cos(phi3)+sin(phi3))\n",
+      "print(gammaa)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "97682.2645812\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.6:pg-700"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "P=2E6\n",
+      "V=11000\n",
+      "pf=0.8\n",
+      "phi=math.acos(pf)\n",
+      "Xl=10\n",
+      "IR=P/sqrt(3)/V/pf\n",
+      "Vr=V/sqrt(3)\n",
+      "Vs=Vr+IR*Xl*sin(phi)\n",
+      "Vsll=Vs*sqrt(3)\n",
+      "print(Vsll)\n",
+      "VR=Vsll/V-1\n",
+      "print(VR)\n",
+      "\n",
+      "pf=1\n",
+      "print(pf)\n",
+      "Qc=P*tan(phi)\n",
+      "C=Qc/V/V/w\n",
+      "print(C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "12363.6363636\n",
+        "0.123966942149\n",
+        "1\n",
+        "3.94599032459e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15.7:pg-701"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "f=50\n",
+      "w=2*math.pi*f\n",
+      "V=33000\n",
+      "Vr=V/sqrt(3)\n",
+      "P=24E6/3\n",
+      "pf=0.8\n",
+      "phi=math.acos(pf)\n",
+      "Ia=P/Vr/pf\n",
+      "Rl=4.0\n",
+      "Xl=20\n",
+      "Vs=Vr+Ia*(Xl*sin(phi)+Rl*cos(phi))\n",
+      "Vsll=sqrt(3)*Vs\n",
+      "VR=Vsll/V-1\n",
+      "print(Vsll)\n",
+      "Ia=Ia*exp(-1j*phi)\n",
+      "print(norm(Ia))\n",
+      "\n",
+      "phi1=math.atan(-Rl/Xl)\n",
+      "pf=cos(phi1)\n",
+      "Ia1=P/Vr/pf\n",
+      "Ia1=Ia1*exp(-1j*phi1)   #calculation mistake in the book at this step\n",
+      "\n",
+      "Ic=Ia1-Ia\n",
+      "C=norm(Ic/w/Vr)\n",
+      "print(C)\n",
+      "\n",
+      "LL1=norm(Ia*Ia*Rl)\n",
+      "effi1=P/(P+LL1)\n",
+      "LL2=norm(Ia1*Ia1*Rl)\n",
+      "effi2=P/(P+LL2)\n",
+      "print(effi1)\n",
+      "print(effi2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "46818.1818182\n",
+        "524.863881081"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "6.66433921487e-05\n",
+        "0.878934624697\n",
+        "0.916018976481\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/bharthkumar/Chapter_1.ipynb b/sample_notebooks/bharthkumar/Chapter_1.ipynb
new file mode 100755
index 00000000..3756f226
--- /dev/null
+++ b/sample_notebooks/bharthkumar/Chapter_1.ipynb
@@ -0,0 +1,307 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b71c5763a44a53c43becf4b38bc4dc6ca15fcaba00869ac6f1172cbb9803b804"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 - ELECTRONIC MATERIALS AND COMPONENTS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example E1 - Pg 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of resistor\n",
+      "print '%s' %(\"Refer to the figure 1.52\")\n",
+      "print '%s' %(\"Hold the resistor as shown in the figure such that tolerance is on your extreme right.\")\n",
+      "print '%s' %(\"Now the value of the resistor is equal to\")\n",
+      "print '%s' %(\"  Red      Black       Blue        Gold\")\n",
+      "print '%s' %(\"   2         0          6         (+/-)5%\")\n",
+      "red=2. #red value\n",
+      "blk=0 #black value\n",
+      "blu=6. #blue value\n",
+      "gld=5. #gold value\n",
+      "value_res=(red*10.+blk)*10.**blu #value of resistor\n",
+      "print '%s %.0f %s %.0f' %(\"\\n The value of resistor is\",value_res,\"ohm (+/-)\",gld)\n",
+      "per_val=0.05*value_res\n",
+      "pos_value_res=value_res+per_val #positive range of resistor\n",
+      "neg_value_res=value_res-per_val #negative range of resistor\n",
+      "print '%s %.0f %s %.0f %s ' %(\"\\n The value of resistor is\",neg_value_res*1e-6,\" Mohm and\",pos_value_res*1e-6,\"Mohm\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refer to the figure 1.52\n",
+        "Hold the resistor as shown in the figure such that tolerance is on your extreme right.\n",
+        "Now the value of the resistor is equal to\n",
+        "  Red      Black       Blue        Gold\n",
+        "   2         0          6         (+/-)5%\n",
+        "\n",
+        " The value of resistor is 20000000 ohm (+/-) 5\n",
+        "\n",
+        " The value of resistor is 19  Mohm and 21 Mohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example E2 - Pg 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the resistor\n",
+      "print '%s' %(\"With the help of colour coding table, one finds\")\n",
+      "print '%s' %(\"  1st_Band    2nd_Band    3rd_Band    4th_Band\")\n",
+      "print(\"    Yellow      Violet      Orange      Gold\")\n",
+      "print '%s' %(\"      4          7         10**3     (+/-)5%\")\n",
+      "yel=4. #yellow value\n",
+      "vio=7. #violet value\n",
+      "org=1e3 #orange value\n",
+      "gld=5. #gold value in %\n",
+      "val_res=(yel*10.+vio)*org\n",
+      "print '%s %.2f %s %.0f' %(\"\\n The value of resistor is\",val_res*1e-3,\"kohm (+/-)\",gld)\n",
+      "gld_ab=0.05 #absolute gold value\n",
+      "per_val=gld_ab*val_res\n",
+      "print '%s %.0f %s'%(\"\\n Now, 5%% of 47k_ohm =\",per_val, \"ohm\")\n",
+      "range_high=val_res+per_val #higher range\n",
+      "range_low=val_res-per_val #lower range\n",
+      "print'%s %.2f %s %.2f %s %.2f %s %.2f %s'%(\"\\n\\n Thus resistance should be within the range\",val_res*1e-3, \"kohm(+/-)\",per_val*1e-3 ,\"Kohm\\n or between\",range_low*1e-3 ,\"kohm and\",range_high*1e-3, \"kohm\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "With the help of colour coding table, one finds\n",
+        "  1st_Band    2nd_Band    3rd_Band    4th_Band\n",
+        "    Yellow      Violet      Orange      Gold\n",
+        "      4          7         10**3     (+/-)5%\n",
+        "\n",
+        " The value of resistor is 47.00 kohm (+/-) 5\n",
+        "\n",
+        " Now, 5%% of 47k_ohm = 2350 ohm\n",
+        "\n",
+        "\n",
+        " Thus resistance should be within the range 47.00 kohm(+/-) 2.35 Kohm\n",
+        " or between 44.65 kohm and 49.35 kohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example E3 - Pg 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the resistor\n",
+      "print '%s' %(\"The specification of the resistor from the color coding table is as follows\")\n",
+      "print '%s' %(\"  1st_Band    2nd_Band    3rd_Band    4th_Band\")\n",
+      "print '%s' %(\"     Gray        Blue        Gold       Silver\")\n",
+      "print '%s' %(\"      8          6        10**(-1)     (+/-)10%\")\n",
+      "gray=8. #gray value\n",
+      "blu=6. #blue value\n",
+      "gld=10.**-1 #gold value\n",
+      "sil=10. #silver value in %\n",
+      "val_res=(gray*10.+blu)*gld\n",
+      "print '%s %.2f %s %.2f' %(\"\\n The value of resistor is\",val_res, \"ohm (+/-)\",sil)\n",
+      "sil_ab=0.1 #absolute gold value\n",
+      "per_val=sil_ab*val_res\n",
+      "print '%s %.2f %s' %(\"\\n Now, 10%% of 8.6 ohm =\",per_val,\" ohm\")\n",
+      "range_high=val_res+per_val #higher range\n",
+      "range_low=val_res-per_val #lower range\n",
+      "print '%s %.2f %s %.2f %s %.2f %s %.2f %s' %(\"\\n\\n Obviously resistance should lie within the range\",val_res,\"ohm(+/-)\",per_val,\"ohm\\n or between\",range_high,\"ohm and\",range_low,\"ohm\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The specification of the resistor from the color coding table is as follows\n",
+        "  1st_Band    2nd_Band    3rd_Band    4th_Band\n",
+        "     Gray        Blue        Gold       Silver\n",
+        "      8          6        10**(-1)     (+/-)10%\n",
+        "\n",
+        " The value of resistor is 8.60 ohm (+/-) 10.00\n",
+        "\n",
+        " Now, 10%% of 8.6 ohm = 0.86  ohm\n",
+        "\n",
+        "\n",
+        " Obviously resistance should lie within the range 8.60 ohm(+/-) 0.86 ohm\n",
+        " or between 9.46 ohm and 7.74 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example E4 - Pg 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the load voltage and load current\n",
+      "print '%s' %(\"Refer to the figure 1.53\")\n",
+      "Vs=2. #supply voltage in V\n",
+      "Rs=1. #resistance in ohm\n",
+      "Is=Vs/Rs\n",
+      "print '%s %.2f %s' %(\"\\n Current Is =\",Is,\" A \\n\")\n",
+      "print '%s' %(\"     Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)\")\n",
+      "print '%s' %(\"    Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L.  From Figure 1.54(b) and using the current-divider concept,one obtains\")\n",
+      "RL=1. #load resistance in ohm\n",
+      "IL=Is*(Rs/(Rs+RL)) #load current using current-divider\n",
+      "VL=IL*RL #load voltage\n",
+      "print '%s %.0f %s' %(\"\\n Load voltage =\",VL,\"V\")\n",
+      "print '%s %.0f %s' %(\"\\n Load current =\",IL,\"A \\n\")\n",
+      "print '%s' %(\"From equation 53(b),using the voltage-divider concept,one obtains\")\n",
+      "VD_vl=Vs*(RL/(RL+Rs)) #load voltage using voltage divider \n",
+      "VD_il=VL/RL #load current\n",
+      "print '%s %.0f %s' %(\"\\n Load voltage =\",VD_vl,\"V\")\n",
+      "print '%s %.0f %s' %(\"\\n Load current =\",VD_il,\"A \\n\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refer to the figure 1.53\n",
+        "\n",
+        " Current Is = 2.00  A \n",
+        "\n",
+        "     Internal resistance remains the same but is now connected in parralel with the current sourceIS,as shown in Figure 1.51(a)\n",
+        "    Now,we connect a load resistance R_L=1 ohm across the terminals of two representations ,and find I_L and V_L.  From Figure 1.54(b) and using the current-divider concept,one obtains\n",
+        "\n",
+        " Load voltage = 1 V\n",
+        "\n",
+        " Load current = 1 A \n",
+        "\n",
+        "From equation 53(b),using the voltage-divider concept,one obtains\n",
+        "\n",
+        " Load voltage = 1 V\n",
+        "\n",
+        " Load current = 1 A \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example E5 - Pg 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the percentage variation in current, current for two extreme values R_L\n",
+      "print '%s' %(\"Refer to the figure 1.55\")\n",
+      "print '%s' %(\"(a) R_L varies from 1 ohm to 10 ohm.\")\n",
+      "print '%s' %(\"Currents for two extreme values of R_L are\")\n",
+      "Vs=10. #supply voltage\n",
+      "RL1=1. #resistance RL1\n",
+      "Rs=100. #source resistance\n",
+      "IL1=(Vs/(RL1+Rs))\n",
+      "RL2=10.\n",
+      "IL2=(Vs/(RL2+Rs))\n",
+      "per_var_cur=((IL1-IL2)/IL1)*100.\n",
+      "print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_cur,\"\\n\")#answer in the text book took a .3 decimal round off value\n",
+      "print '%s' %(\"   Now,load voltage for the two extreme values of R_L are\")\n",
+      "VL1=IL1*RL1\n",
+      "VL2=IL2*RL2\n",
+      "per_var_vol=((VL2-VL1)/VL2)*100.\n",
+      "print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_vol,\"\\n\")\n",
+      "\n",
+      "print '%s' %(\"(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))\")\n",
+      "print '%s' %(\"Currents for the two extreme values R_L are\")\n",
+      "RL11=1000.\n",
+      "IL11=(Vs/(RL11+Rs))\n",
+      "RL22=10000.\n",
+      "IL22=(Vs/(RL22+Rs)) #mistake in book value\n",
+      "per_var_cur11=((IL11-IL22)/IL11)*100.\n",
+      "print '%s %.2f %s' %(\"\\n Percentage variation in current =\",per_var_cur11,\"\\n\") #mistake in book value\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refer to the figure 1.55\n",
+        "(a) R_L varies from 1 ohm to 10 ohm.\n",
+        "Currents for two extreme values of R_L are\n",
+        "\n",
+        " Percentage variation in current = 8.18 \n",
+        "\n",
+        "   Now,load voltage for the two extreme values of R_L are\n",
+        "\n",
+        " Percentage variation in current = 89.11 \n",
+        "\n",
+        "(b) R_L varies from 1 k-ohm to 10 k-ohm(Figure 1.55(b))\n",
+        "Currents for the two extreme values R_L are\n",
+        "\n",
+        " Percentage variation in current = 89.11 \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/harikagunturu/Chapter_4_Angle_Modulation.ipynb b/sample_notebooks/harikagunturu/Chapter_4_Angle_Modulation.ipynb
new file mode 100755
index 00000000..de7d514c
--- /dev/null
+++ b/sample_notebooks/harikagunturu/Chapter_4_Angle_Modulation.ipynb
@@ -0,0 +1,666 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Angle Modulation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.1.A page.no: 286"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "12000 the new deviation( in Hz)\n"
+     ]
+    }
+   ],
+   "source": [
+    "Freq_dev=6; #Frequency Deviation in kHz\n",
+    "Vm=3; #Modulating Voltage in V\n",
+    "Dev=Freq_dev*10**3/Vm; \n",
+    "# for Vm=6V\n",
+    "Vm=6;\n",
+    "Freq_dev_new=Dev*Vm;\n",
+    "print Freq_dev_new,\"the new deviation( in Hz)\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.1 page.no: 287"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Instantaneous Frequency(in Hz) at (t=0.4 ms)N =  100290.574948\n",
+      "Maximum Phase Deviation (in rad) =  3\n",
+      "MAximum Frequency Deiation (in Hz)=  300.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,cos\n",
+    "\n",
+    "t1=0.4;# time in ms\n",
+    "Ang_Freq =2*pi*10**5 +3*2*pi*100*cos(2*pi*100*(t1*10**(-3)));\n",
+    "Freq=Ang_Freq/(2*pi);\n",
+    "#change in answer due to calculation error in book\n",
+    "print \"Instantaneous Frequency(in Hz) at (t=0.4 ms)N = \",Freq\n",
+    "Max_pha_Dev=3; #max(3sin(2∗pi∗100t))\n",
+    "print \"Maximum Phase Deviation (in rad) = \",Max_pha_Dev\n",
+    "Max_fre_Dev=6*pi*100;  #max(6∗pi∗100∗cos(2∗pi∗100t))\n",
+    "print \"MAximum Frequency Deiation (in Hz)= \",Max_fre_Dev/(2*pi)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.2.A page.no: 287"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Power Dissipated (in W) is  9.375\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi,sqrt\n",
+    "\n",
+    "Wc=8*10**(8);# Angular Frequency of Carrier Signal\n",
+    "fc=Wc/(2*pi);\n",
+    "Wm=1300;#Angular Frequency of Message Signal\n",
+    "fm=Wm/(2*pi);\n",
+    "B=3;#Modulation Index\n",
+    "R=12;\n",
+    "Vc_rms=15/sqrt(2);\n",
+    "Max_dev=B*fm;\n",
+    "Power=Vc_rms**(2)/R;\n",
+    "print \"Power Dissipated (in W) is \",Power"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.2 page.no: 287"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Peak Frequency Deviation(in Hz) is  12000\n",
+      "modulation index  8.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "a=3;#amplitude in volts\n",
+    "Dev_sen=4;# deviation sensitivity in KHz/volts\n",
+    "fm=1.5;# frequency modulating signal in KHz\n",
+    "f=Dev_sen*10**(3)*3;#peak frequency deviation\n",
+    "B=f/(fm*10**3);\n",
+    "print \"Peak Frequency Deviation(in Hz) is \",f\n",
+    "print \"modulation index \",B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.3.A page.no: 289"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Bandwidth (in Hz) is  72000\n"
+     ]
+    }
+   ],
+   "source": [
+    "fm=3; #Modulating Frequency in kHZ\n",
+    "Max_Dev=18; #MAximum Deviation in kHz\n",
+    "B=Max_Dev/fm; # modulation index 7\n",
+    "J=12;#from Bessel Table , for B=6\n",
+    "Bw=fm*J*2*10**(3);\n",
+    "print \"The Bandwidth (in Hz) is \",Bw"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.3 page.no: 289"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Peak Phase Deviation( in rad)  8.75\n"
+     ]
+    }
+   ],
+   "source": [
+    "Dev_sen=3.5 # Deviation Sensitivity in rad/volt\n",
+    "a=2.5; #amplitude in volts\n",
+    "B=a*Dev_sen; # Peak Phase Deviation\n",
+    "print \"Peak Phase Deviation( in rad) \",B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.4.A page.no: 290"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Maximum Frequency Deviation (in Hz) is  18000\n",
+      "Modulation Index is  5.99985864877\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "\n",
+    "Wm=18850;#Angular Frequency of message signal\n",
+    "fm=Wm/(2*pi);\n",
+    "a=3;# amplitude of message signal\n",
+    "Dev_sen=6;#Deviation Sensitivity in kHz/V\n",
+    "Max_Freq_Dev=a*Dev_sen*10**(3);\n",
+    "B=Max_Freq_Dev/(fm);\n",
+    "print \"Maximum Frequency Deviation (in Hz) is \",Max_Freq_Dev\n",
+    "print \"Modulation Index is \",B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.4 page.no: 291"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Deviation Sensitivity(in kHz/V)  1333\n",
+      "Modulation Index is  4\n",
+      "Deviation Sensitivity for 5V (in Hz)  6665\n",
+      "Modulation index 6\n",
+      "Deviation Sensitivity for 10V (in Hz)  13330\n",
+      "Modulation index is  33\n"
+     ]
+    }
+   ],
+   "source": [
+    "a=3; #amplitude in Volts\n",
+    "Dev=4;# Deviation in kHz\n",
+    "fm=1;# modulating frequency in kHz\n",
+    "Dev_sen=Dev*10**(3)/a; #Deviation Sensitivity\n",
+    "B=Dev/fm; # Modulation Index\n",
+    "print \"Deviation Sensitivity(in kHz/V) \",Dev_sen\n",
+    "print \"Modulation Index is \",B\n",
+    "#a)\n",
+    "a=5;\n",
+    "Dev_sen_1=a*Dev_sen;\n",
+    "B=Dev_sen_1/(fm*10**(3));\n",
+    "print \"Deviation Sensitivity for 5V (in Hz) \",Dev_sen_1\n",
+    "print \"Modulation index\",B\n",
+    "#b)\n",
+    "a=10;\n",
+    "fm=400;\n",
+    "Dev_sen_2=a*Dev_sen;\n",
+    "B=Dev_sen_2/fm;\n",
+    "print \"Deviation Sensitivity for 10V (in Hz) \",Dev_sen_2\n",
+    "print \"Modulation index is \",B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.5.A page.no: 291"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "for B=2, The number of significant frequencies are 6\n",
+      "They are J1,J2,J3,J4,J5 and J6\n",
+      "Their amplitudes with carriers are \n",
+      "they are (in V) 1.792 4.616 2.824 1.032 0.272 0.056 0.008\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"for B=2, The number of significant frequencies are 6\"\n",
+    "print \"They are J1,J2,J3,J4,J5 and J6\"\n",
+    "print \"Their amplitudes with carriers are \"\n",
+    "J0= 0.224*8;\n",
+    "J1= 0.577*8;\n",
+    "J2= 0.353*8;\n",
+    "J3= 0.129*8;\n",
+    "J4= 0.034*8;\n",
+    "J5= 0.007*8;\n",
+    "J6= 0.001*8;\n",
+    "print\"they are (in V)\",J0,J1,J2,J3,J4,J5,J6"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.5 page.no: 292"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Bandwidth required (in Hz)  24000\n",
+      "According to Carsons rule , Bandwidth (in Hz)  36000\n"
+     ]
+    }
+   ],
+   "source": [
+    "fm=3; #Modulating Frequency in kHZ\n",
+    "Max_dev=15;# Maximum Deviatin in kHZ\n",
+    "B=Max_dev/fm; \n",
+    "J=8; # Bessel table , the highest J coefficient\n",
+    "BW=J*fm*10**(3);#Bandwidth in kHz\n",
+    "BW1=2*(fm+Max_dev)*10**(3);# According to carson rule , BAndwidth\n",
+    "print \"Bandwidth required (in Hz) \",BW\n",
+    "print \"According to Carsons rule , Bandwidth (in Hz) \",BW1"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.6.A page.no: 292"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Minimum Bandwidth (in Hz) is  72000\n",
+      "Approximate Minimum Bandwidth is  36000\n"
+     ]
+    }
+   ],
+   "source": [
+    "Max_Freq_Dev=12; #Maximum Frequency Deviation in kHZ\n",
+    "fm=6; #Modulating frquency in kHz\n",
+    "B=Max_Freq_Dev/fm;# Modulation index 7\n",
+    "J=6;#From Bessel Table , for B=2\n",
+    "Bw=2*J*6*10**(3);\n",
+    "BW_carson=2*(fm + Max_Freq_Dev)*10**(3);\n",
+    "print \"Minimum Bandwidth (in Hz) is \",Bw\n",
+    "print \"Approximate Minimum Bandwidth is \",BW_carson"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.6.A page.no: 283"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "For B=5 from the Bessel table ,The Bessel Function is taken upto J9\n",
+      "Hence the average power of the modulated signal (in W) is  9.936\n",
+      "Hence, the average power of the modulated signal is equal to \n",
+      "unmodulated carrier power\n"
+     ]
+    }
+   ],
+   "source": [
+    "a=10; #Amplitude in V\n",
+    "Pt=a*(0.18**2 +2*(0.33**2+0.05**2+0.36**2+0.39**2+0.26**2+0.13**2+0.05**2+0.02**2+0.01**2))\n",
+    "print \"For B=5 from the Bessel table ,The Bessel Function is taken upto J9\"\n",
+    "print \"Hence the average power of the modulated signal (in W) is \",Pt\n",
+    "print \"Hence, the average power of the modulated signal is equal to \"\n",
+    "print \"unmodulated carrier power\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.7.A page.no: 294"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Unmodulated Power Carrier ( in W) =  1\n",
+      "Total Power in modulated wave(in W)=  1.06767573333\n",
+      "Power in the modulated wave is equal to \n",
+      "power in the unmodulated wave \n"
+     ]
+    }
+   ],
+   "source": [
+    "a=8;# amplitude in V\n",
+    "r=30; # resistance in ohms\n",
+    "Pc_unmodulated=a**2/(2*r);\n",
+    "Pt=1.792**2/(2*30)+2*(4.616)**2/(2*30)+2*(2.824**2)/(2*30) +2*(1.032) **2/(2*30) +2*(0.272) **2/(2*30) +2*(0.056)**2/(2*30)+2*(0.008)**2/(2*30);\n",
+    "# change in answer due to approximations in the book\n",
+    "print \"Unmodulated Power Carrier ( in W) = \",Pc_unmodulated\n",
+    "print \"Total Power in modulated wave(in W)= \",Pt\n",
+    "print \"Power in the modulated wave is equal to \"\n",
+    "print \"power in the unmodulated wave \" \n",
+    "#\"Small error due to rounded off values in Bessel functions\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.7 page.no: 295"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the Phase Modulation Function =  12.0*sin(3000.0*pi*t)\n",
+      "The Modulated Wave Function =  12.0*sin(3000.0*pi*t) + 8*cos(20000*pi*t)\n"
+     ]
+    }
+   ],
+   "source": [
+    "from sympy import symbols,sin,cos\n",
+    "\n",
+    "t,pi=symbols('t,pi') \n",
+    "Pha_dev=3.; #Phase Deviation constant in rad/V 6\n",
+    "# Phase Modulation Function\n",
+    "Pha_function=Pha_dev*4*sin(2.*pi*1.5*10**3*t);\n",
+    "Mod_wave=8*cos(2*pi*10**4*t)+Pha_function\n",
+    "print \"the Phase Modulation Function = \",Pha_function\n",
+    "print \"The Modulated Wave Function = \",Mod_wave"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.8 page.no: 295"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The outputs of the balanced modulator for these parameters\n",
+      "are same as the inputs \n",
+      "They remain unaltered \n",
+      "At the output of the Multiplier , \n",
+      "Fc(in kHz)=  9600 , Fm(in kHz)=  10 , B=  6.0\n",
+      "Frequency Deviation ( in kHz)=  60\n"
+     ]
+    }
+   ],
+   "source": [
+    "initial_Freq_Dev=5; # frequency in kHz\n",
+    "B_initial=0.5; #modulation index\n",
+    "fm_initial=10;# message signal frequency in kHz\n",
+    "fc_initial=800; # carrier frequency in kHz\n",
+    "print \"The outputs of the balanced modulator for these parameters\"\n",
+    "print \"are same as the inputs \"\n",
+    "print \"They remain unaltered \"\n",
+    "#at the output of the multiplier 14\n",
+    "m=12;# multiplication factor\n",
+    "final_Freq_Dev=initial_Freq_Dev*m;\n",
+    "B_final=0.5*m;\n",
+    "fm_final=10; #modulating signal remains unaltered\n",
+    "fc_final=800*m;\n",
+    "print \"At the output of the Multiplier , \"\n",
+    "print \"Fc(in kHz)= \",fc_final,\", Fm(in kHz)= \",fm_final,\", B= \",B_final\n",
+    "print \"Frequency Deviation ( in kHz)= \",final_Freq_Dev"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.9.A page.no: 296"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a) MAster Oscillator Centre Frequency(in MHz) =  4.008\n",
+      "b) Frequency Deviation at the output of modulator(in KHz)=  2.4\n",
+      "c)Devaition ratio at the output of modulator  0.24\n",
+      "d)deviation ratio at power amplifier 6.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "ft=100.2; #final carrier frequency in MHz\n",
+    "Freq_Dev_ft=60.;# Frequency Deviation in KHz at power amplifier\n",
+    "fm=10.;#modulating frequency in KHz\n",
+    "m=25.;#multiplication factor\n",
+    "#a)\n",
+    "fc=ft/25.;\n",
+    "#b)\n",
+    "Freq_Dev=Freq_Dev_ft/25;\n",
+    "#c)\n",
+    "B=Freq_Dev/fm;\n",
+    "#d)\n",
+    "Bt=B*m;\n",
+    "print \"a) MAster Oscillator Centre Frequency(in MHz) = \",fc\n",
+    "print \"b) Frequency Deviation at the output of modulator(in KHz)= \",Freq_Dev\n",
+    "print \"c)Devaition ratio at the output of modulator \",B\n",
+    "print \"d)deviation ratio at power amplifier\",Bt"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 4.10.A page.no: 297"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "a) Frequency Deviation(in Hz)=  5954.92965855\n",
+      "b) Devaition Ratio=  5.95492965855\n",
+      "c) Phase Deviation( in rad)=  8\n",
+      "d) Bandwidth( in Hz)=  13909.8593171\n"
+     ]
+    }
+   ],
+   "source": [
+    "from math import pi\n",
+    "\n",
+    "#f(t)=5cos(Wc∗t+3sin(2000∗t)+5sin(2000∗pi∗t)) 5\n",
+    "fm=2000*pi/(2*pi); #bandwidth is the highest frequency component\n",
+    "#a) \n",
+    "Freq_dev=(6000+10000*pi)/(2*pi); 11\n",
+    "#b)\n",
+    "B=Freq_dev/fm; \n",
+    "#c)\n",
+    "Phase_dev=8;#Highest value of[3sin(2000t)+5sin(2000∗ pi∗t)]\n",
+    "#d)\n",
+    "Bw= 2*(fm+Freq_dev);\n",
+    "print \"a) Frequency Deviation(in Hz)= \",Freq_dev\n",
+    "print \"b) Devaition Ratio= \",B\n",
+    "print \"c) Phase Deviation( in rad)= \",Phase_dev\n",
+    "print \"d) Bandwidth( in Hz)= \",Bw"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/harishsahu/Chapter2.ipynb b/sample_notebooks/harishsahu/Chapter2.ipynb
new file mode 100755
index 00000000..d4e7a3ad
--- /dev/null
+++ b/sample_notebooks/harishsahu/Chapter2.ipynb
@@ -0,0 +1,1124 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4425378c6999e9724676588b0097c2038f3833cd503390f3d7cf7bb3508521f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2-Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input from given graph\n",
+      "##calculation of initial accleration\n",
+      "ia=18/4.\n",
+      "## calculation of final accleration\n",
+      "fa=-18/10.\n",
+      "decel=-(fa)##calculation of deceleration\n",
+      "##calculation of total distance covered\n",
+      "d=0.5*(4.*18.)+(8.*18.)+0.5*(10.*18.)##area under velocity time graph\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"\\n the initial acceleration is \",ia,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the final acceleration is \",decel,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the distance covered is is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " the initial acceleration is  4.50  m/s^2\n",
+        "\n",
+        " the final acceleration is  1.80  m/s^2\n",
+        "\n",
+        " the distance covered is is  270.00  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v=0. ##car stops => final velocity=0\n",
+      "u=29. ##initial velocity\n",
+      "t=11. ##time \n",
+      "##calculation of acceleration\n",
+      "a=(v-u)/t##eqn of uniformly accelerated body\n",
+      "##calculating distance travelled during this period\n",
+      "d=(v+u)*t*0.5##eqn of uniformly accelerated body\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the accleration is \",a,\" ms^-2 \")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe distance travelled is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the accleration is  -2.64  ms^-2 \n",
+        "\n",
+        "the distance travelled is  159.50  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v=120. ##velocity\n",
+      "a=75 ##accleration\n",
+      "##ca.lculation of time\n",
+      "t=2.*v/(a*math.cos(45/57.3))##eqn of uniformly accelerated body\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the time taken is \",t,\" s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the time taken is  4.53  s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "f1=50.\n",
+      "f2=50.\n",
+      "##calculation of net force\n",
+      "f=f1+f2 ## the two forces act in same direction\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the resultant force is \",f,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the resultant force is  100.00  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "vc=25. ##velocity of car\n",
+      "va=10. ##velocity of wind\n",
+      "va1=15. ##velocity of wind westward\n",
+      "##calculation\n",
+      "v1=vc+va##resultant velocity for a tail of wind\n",
+      "v2=vc-va##when wind blows westward at 10 m/s^resultant velocity \n",
+      "v3=vc-va1##resultant velocity when wind blows westward at 15m/s^2\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"1. the resultant velocity of wind is \",v1,\" ms^-1 eastwards \")\n",
+      "print\"%s %.2f %s\"%(\"\\n2. the resultant velocity of wind is \",v2,\" ms^-1 westwards \")\n",
+      "print\"%s %.2f %s\"%(\"\\n3. the resultant velocity of wind is \",v3,\"  ms^-1westwards \")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. the resultant velocity of wind is  35.00  ms^-1 eastwards \n",
+        "\n",
+        "2. the resultant velocity of wind is  15.00  ms^-1 westwards \n",
+        "\n",
+        "3. the resultant velocity of wind is  10.00   ms^-1westwards \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7-pg31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=30. ##velocity of speedboat\n",
+      "vw=40. ##velocity of wind\n",
+      "##calculation\n",
+      "x=(30./40.)##angle between original velocity of boat and resultant velocity\n",
+      "y=math.atan(x)*(57.3)##applying trigonometry\n",
+      "b=90.+y##bearing of boat\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the bearing of speedboat is \",b,\" deg\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the bearing of speedboat is  126.87  deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8-pg32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#input\n",
+      "f1=6. ##tension on string AB\n",
+      "f2=6. ##tension on string BC\n",
+      "##calculation of tension\n",
+      "t=2.*f1*math.sin(55/57.3)## the resultant tension is the diagonal of rhombus formed\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"/n the resultant tension is \",t,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "/n the resultant tension is  9.83  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10-pg33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input magnitude of forces\n",
+      "f1=40.\n",
+      "f2=50.\n",
+      "##calculation\n",
+      "d=50**2+40**2+2.*50.*40.*math.cos(50./57.3)##finding the diagonal\n",
+      "r=50**2+40**2+2.*50.*(40.)*math.cos(130./57.3)##reversing the side and finding diagonlprint\"%s %.2f %s\"%(\"the resultant is %3.3f\",d1)\n",
+      "r1=math.sqrt(r)##resultant sum\n",
+      "d1=math.sqrt(d)## resultant when smaller force is subtracted from larger\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"1. the resultant sum is \",d1,\" N\")\n",
+      "print\"%s %.2f %s\"%(\"\\n 2. the resultant when smaller force is subtracted from larger is \",r1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. the resultant sum is  81.68  N\n",
+        "\n",
+        " 2. the resultant when smaller force is subtracted from larger is  39.11  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=380.##velocity\n",
+      "##calculation\n",
+      "vh=v*math.cos(60./57.3)##horizontal component\n",
+      "vv=v*math.sin(60./57.3)##vertical component\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the horizontal component is \",vh,\" ms**-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe vertical component is \",vv,\" ms**-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the horizontal component is  190.03  ms**-1\n",
+        "\n",
+        "the vertical component is  329.07  ms**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "fc=50.##force applied by magnet\n",
+      "x=90.-20. ##angle of force\n",
+      "##calculation\n",
+      "fb=fc*math.sin(70./57.3)##force due to b\n",
+      "fa=fc*math.cos(70./57.3)##force due to a\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the force due to b is \",fb,\" N\")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe force due to b is \",fa,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the force due to b is  46.98  N\n",
+        "\n",
+        "the force due to b is  17.11  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13-pg35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m1=1.\n",
+      "v1=25.\n",
+      "m2=2.\n",
+      "v2=0.\n",
+      "##calculation\n",
+      "v=(m1*v1)+(m2*v2)##applying princilpe of conservation of linear momentum\n",
+      "v4=v/(m1+m2)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the velocity with which both will move is \",v4,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the velocity with which both will move is  8.33  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex14-pg35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m1=1.##mass of object 1\n",
+      "v1=25.##velocity of object 1\n",
+      "m2=2.##mass of object 2\n",
+      "v2=0.##body at rest,velocity =0\n",
+      "v3=10.\n",
+      "##caclulation\n",
+      "u=((m1*v1)+(m2*v2)-(m2*v3))/2.##applying princilpe of conservation of linear momentum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"\\n the value of u is \",-u,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " the value of u is  -2.50  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15-pg36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=2.##mass\n",
+      "r=4.##radius\n",
+      "v=6.##uniform speed\n",
+      "##calculation\n",
+      "w=v/r##angular velocity\n",
+      "f=m*r*w*w##centripetal force\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the angular velocity is \",w,\" rads^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the centripetal force is \",f,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the angular velocity is  1.50  rads^-1\n",
+        "\n",
+        " the centripetal force is  18.00  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex16-pg37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=140.##mass\n",
+      "v=8.##speed\n",
+      "r=5.##radius\n",
+      "g=9.8##acceleration due to gravity\n",
+      "##calculation\n",
+      "t=((m*v**2/5.)**2)+(140.*9.8)**2 ##applying parallelogram of vectors\n",
+      "t1=math.sqrt(t)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the tension in arm is \",t1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the tension in arm is  2256.91  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex17-pg38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=15.##velocity\n",
+      "m=70.##mass\n",
+      "r=50.##radius\n",
+      "##calculation\n",
+      "x=v*v/(r*10.)##applying parallelogram of vectors,then for equilibrium\n",
+      "y=math.atan(x)*57.3\n",
+      "r1=(m*10.)/math.cos(y/57.3)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the inclination is \",y,\" deg\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the reaction is \",r1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the inclination is  24.23  deg\n",
+        "\n",
+        " the reaction is  767.61  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex18-pg39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "r=5500.##radius\n",
+      "g1=6.7*10**-11\n",
+      "g=7##acceleration due to gravity\n",
+      "##calculation of mean density\n",
+      "p=3.*g/(4.*math.pi*r*10**3*g1)##mean density\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the mean density of planet is \",p,\" kgm^-3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean density of planet is  4534.94  kgm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex19-pg40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=5.*10**24##mass of earth\n",
+      "g1=6.7*10**-11\n",
+      "##calculation\n",
+      "r=((6.7*10**-11.*5.*10**24*(3600.*24.)**2)/(4.*math.pi**2))**(1./3.)##orbit radius\n",
+      "v=(2.*math.pi*r)/(3600.*24.)##linear velocity\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the orbit radius is \",r,\"\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the linear velocity is \",v,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the orbit radius is  39863080.05 \n",
+        "\n",
+        " the linear velocity is  2898.92  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex20-pg41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=3.*10**5##orbit speed\n",
+      "r=4.6*10**20##distance\n",
+      "g1=6.7*10**-11\n",
+      "##calculation of mass\n",
+      "m=v*v*r/g1 ##Newtons law\n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the mass is \",m,\" kg\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass is  6.18e+41  kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex21-pg42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=0.6##speed\n",
+      "m=0.3##mass\n",
+      "##calculation\n",
+      "e=0.75*m*v*v##total kinetic energy is kinetic energy+moment of inertia\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the total kinetic energy is \",e,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the total kinetic energy is  0.08  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex22-pg43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "t1=34.\n",
+      "u=0.##starts from rest\n",
+      "x=3.##distance to move\n",
+      "##calculation\n",
+      "t=(3.*3./(10.*math.sin(t1)))**0.5##from law of conservation of energy\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the time taken is \",t,\" s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the time taken is  1.30  s\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex23-pg43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "i1=53.##inertia when it spins with panels carrying solar cells\n",
+      "i2=37.##inertia about axis of rotation\n",
+      "##calculation\n",
+      "r=i1/i2##law of conservation of angular momentum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the ratio of angular velocities is\",r,\"\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of angular velocities is 1.43 \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex25-pg45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "f=9.##frequency\n",
+      "x=0.##at midpoint of stroke x=0\n",
+      "##calculation\n",
+      "t=1./f\n",
+      "a=-4.*math.pi**2*f**2*x##acceleration for shm\n",
+      "v=2.*math.pi*f*0.05##velocity for shm\n",
+      "a1=-4.*math.pi**2*9**2*0.05##acceleration at amplitude\n",
+      "v1=0.##velocity at amplitude is 0\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the period of oscillation is \",t,\" s^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the velocity  at midpoint of stroke is \",v,\" ms^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the acceleration at midpoint of stroke is \",a,\" ms^-2\")\n",
+      "\n",
+      "print\"%s %.2f %s\"%(\"\\n the velocity at amplitude is \",v1,\"  ms^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the acceleration at amplitude is \",a1,\" ms^-2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the period of oscillation is  0.11  s^-1\n",
+        "\n",
+        " the velocity  at midpoint of stroke is  2.83  ms^-1\n",
+        "\n",
+        " the acceleration at midpoint of stroke is  -0.00  ms^-2\n",
+        "\n",
+        " the velocity at amplitude is  0.00   ms^-1\n",
+        "\n",
+        " the acceleration at amplitude is  -159.89  ms^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex26-pg47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "g=10.\n",
+      "t=0.3##period of shm\n",
+      "##calculation\n",
+      "x=g*t**2/(4.*math.pi**2)##for shm maximum amplitude\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the maximum amplitude for bead to be in contact is \",x,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum amplitude for bead to be in contact is  0.02  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex27-pg48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "p1=2.3##period of pendulum\n",
+      "p2=3.1##period when pendulum is lengthened\n",
+      "##calculation\n",
+      "g=4.*math.pi**2/(p2**2-p1**2)##acceleration of free fall\n",
+      "l=p1**2*g/(4.*math.pi**2)##length of pendulum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the acceleration of free fall is \",g,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the length of pendulum is \",l,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the acceleration of free fall is  9.14  m/s^2\n",
+        "\n",
+        " the length of pendulum is  1.22  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex28-pg49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##INPUT DATA\n",
+      "f=55. ##frequency\n",
+      "a=7.*10**-3 ##amplitude\n",
+      "\n",
+      "\n",
+      "##calculation\n",
+      "a=(-2.*math.pi*f)**2*a\n",
+      "\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the acceleration of the body when it is at its maximum displacement from its zero position is \",a,\" ms^-2\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the acceleration of the body when it is at its maximum displacement from its zero position is  835.96  ms^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex29-pg50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "f=55.##frequency\n",
+      "amp=7.*10**-3##amplitude\n",
+      "m=1.2##mass\n",
+      "##calculation\n",
+      "e=0.5*m*4.*math.pi**2*f**2*amp**2##maximum pe occurs at zero position\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the maximum pe is \",e,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum pe is  3.51  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex30-pg51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "l=6.5##length\n",
+      "m=0.06##mass of wire\n",
+      "m1=10##mass attached\n",
+      "g=9.8##acceleration due to gravity\n",
+      "e=2.1*10**11##youngs modulus\n",
+      "ro=8.*10**3##density of steel\n",
+      "##calculation\n",
+      "e1=m1*g*ro*l*l/(e*m)##extension caused \n",
+      "pe=0.5*g*m1*e1##potential energy \n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the extension caused is \",e1,\" m\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the potential energy is \",pe,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the extension caused is  2.63e-03  m\n",
+        "\n",
+        " the potential energy is  0.13  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex31-pg52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "w=250.*10**3\n",
+      "s=0.00003##strain\n",
+      "a=0.04##area\n",
+      "w1=320.*10**3\n",
+      "##calculation\n",
+      "e=w/(a*s)##youngs module\n",
+      "st=w1/a##stress\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the youngs modulus is \",e,\" N/m^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the stress is \",st,\" N/m^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the youngs modulus is  208333333333.33  N/m^2\n",
+        "\n",
+        " the stress is  8000000.00  N/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex32-pg53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=40.##mass\n",
+      "g=9.8##acceleration due to gravity\n",
+      "E=2*10**11##youngs modulus\n",
+      "##calculation\n",
+      "t1=m*g/5.##principle of momentum\n",
+      "t2=4*m*g/5. ##principle of momentum\n",
+      "d=4.*(t2-t1)/(4.*math.pi*10**-6*E)##difference in length\n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the difference is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the difference is  3.74e-04  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/hemlarathod/Chapter2.ipynb b/sample_notebooks/hemlarathod/Chapter2.ipynb
new file mode 100755
index 00000000..6530cd52
--- /dev/null
+++ b/sample_notebooks/hemlarathod/Chapter2.ipynb
@@ -0,0 +1,778 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:9826abe74c775578903ec0e922c705aacf445defe2dc3badb10ce4727f434663"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2-Compressible Flow with Friction and Heat: A Review"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what is the gas constant of air and density of air\n",
+      "import math\n",
+      "#intilization variable\n",
+      "p=3*10**6 ; #pressure in Pa\n",
+      "t=298. ; #temperatue in kelvin\n",
+      "mw= 29.; #molecular weight in kg/mol\n",
+      "ru=8314.; #universal constant in J/kmol.K\n",
+      "r=ru/mw ;\n",
+      "#using perfect gas law to get density:\n",
+      "rho=p/(r*t) ;\n",
+      "print'%s %.2f %s'%('Gas constant of air in',r,'J/kg.K')\n",
+      "print'%s %.1f %s'%('Density of air in',rho,'kg/m^3')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gas constant of air in 286.69 J/kg.K\n",
+        "Density of air in 35.1 kg/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find out the exit temperature and exit density by various methods \n",
+      "import math\n",
+      "t1=288.; #inlet temperture in Kelvin\n",
+      "p1=100*10**3; #inlet pressure in Pa\n",
+      "p2=1*10**6 #exit pressure in Pa\n",
+      "gma=1.4; #gamma.\n",
+      "rg=287.; #gas constant in J/kg.K\n",
+      "t2=t1*(p2/p1)**((gma-1)/gma);   #exit temperature \n",
+      "print'%s %.5f %s'%('Exit temperature in',t2,'K')\n",
+      "#first method to find exit density:\n",
+      "#application of perfect gas law at exit\n",
+      "rho=p2/(rg*t2); #rho= exit density.\n",
+      "print'%s %.7f %s'%('exit density at by method 1 in',rho,'kg/m^3')\n",
+      "#method 2: using isentropic relation between inlet and exit density.\n",
+      "rho1=p1/(rg*t1); #inlet density.\n",
+      "rho=rho1*(p2/p1)**(1/gma);\n",
+      "print'%s %.2f %s'%('exit density by method 2 in',rho,'kg/m^3')\n",
+      "\n",
+      "                                                     "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit temperature in 556.04095 K\n",
+        "exit density at by method 1 in 6.2663021 kg/m^3\n",
+        "exit density by method 2 in 6.27 kg/m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what is the rate of mass flow through exit \n",
+      "import math\n",
+      "d1=1.2 #inlet 1 density in kg/m^3.\n",
+      "u1=25. # inlet 1 veocity in m/s.\n",
+      "a1=0.25 #inlet 1 area in m^2.\n",
+      "d2=0.2 #inlet 2 density in kg/m^3.\n",
+      "u2=225. #inlet 2 velocity in m/s.\n",
+      "a2=0.10 #inlet 2 area in m^2.\n",
+      "m1=d1*a1*u1; #rate of mass flow entering inlet 1.\n",
+      "m2=d2*u2*a2; #rate of mass flow entering inlet 2.\n",
+      "#since total mass in=total mass out,\n",
+      "m3=m1+m2; #m3=rate of mass flow through exit.\n",
+      "print'%s %.f %s'%('Rate of mass flow through exit in',m3,' kg/s')\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rate of mass flow through exit in 12  kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what is the axial force needed to support the plate and lateral force needed to support the plate\n",
+      "import math\n",
+      "u1=2 #speed of water going on the plate. X-component in m/s.\n",
+      "v1=0 #speed of water going on the plate. Y-component in m/s.\n",
+      "u2=1 #speed of water going on the plate. X-component in m/s.\n",
+      "v2=1.73 #speed of water going on the plate Y-coponent in m/s.\n",
+      "m=0.1 #rate of flow of mass of the water on the plate in kg/s.\n",
+      "#Using Newton's second law.\n",
+      "Fx=m*(u2-u1); #X-component of force exerted by water\n",
+      "print'%s %.1f %s'%('Axial force needed to support the plate in',Fx,'N')\n",
+      "Fy=m*(v2-v1); #Y-component of force exerted by water.\n",
+      "print'%s %.3f %s'%('Lateral force needed to support the plate in',Fy,'N')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Axial force needed to support the plate in -0.1 N\n",
+        "Lateral force needed to support the plate in 0.173 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the Exit total and static temperature \n",
+      "m=50 #mass flow rate in kg/s.\n",
+      "T1=298 #inlet temperature in K.\n",
+      "u1=150 #inlet velocity in m/s.\n",
+      "cp1=1004 #specific heat at constant pressure of inlet in J/kg.K.\n",
+      "gm=1.4 #gamma.\n",
+      "u2=400 # exit velocity in m/s.\n",
+      "cp2=1243. #specific heat at constant pressure of exit in J/kg.K.\n",
+      "q=42*10**6 #heat transfer rate in control volume in Watt.\n",
+      "me=-100*10**3 #mechanical power in Watt.\n",
+      "#first calculate total enthalpy at the inlet:\n",
+      "ht1=cp1*T1+(u1**2)/2; #ht1=Total inlet enthalpy.\n",
+      "#now applying conservation of energy equation:\n",
+      "ht2=ht1+((q-me)/m) #ht2=Total enthalpy at exit.\n",
+      "Tt2=ht2/cp2; #Tt2=Total exit temperature.\n",
+      "T2=Tt2-((u2**2)/(2*cp2)); #T2=static exit temperature.\n",
+      "print'%s %.5f %s'%('Exit total temperature in',Tt2,'K')\n",
+      "print'%s %.4f %s'%('Exit static temperature in',T2,'K')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Exit total temperature in 927.14562 K\n",
+        "Exit static temperature in 862.7852 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6-pg65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#intilization variable\n",
+      "import math\n",
+      "d=0.2 #Diameter in meters.\n",
+      "M1=0.2 #inlet Mach no.\n",
+      "p1=100*10**3 #inlet pressure in Pa\n",
+      "Tt1=288. #total inlet temperature in K\n",
+      "q=100*10**3 #rate of heat transfer to fluid in Watt.\n",
+      "rg=287. #Gas constant in J/kg.K.\n",
+      "gm=1.4 #gamma\n",
+      "#(a)inlet mass flow:\n",
+      "m=((gm/rg)**(1./2.))*(p1/(Tt1)**(1./2.))*3.14*(d*d)/4.*(M1/(1.+((gm-1.)/2.)*(M1**2.))**((gm+1.)/(2.*(gm-1.))));\n",
+      "\n",
+      "#(b)\n",
+      "qm=q/m; #Heat per unit mass.\n",
+      "#Tt1/Tcr=0.1736, pt1/Pcr=1.2346, ((Delta(s)/R)1=6.3402,p1/Pcr=2.2727)\n",
+      "Tcr=Tt1/0.1736;\n",
+      "\n",
+      "Pcr=p1/2.2727;\n",
+      "#From energy equation:\n",
+      "cp=(gm/(gm-1.))*rg;\n",
+      "Tt2=Tt1+(q/cp);\n",
+      "q1cr=cp*(Tcr-Tt1)/1000.;\n",
+      "M2=0.22;\n",
+      "#From table : pt2/Pcr=1.2281, (Delta(s)/R)2=5.7395, p2/Pcr=2.2477.\n",
+      "#The percent total pressure drop is (((pt1/Pcr)-(pt2/Pcr))/(pt1/Pcr))*100.\n",
+      "p2=2.2477*Pcr;\n",
+      "dp=((1.2346-1.2281)/1.2346)*100;\n",
+      "#Entropy rise is the difference between (delta(s)/R)1 and (delta(s)/R)2.\n",
+      "ds=6.3402-5.7395;\n",
+      "#Static pressure drop in duct due to heat transfer is\n",
+      "dps=((p1/Pcr)-(p2/Pcr))*Pcr/1000.;\n",
+      "print'%s %.7f %s'%('Mass flow rate through duct in',m,'kg/s')\n",
+      "print'%s %.4f %s'%('Critical heat flux that would choke the duct for the M1 in',q1cr,'kJ/kg')\n",
+      "print'%s %.2f %s'%('The exit Mach No.',M2,'')\n",
+      "print'%s %.7f %s'%('The percent total pressure loss',dp,'%')\n",
+      "print'%s %.4f %s'%('The entropy rise',ds,'')\n",
+      "print'%s %.7f %s'%('The static pressure drop in ',dps,'kPa')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate through duct in 2.5235091 kg/s\n",
+        "Critical heat flux that would choke the duct for the M1 in 1377.1556 kJ/kg\n",
+        "The exit Mach No. 0.22 \n",
+        "The percent total pressure loss 0.5264863 %\n",
+        "The entropy rise 0.6007 \n",
+        "The static pressure drop in  1.1000132 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7-pg67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what is total exit temperautre if exit is choked and maximum heat released and fule to air ratio to thermally choke the combustor exit and total pressure loss\n",
+      "#intilization variable\n",
+      "import math\n",
+      "M1=3.0 ##Mach no. at inlet\n",
+      "pt1=45*10**3 ##Total pressure t inlet in Pa\n",
+      "Tt1=1800 ##Total temperature at inlet in K\n",
+      "hv=12000 ##Lower heating value of hydrogen kJ/kg\n",
+      "gm=1.3 ##gamma\n",
+      "R=0.287 ##in kJ/kg.K\n",
+      "##Using RAYLEIGH table for M1=3.0 and gamma=1.3, we get Tt1/Tcr=0.6032, pt1/Pcr=4.0073.\n",
+      "Tcr=Tt1/0.6032\n",
+      "Pcr=pt1/4.0073\n",
+      "##if exit is choked, Tt2=Tcr\n",
+      "Tt2=Tt1/0.6032;\n",
+      "cp=gm*R/(gm-1);\n",
+      "##Energy balance across burner:\n",
+      "Q1cr=cp*(Tcr-Tt1);\n",
+      "f=(Q1cr/120000);\n",
+      "##total pressure loss:\n",
+      "dpt=1-Pcr/pt1;\n",
+      "print'%s %.4f %s'%('Total exit temperature if exit is choked in',Tt2,'K')\n",
+      "print'%s %.4f %s'%('Maximum heat released per unit mass of air in',Q1cr, 'kJ/kg')\n",
+      "print'%s %.7f %s'%('fuel-to-air ratio to thermally choke the combustor exit',f,'')\n",
+      "print'%s %.7f %s'%('Total pressure loss (in fraction)',dpt,'')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total exit temperature if exit is choked in 2984.0849 K\n",
+        "Maximum heat released per unit mass of air in 1472.6069 kJ/kg\n",
+        "fuel-to-air ratio to thermally choke the combustor exit 0.0122717 \n",
+        "Total pressure loss (in fraction) 0.7504554 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8-pg67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the new inlet mach no and spilled flow at the inlet\n",
+      "#initilization variable \n",
+      "import math\n",
+      "Tt1=50.+460. ##Converting the inlet temp. to the absolute scale i.e. in degree R\n",
+      "M1=0.5 ##Initial inlet Mach no.\n",
+      "pt1=14.7 ##Units in psia\n",
+      "gm=1.4 ##gamma\n",
+      "R=53.34 ##units in ft.lbf/lbm.degree R\n",
+      "Tcr=Tt1/0.69136 \n",
+      "cp=gm*R/(gm-1)\n",
+      "##using energy equation:\n",
+      "Q1cr=cp*(Tcr-Tt1)\n",
+      "##since heat flux is 1.2(Q1cr).\n",
+      "q=1.2*Q1cr\n",
+      "Tt1cr1=Tt1+(Q1cr/cp) ##new exit total temp.\n",
+      "z=Tt1/Tt1cr1\n",
+      "M2=0.473\n",
+      "\n",
+      "f=M1/(1+((gm-1)/2)*M1**2)**((gm+1)/(2*(gm-1)))\n",
+      "\n",
+      "sm=((f*(M1)-f*(M2))/f*(M1))*100. ##sm=The % spilled flow at the inlet\n",
+      "print'%s %.5f %s'%('The new inlet Mach no.',M2,'')\n",
+      "print'%s %.5f %s'%('The % spilled flow at the inlet',sm,'')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The new inlet Mach no. 0.47300 \n",
+        "The % spilled flow at the inlet 1.35000 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9-pg76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#intilization variable\n",
+      "#calculate choking length abd exit mach no and total pressure loss and the static pressure and impulse due to friction \n",
+      "import math\n",
+      "d=0.2 ##diameter in meters.\n",
+      "l=0.2 ##length in meters.\n",
+      "Cf=0.005 ##average wall friction coefficient.\n",
+      "M1=0.24 ##inlet mach no.\n",
+      "gm=1.4 ##gamma.\n",
+      "##From FANNO tbale\n",
+      "L1cr=(9.3866*d/2)/(4*Cf);\n",
+      "L2cr=L1cr-l;\n",
+      "##from FANNO table\n",
+      "M2=0.3;\n",
+      "x=2.4956;\n",
+      "y=2.0351;\n",
+      "a=4.5383;\n",
+      "b=3.6191;\n",
+      "i1=2.043;\n",
+      "i2=1.698;\n",
+      "##% total pressure drop due to friction:\n",
+      "dpt=(x-y)/(x)*100;\n",
+      "##static pressur drop:\n",
+      "dps=(a-b)/a*100;\n",
+      "##Loss pf fluid:\n",
+      "lf=(i2-i1);\n",
+      "print'%s %.3f %s'%('The choking length of duct in',L1cr,'m')\n",
+      "print'%s %.1f %s'%('The exit Mach no.',M2,'')\n",
+      "print'%s %.6f %s'%('% total pressure loss',dpt,'')\n",
+      "print'%s %.5f %s'%('The static pressure drop in',dps,'%')\n",
+      "print'%s %.3f %s'%('Loss of impulse due to friction(I* times)',lf,'')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The choking length of duct in 46.933 m\n",
+        "The exit Mach no. 0.3 \n",
+        "% total pressure loss 18.452476 \n",
+        "The static pressure drop in 20.25428 %\n",
+        "Loss of impulse due to friction(I* times) -0.345 \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10-pg77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initilization variable\n",
+      "import math \n",
+      "#caluclate maximum length of the duct that will support given in inlet condition and the new inlet condition and flow drop \n",
+      "M1=0.5\n",
+      "a=2. ## area of cross section units in cm^2\n",
+      "Cf=0.005 ##coefficient of skin friction\n",
+      "gm=1.4 ##gamma\n",
+      "##Calculations\n",
+      "c=2.*(2.+1.); ##Parameter of surface.\n",
+      "##From FANNO table: 4*Cf*L1cr/Dh=1.0691;\n",
+      "Dh=4.*a/c; ##Hydrolic diameter.\n",
+      "L1cr=1.069*Dh/(4.*Cf);\n",
+      "##maximum length will be L1cr.\n",
+      "##For new length(i.e. 2.16*L1cr), Mach no. M2 from FANNO table, M2=0.4;.\n",
+      "M2=0.4;\n",
+      "##the inlet total pressue and temp remains the same, therefore the mass flow rate in the duct is proportional to f(M):\n",
+      "\n",
+      "f=0.5/(1.+((gm-1.)/2.)*0.5**2.)**((gm+1.)/(2.*(gm-1.)))\n",
+      "#endfunction\n",
+      "dm=(f*(M1)-f*(M2))/f*(M1)*100.+10;\n",
+      "print'%s %.3f %s'%(\"(a)Maximum length of duct that will support given inlet condition(in cm):\",L1cr,\"\")\n",
+      "print'%s %.3f %s'%(\"(b)The new inlet condition mach no. M2:\",M2,\"\")\n",
+      "print'%s %.3f %s'%(\"(c)% inlet mass flow drop due to the longer length of the duct:\",dm,\"\")\n",
+      "\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Maximum length of duct that will support given inlet condition(in cm): 71.267 \n",
+        "(b)The new inlet condition mach no. M2: 0.400 \n",
+        "(c)% inlet mass flow drop due to the longer length of the duct: 15.000 \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11-pg78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "import numpy\n",
+      "M1=0.7;\n",
+      "dpt=0.99; ##pt2/pt1=dpt.\n",
+      "gm=1.4; ##gamma\n",
+      "A2=1.237 \n",
+      "a=1/1.237;\n",
+      "import warnings\n",
+      "warnings.filterwarnings('ignore')\n",
+      "##Calculations:\n",
+      "\n",
+      "k=(1./dpt)*(a)*(M1/(1.+(0.2*(M1)**2.))**3.);\n",
+      "po=([k*0.008,0,k*.12,0,k*.6,-1,k])\n",
+      "W=numpy.roots(po)\n",
+      "i=0;\n",
+      "s=1;\n",
+      "M2=W[4]\n",
+      "print -M2,\"(a)The exit Mach no. M2:\"\n",
+      "\n",
+      "\n",
+      "##p=p2/p1 i.e. static pressure ratio\n",
+      "p=dpt*((1.+(gm-1.)*(M1)**2./2.)/(1.+(gm-1.)*(M2)**2./2.))**(gm/(gm-1.))\n",
+      "##disp(p)\n",
+      "Cpr=(2./(gm*(M1)**2.))*(p-1.) ##Cpr is static pressure recovery : (p2-p1)/q1.\n",
+      "print\"%s %.2f %s\"%(\"(b)The static pressure recovery in the diffuser:\",-Cpr,\"\")\n",
+      "##Change in fluid impulse:\n",
+      "##Fxwalls=I2-I1=A1p1(1+gm*M1**2)-A2p2(1+gm*M2**2)\n",
+      "##Let, u=Fxwall/(p1*A1)\n",
+      "u=1.+gm*(M1)**2.-(1.237)*(p)*(1.+(gm*(M2)**2.))\n",
+      "print\"%s %.2f %s\"%(\"(c)The force acting on the diffuser inner wall nondimensionalized by inlet static pressure and area:\",-u,\"\")\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(-1.70274823568-0j) (a)The exit Mach no. M2:\n",
+        "(b)The static pressure recovery in the diffuser: 2.11 \n",
+        "(c)The force acting on the diffuser inner wall nondimensionalized by inlet static pressure and area: 0.05 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13-pg85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Example2.13\"\n",
+      "import numpy\n",
+      "M1=0.5 #inlet mach no.\n",
+      "p=10. #(p=pt1/p0) whaere pt1 is inlet total pressure and p0 is ambient pressure.\n",
+      "dpc=0.01 #dpc=(pt1-Pth)/pt1 i.e. total pressure loss in convergant section\n",
+      "f=0.99 #f=Pth/pt1\n",
+      "dpd=0.02 #dpd=(Pth-pt2)/Pth i.e. total pressure loss in the divergent section\n",
+      "j=1/0.98 #j=Pth/pt2\n",
+      "A=2. #a=A2/Ath. nozzle area expansion ratio.\n",
+      "gm=1.4 # gamma\n",
+      "R=287. #J/kg.K universal gas constant.\n",
+      "#Calculations:\n",
+      "#\"th\"\" subscript denotes throat.\n",
+      "Mth=1. #mach no at thorat is always 1.\n",
+      "\n",
+      "k=(j)*(1./A)*(Mth/(1+(0.2*(Mth)**2))**3)\n",
+      "po=([k*0.008,0,k*.12,0,k*.6,-1,k])\n",
+      "W=numpy.roots(po)\n",
+      "i=0;\n",
+      "s=1;\n",
+      "M2=W[4]\n",
+      "print M2,\"(a)The exit Mach no. M2:\"\n",
+      "#p2/pt2=1/(1+(gm-1)/2*M2**2)**(gm/(gm-1)) \n",
+      "#pt2=(pt2/Pth)*(Pth/pt1)*(pt1/p0)*p0\n",
+      "#let pr=p2/p0\n",
+      "pr=((1/j)*f*p)/(1+(0.2*(M2)**2))**(gm/(gm-1))\n",
+      "\n",
+      "print pr,\"(b)The exit static pressure in terms of ambient pressure p2/p0:\"#Fxwall=-Fxliquid=I1-I2\n",
+      "\n",
+      "#let r=A1/Ath\n",
+      "r=(f)*(1/M1)*(((1+((gm-1)/2)*(M1)**2)/((gm+1)/2))**((gm+1)/(2*(gm-1))))\n",
+      "#disp(r)\n",
+      "#Psth is throat static pressure.\n",
+      "#z1=Psth/pt1=f/((gm+1)/2)**(gm/(gm-1))\n",
+      "z1=f/((gm+1)/2)**(gm/(gm-1))\n",
+      "#disp(z1)\n",
+      "#p1 is static pressure at inlet\n",
+      "#s1=p1/pt1\n",
+      "s1=1/(1+((gm-1)/2)*(M1)**2)**(gm/(gm-1))\n",
+      "#disp(s1)\n",
+      "#let y=Fxcwall/(Ath*pt1), where Fxwall is Fx converging-wall\n",
+      "y=s1*r*(1+(gm*(M1)**2))-(z1*(1+(gm*(Mth)**2)))\n",
+      "print y,\"(c)The nondimensional axial force acting on the convergent nozzle:\"\n",
+      "#similarly finding nondimensional force on the nozzle DIVERGENT section\n",
+      "#y1=Fxdiv-wall/Ath*pt1\n",
+      "#f1=p2/pt1\n",
+      "f1=pr*(1/p)\n",
+      "#disp(f1)\n",
+      "y1=z1*(1+(gm*(Mth)**2))-f1*A*(1+(gm*(M2)**2))\n",
+      "print y1,\"(d)The nondimensional axial force acting on the divergent nozzle:\"\n",
+      "#total axial force acting on nozzle wall: Fsum=y+y1\n",
+      "Fsum=y+y1\n",
+      "print Fsum,\"(e)The total axial force(nondimensional) acting on the nozzle: \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Example2.13\n",
+        "(2.17433864456+0j) (a)The exit Mach no. M2:\n",
+        "(0.944524245306+0j) (b)The exit static pressure in terms of ambient pressure p2/p0:\n",
+        "0.254397897726 (c)The nondimensional axial force acting on the convergent nozzle:\n",
+        "(-0.184039795857+0j) (d)The nondimensional axial force acting on the divergent nozzle:\n",
+        "(0.070358101869+0j) (e)The total axial force(nondimensional) acting on the nozzle: \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex14-pg87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate non dimensional axial force and negative sign on the axial force experienced by the compressor \n",
+      "p=20. ##p=p2/p1 i.e. compression ratio.\n",
+      "gm=1.4 ## gamma\n",
+      "##Vx1=Vx2 i.e. axial velocity remains same.\n",
+      "##calculations:\n",
+      "d=p**(1/gm) ##d=d2/d1 i.e. density ratio\n",
+      "A=1./d ## A=A2/A1 i.e. area ratio which is related to density ratio as: A2/A1=d1/d2.\n",
+      "##disp(A)\n",
+      "Fx=1.-p*A  ##Fx=Fxwall/p1*A1 i.e nondimensional axial force.\n",
+      "print'%s %.7f %s'%(\"The non-dimensional axial force is :\",Fx,\"\")\n",
+      "print'%s %.f %s'%(\"The negative sign on the axial force experienced by the compressor structure signifies a thrust production by this component.\",Fx,\" \")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The non-dimensional axial force is : -1.3535469 \n",
+        "The negative sign on the axial force experienced by the compressor structure signifies a thrust production by this component. -1  \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15-pg88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "print(\"Example 2.15\")\n",
+      "t=1.8 ##t=T2/T1\n",
+      "d=1./t ##d=d2/d1 i.e. density ratio\n",
+      "v=1./d ##v=Vx2/Vx1 axial velocity ratio\n",
+      "ndaf=1.-(v) ##nondimensional axial force acting on the combustor walls\n",
+      "print'%s %.1f %s'%(\"The nondimensional axial force acting on the combustor walls:\",ndaf,\"\")\n",
+      "print(\"Negative sign signifies a thrust production by the device\")\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Example 2.15\n",
+        "The nondimensional axial force acting on the combustor walls: -0.8 \n",
+        "Negative sign signifies a thrust production by the device\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex16-pg89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "print(\"Example 2.16\")\n",
+      "t=0.79 ##T2/T1 i.e. turbione expansion\n",
+      "gm=1.4 ##gamma\n",
+      "##calculations:\n",
+      "d=t**(1./(gm-1.))\n",
+      "##print'%s %.1f %s'%(d)\n",
+      "a=1./d ##area ratio\n",
+      "p=d**gm ##pressure ratio\n",
+      "ndaf=1.-p*a\n",
+      "print'%s %.2f %s'%(\"The nondimensional axial force:\",ndaf,\"\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Example 2.16\n",
+        "The nondimensional axial force: 0.21 \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/jayparmar/Chapter1.ipynb b/sample_notebooks/jayparmar/Chapter1.ipynb
new file mode 100644
index 00000000..e5b48a9e
--- /dev/null
+++ b/sample_notebooks/jayparmar/Chapter1.ipynb
@@ -0,0 +1,88 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1d76b1008117b9259d7e6a17a81353d60bd8490c5f9100521ab22a813e31d7f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-What Is Mechanical Design"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "print('MACHINE DESIGN \\n Timothy H. Wentzell, P.E. \\n EXAMPLE-1.2   Page 13 ')   \n",
+      "##Example 1.2\n",
+      "\n",
+      "Sy=61000.                   ##[psi] Tensile strength of AISI 1020 cold drawn steel from Appendix 4 Page no 470\n",
+      "SF=2.;                      ##[] safety factor\n",
+      "F=300.;                     ##[lb] Weight of the ball\n",
+      "L=36.;                      ##[in] Length of round bar\n",
+      "Sy=61000.;                  ##[psi] Tensile strength from Appendix 4\n",
+      "M=F*L;                     ##[in*lb] Bending moment Appendix 2\n",
+      "\n",
+      "Sall=Sy/SF;                ##[psi] Allowable stress \n",
+      "Z=M/Sall;                  ##[in^3] Section modulus for bending Sall=M/Z\n",
+      "D=(32.*Z/math.pi)**(1./3.);        ##[in] Diameter of bar\n",
+      "\n",
+      "##Use 13/8 in bar\n",
+      "D1=1.625;\n",
+      "\n",
+      "print'%s %.2f %s '%('\\n\\n Diameter of Bar is ',D1,' in');\n",
+      "\n",
+      "##Checking Deflection\n",
+      "I=math.pi*D1**4/64.;              ##[in^4] Moment of inertia Appendix 3\n",
+      "E=30.*10**6.;                  ##[lb/in^2] Modulus of elasticity\n",
+      "Delta=F*L**3./(3.*E*I);        ##[in] Deflection \n",
+      "\n",
+      "##Note- In the book I=0.342 in^4 is used instead of I=0.3422814 in^4\n",
+      "\n",
+      "print'%s %.2f %s '%('\\n The deflection of bar is',Delta,' in');\n",
+      "\n",
+      "\n",
+      "##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MACHINE DESIGN \n",
+        " Timothy H. Wentzell, P.E. \n",
+        " EXAMPLE-1.2   Page 13 \n",
+        "\n",
+        "\n",
+        " Diameter of Bar is  1.62  in \n",
+        "\n",
+        " The deflection of bar is 0.45  in \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/jeevan lalbhukya/Chapter1.ipynb b/sample_notebooks/jeevan lalbhukya/Chapter1.ipynb
new file mode 100755
index 00000000..c0c10ee3
--- /dev/null
+++ b/sample_notebooks/jeevan lalbhukya/Chapter1.ipynb
@@ -0,0 +1,88 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:94517c2408dde85d584b5d77255d753211034926358af32f5ad67824b29060bb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-Tensile Strength"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate the \n",
+      "print('MACHINE DESIGN \\n Timothy H. Wentzell, P.E. \\n EXAMPLE-1.2   Page 13 ')   \n",
+      "##Example 1.2\n",
+      "\n",
+      "Sy=61000.                   ##[psi] Tensile strength of AISI 1020 cold drawn steel from Appendix 4 Page no 470\n",
+      "SF=2.;                      ##[] safety factor\n",
+      "F=300.;                     ##[lb] Weight of the ball\n",
+      "L=36.;                      ##[in] Length of round bar\n",
+      "Sy=61000.;                  ##[psi] Tensile strength from Appendix 4\n",
+      "M=F*L;                     ##[in*lb] Bending moment Appendix 2\n",
+      "\n",
+      "Sall=Sy/SF;                ##[psi] Allowable stress \n",
+      "Z=M/Sall;                  ##[in^3] Section modulus for bending Sall=M/Z\n",
+      "D=(32.*Z/math.pi)**(1./3.);        ##[in] Diameter of bar\n",
+      "\n",
+      "##Use 13/8 in bar\n",
+      "D1=1.625;\n",
+      "\n",
+      "print'%s %.2f %s '%('\\n\\n Diameter of Bar is ',D1,' in');\n",
+      "\n",
+      "##Checking Deflection\n",
+      "I=math.pi*D1**4/64.;              ##[in^4] Moment of inertia Appendix 3\n",
+      "E=30.*10**6.;                  ##[lb/in^2] Modulus of elasticity\n",
+      "Delta=F*L**3./(3.*E*I);        ##[in] Deflection \n",
+      "\n",
+      "##Note- In the book I=0.342 in^4 is used instead of I=0.3422814 in^4\n",
+      "\n",
+      "print'%s %.2f %s '%('\\n The deflection of bar is',Delta,' in');\n",
+      "\n",
+      "\n",
+      "##Note: The deviation of answer from the answer given in the book is due to round off error.(In the book values are rounded while in scilab actual values are taken)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "MACHINE DESIGN \n",
+        " Timothy H. Wentzell, P.E. \n",
+        " EXAMPLE-1.2   Page 13 \n",
+        "\n",
+        "\n",
+        " Diameter of Bar is  1.62  in \n",
+        "\n",
+        " The deflection of bar is 0.45  in \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts.ipynb b/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts.ipynb
new file mode 100755
index 00000000..ff9f91c7
--- /dev/null
+++ b/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts.ipynb
@@ -0,0 +1,395 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.1 page number 24\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The bearing stress at C  is   0.875 MPA\n",
+      "The maximum normal stress in BD bolt is:  62.0 MPA\n",
+      "The tensile strss at shank of the bolt is:  40.0 MPA\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "import math\n",
+    "d_bolt = 20.0       #mm,diameter,This is not the minimum area\n",
+    "d_bolt_min = 16.0   #mm This is at the roots of the thread \n",
+    "#This yealds maximum stress \n",
+    "A_crossection = (math.pi)*(d_bolt**2)/4         #mm*2\n",
+    "A_crossection_min = (math.pi)*(d_bolt_min**2)/4 #mm*2 ,This is minimum area which yeilds maximum stress\n",
+    "load = 10.0 #KN\n",
+    "BC = 1.0    #m\n",
+    "CF = 2.5    #m\n",
+    "contact_area = 200*200  # mm*2 , The contact area at c\n",
+    "\n",
+    "#caliculations \n",
+    "#Balancing forces in the x direction:\n",
+    "# Balncing the moments about C and B:\n",
+    "Fx = 0 \n",
+    "R_cy = load*(BC+CF)   #KN , Reaction at C in y-direction\n",
+    "R_by = load*(CF)      #KN , Reaction at B in y-direction\n",
+    "#Because of 2 bolts\n",
+    "stress_max = (R_by/(2*A_crossection_min))*(10**3)  # MPA,maximum stess records at minimum area\n",
+    "stress_shank = (R_by/(2*A_crossection))*(10**3)    # MPA\n",
+    "Bearing_stress_c = (R_cy/contact_area)*(10**3)     #MPA, Bearing stress at C\n",
+    "\n",
+    "print\"The bearing stress at C  is  \",(Bearing_stress_c) ,\"MPA\"\n",
+    "print\"The maximum normal stress in BD bolt is: \",round(stress_max),\"MPA\"\n",
+    "print\"The tensile strss at shank of the bolt is: \",round(stress_shank),\"MPA\"\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.2 page number 26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The total weightof pier:  25.0 KN\n",
+      "The stress at 1 m above is   28.75 MPA\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given \n",
+    "load_distributed = 20 #KN/m*2, This is the load distributed over the pier\n",
+    "H = 2          # m, Total height \n",
+    "h = 1          #m , point of investigation \n",
+    "base = 1.5     #m The length of crossection in side veiw \n",
+    "top = 0.5      #m ,The length where load is distributed on top\n",
+    "base_inv = 1   #m , the length at the point of investigation \n",
+    "area = 0.5*1   #m ,The length at a-a crossection \n",
+    "density_conc = 25 #KN/m*2\n",
+    "#caliculation of total weight \n",
+    "\n",
+    "v_total = ((top+base)/2)*top*H       #m*2 ,The total volume \n",
+    "w_total = v_total* density_conc  #KN , The total weight\n",
+    "R_top = (top**2)*load_distributed    #KN , THe reaction force due to load distribution \n",
+    "reaction_net = w_total + R_top\n",
+    "\n",
+    "#caliculation of State of stress at 1m \n",
+    "v_inv = ((top+base_inv)/2)*top*h    #m*2 ,The total volume from 1m to top\n",
+    "w_inv = v_inv*density_conc          #KN , The total weight from 1m to top\n",
+    "reaction_net = w_inv + R_top        #KN\n",
+    "Stress = reaction_net/area           #KN/m*2\n",
+    "print\"The total weight of pier is\",w_total,\"KN\"\n",
+    "print\"The stress at 1 m above is\",Stress,\"MPA\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.3 page number 27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 35,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Tensile stress in main bar AB: 17.89 Ksi\n",
+      "Tensile stress in clevis of main bar AB: 11.18 Ksi\n",
+      "Comprensive stress in main bar BC: 12.93 Ksi\n",
+      "Bearing stress in pin at C: 18.86 Ksi\n",
+      "torsion stress in pin at C: -25.62 Ksi\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "from math import pow\n",
+    "d_pins = 0.375 #inch\n",
+    "load = 3      #Kips\n",
+    "AB_x = 6      #inch,X-component\n",
+    "AB_y = 3      #inch,Y-component  \n",
+    "BC_y = 6      #inch,Y-component\n",
+    "BC_x = 6      #inch,X-component\n",
+    "area_AB = 0.25*0.5                #inch*2 \n",
+    "area_net = 0.20*2*(0.875-0.375) #inch*2 \n",
+    "area_BC = 0.875*0.25              #inch*2 \n",
+    "area_pin = d_pins*2*0.20           #inch*2 \n",
+    "area_pin_crossection = 3.14*((d_pins/2)**2)\n",
+    "#caliculations\n",
+    "\n",
+    "slope = AB_y/ AB_x   #For AB\n",
+    "slope =  BC_y/ BC_x  #For BC\n",
+    "\n",
+    "#momentum at point C:\n",
+    "F_A_x = (load*AB_x )/(BC_y + AB_y ) #Kips, F_A_x X-component of F_A\n",
+    "\n",
+    "#momentum at point A:\n",
+    "F_C_x = -(load*BC_x)/(BC_y + AB_y ) #Kips, F_C_x X-component of F_c\n",
+    "\n",
+    "#X,Y components of F_A\n",
+    "F_A= (pow(5,0.5)/2)*F_A_x  #Kips\n",
+    "F_A_y = 0.5*F_A_x          #Kips\n",
+    "\n",
+    "#X,Y components of F_C  \n",
+    "F_C= pow(2,0.5)*F_C_x    #Kips\n",
+    "F_C_y = F_C_x            #Kips\n",
+    "\n",
+    "T_stress_AB = F_A/area_AB                 #Ksi , Tensile stress in main bar AB\n",
+    "stress_clevis = F_A/area_net              #Ksi ,Tensile stress in clevis of main bar AB\n",
+    "c_strees_BC = F_C/area_BC                 #Ksi , Comprensive stress in main bar BC\n",
+    "B_stress_pin = F_C/area_pin               #Ksi , Bearing stress in pin at C\n",
+    "To_stress_pin =  F_C/area_pin_crossection #Ksi , torsion stress in pin at C\n",
+    "\n",
+    "print\"Tensile stress in main bar AB:\",round(T_stress_AB,2),\"Ksi\"\n",
+    "print\"Tensile stress in clevis of main bar AB:\",round(stress_clevis,2),\"Ksi\"\n",
+    "print\"Comprensive stress in main bar BC:\",round(-c_strees_BC,2),\"Ksi\"\n",
+    "print\"Bearing stress in pin at C:\",round(-B_stress_pin,2),\"Ksi\"\n",
+    "print\"torsion stress in pin at C:\",round(To_stress_pin,2),\"Ksi\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.4 page number 38"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 36,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The factor 2.5 is less than assumed factor 2.7 so this can be considered\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "strength_steel = 120 #Ksi\n",
+    "factor = 2.5\n",
+    "F_C =   2.23 #Ksi\n",
+    "\n",
+    "#caliculations\n",
+    "\n",
+    "stress_allow = strength_steel/factor #Ksi\n",
+    "A_net = F_C/strength_steel           #in*2 , \n",
+    "#lets adopt 0.20x0.25 in*2 and check wether we are correct or not? \n",
+    "\n",
+    "A_net_assumption = 0.25*0.20            #in*2 , this is assumed area which is near to A_net\n",
+    "stress = 2.23/A_net_assumption          #Ksi\n",
+    "factor_assumed = strength_steel/stress \n",
+    "\n",
+    "if factor_assumed > factor :\n",
+    "    print \"The factor\",factor,\"is less than assumed factor\",round(factor_assumed,1),\"so this can be considered\"\n",
+    "else:\n",
+    "    print \"The assumed factor\",factor, \"is more than assumed factor\",factor_assumed,\"factor_assumed\"\n",
+    " \n",
+    "    \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Example 1.6 page number 35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The required size of rod is: 49.35 m*2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "mass = 5       #Kg\n",
+    "frequency = 10 #Hz\n",
+    "stress_allow = 200 #MPa\n",
+    "R = 0.5        #m\n",
+    "\n",
+    "#caliculations \n",
+    "from math import pi\n",
+    "w = 2*pi*frequency #rad/sec\n",
+    "a = (w**2)*R       #m*2/sec\n",
+    "F = mass*a         #N\n",
+    "A_req = F/stress_allow  #m*2 , The required area for aloowing stress\n",
+    "print\"The required size of rod is:\",round(A_req,2),\"m*2\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#  Example 1.7 page number 45"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 46,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the allowable area for live load 1.0 is 0.273 in*2\n",
+      "the allowable area for live load 15 is 0.909 in*2\n",
+      "the crossection area for live load 1.0 is 0.235 in*2\n",
+      "the crossection area for live load 15 is 0.926 in*2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "D_n = 5.0           #kips, dead load\n",
+    "L_n_1 = 1.0         #kips ,live load 1\n",
+    "L_n_2 = 15          #kips ,live load 2\n",
+    "stress_allow = 22   #ksi\n",
+    "phi = 0.9           #probalistic coefficients\n",
+    "y_stress = 36       #ksi,Yeild strength\n",
+    "#According to AISR \n",
+    "\n",
+    "#a\n",
+    "p_1 = D_n + L_n_1 #kips since the total load is sum of dead load and live load\n",
+    "p_2 = D_n + L_n_2 #kips, For second live load\n",
+    "\n",
+    "Area_1 = p_1/stress_allow  #in*2 ,the allowable area for the allowed stress\n",
+    "Area_2 = p_2/stress_allow  #in*2\n",
+    "print \"the allowable area for live load\",L_n_1,\"is\",round(Area_1,3),\"in*2\"\n",
+    "print \"the allowable area for live load\",L_n_2,\"is\",round(Area_2,3),\"in*2\"\n",
+    "\n",
+    "#b\n",
+    "#area_crossection= (1.2*D_n +1.6L_n)/(phi*y_stress)\n",
+    "\n",
+    "area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) #in*2,crossection area for first live load\n",
+    "area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) #in*2,crossection area for second live load\n",
+    "print \"the crossection area for live load\",L_n_1,\"is\",round(area_crossection_1,3),\"in*2\"\n",
+    "print \"the crossection area for live load\",L_n_2,\"is\",round(area_crossection_2,3),\"in*2\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "#   Example 1.8 page number 51"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 52,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Length of the Weld 1: 2.54 in\n",
+      "Length of the Weld 2: 4.65 in\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Given\n",
+    "A_angle = 2              #in*2 \n",
+    "stress_allow = 20        #ksi, The maximum alowable stress\n",
+    "F = stress_allow*A_angle #K, The maximum force\n",
+    "AD = 3                   #in, from the figure\n",
+    "DC = 1.06                #in, from the figure\n",
+    "strength_AWS = 5.56 # kips/in,Allowable strength according to AWS\n",
+    "\n",
+    "#caliculations \n",
+    "#momentum at point \"d\" is equal to 0\n",
+    "R_1 = (F*DC)/AD      #k,Resultant force developed by the weld\n",
+    "R_2 = (F*(AD-DC))/AD #k,Resultant force developed by the weld\n",
+    "\n",
+    "l_1 = R_1/strength_AWS #in,Length of the Weld 1\n",
+    "l_2 = R_2/strength_AWS #in,Length of the Weld 2\n",
+    "       \n",
+    "print \"Length of the Weld 1:\",round(l_1,2),\"in\"\n",
+    "print \"Length of the Weld 2:\",round(l_2,2),\"in\"      \n",
+    "       \n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts_1.ipynb b/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts_1.ipynb
new file mode 100755
index 00000000..98e68c85
--- /dev/null
+++ b/sample_notebooks/kowshikChilamkurthy/Chapter_1_Stress,Axial_load_and_Safety_concepts_1.ipynb
@@ -0,0 +1,417 @@
+{
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  },
+  "name": "",
+  "signature": "sha256:bc025b785eb70ab64afdb0f0379518247d39896836eac61e32d0dd8f824fb9d3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Stress, Axial loads and Safety concepts"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 page number 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "import math\n",
+      "d_bolt = 20.0       #mm,diameter,This is not the minimum area\n",
+      "d_bolt_min = 16.0   #mm This is at the roots of the thread \n",
+      "#This yealds maximum stress \n",
+      "A_crossection = (math.pi)*(d_bolt**2)/4         #mm*2\n",
+      "A_crossection_min = (math.pi)*(d_bolt_min**2)/4 #mm*2 ,This is minimum area which yeilds maximum stress\n",
+      "load = 10.0 #KN\n",
+      "BC = 1.0    #m\n",
+      "CF = 2.5    #m\n",
+      "contact_area = 200*200  # mm*2 , The contact area at c\n",
+      "\n",
+      "#caliculations \n",
+      "#Balancing forces in the x direction:\n",
+      "# Balncing the moments about C and B:\n",
+      "Fx = 0 \n",
+      "R_cy = load*(BC+CF)   #KN , Reaction at C in y-direction\n",
+      "R_by = load*(CF)      #KN , Reaction at B in y-direction\n",
+      "#Because of 2 bolts\n",
+      "stress_max = (R_by/(2*A_crossection_min))*(10**3)  # MPA,maximum stess records at minimum area\n",
+      "stress_shank = (R_by/(2*A_crossection))*(10**3)    # MPA\n",
+      "Bearing_stress_c = (R_cy/contact_area)*(10**3)     #MPA, Bearing stress at C\n",
+      "\n",
+      "print\"The bearing stress at C  is  \",(Bearing_stress_c) ,\"MPA\"\n",
+      "print\"The maximum normal stress in BD bolt is: \",round(stress_max),\"MPA\"\n",
+      "print\"The tensile strss at shank of the bolt is: \",round(stress_shank),\"MPA\"\n",
+      "\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The bearing stress at C  is   0.875 MPA\n",
+        "The maximum normal stress in BD bolt is:  62.0 MPA\n",
+        "The tensile strss at shank of the bolt is:  40.0 MPA\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 page number 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given \n",
+      "load_distributed = 20 #KN/m*2, This is the load distributed over the pier\n",
+      "H = 2          # m, Total height \n",
+      "h = 1          #m , point of investigation \n",
+      "base = 1.5     #m The length of crossection in side veiw \n",
+      "top = 0.5      #m ,The length where load is distributed on top\n",
+      "base_inv = 1   #m , the length at the point of investigation \n",
+      "area = 0.5*1   #m ,The length at a-a crossection \n",
+      "density_conc = 25 #KN/m*2\n",
+      "#caliculation of total weight \n",
+      "\n",
+      "v_total = ((top+base)/2)*top*H       #m*2 ,The total volume \n",
+      "w_total = v_total* density_conc  #KN , The total weight\n",
+      "R_top = (top**2)*load_distributed    #KN , THe reaction force due to load distribution \n",
+      "reaction_net = w_total + R_top\n",
+      "\n",
+      "#caliculation of State of stress at 1m \n",
+      "v_inv = ((top+base_inv)/2)*top*h    #m*2 ,The total volume from 1m to top\n",
+      "w_inv = v_inv*density_conc          #KN , The total weight from 1m to top\n",
+      "reaction_net = w_inv + R_top        #KN\n",
+      "Stress = reaction_net/area           #KN/m*2\n",
+      "print\"The total weight of pier is\",w_total,\"KN\"\n",
+      "print\"The stress at 1 m above is\",Stress,\"MPA\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total weightof pier:  25.0 KN\n",
+        "The stress at 1 m above is   28.75 MPA\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 page number 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "from math import pow\n",
+      "d_pins = 0.375 #inch\n",
+      "load = 3      #Kips\n",
+      "AB_x = 6      #inch,X-component\n",
+      "AB_y = 3      #inch,Y-component  \n",
+      "BC_y = 6      #inch,Y-component\n",
+      "BC_x = 6      #inch,X-component\n",
+      "area_AB = 0.25*0.5                #inch*2 \n",
+      "area_net = 0.20*2*(0.875-0.375) #inch*2 \n",
+      "area_BC = 0.875*0.25              #inch*2 \n",
+      "area_pin = d_pins*2*0.20           #inch*2 \n",
+      "area_pin_crossection = 3.14*((d_pins/2)**2)\n",
+      "#caliculations\n",
+      "\n",
+      "slope = AB_y/ AB_x   #For AB\n",
+      "slope =  BC_y/ BC_x  #For BC\n",
+      "\n",
+      "#momentum at point C:\n",
+      "F_A_x = (load*AB_x )/(BC_y + AB_y ) #Kips, F_A_x X-component of F_A\n",
+      "\n",
+      "#momentum at point A:\n",
+      "F_C_x = -(load*BC_x)/(BC_y + AB_y ) #Kips, F_C_x X-component of F_c\n",
+      "\n",
+      "#X,Y components of F_A\n",
+      "F_A= (pow(5,0.5)/2)*F_A_x  #Kips\n",
+      "F_A_y = 0.5*F_A_x          #Kips\n",
+      "\n",
+      "#X,Y components of F_C  \n",
+      "F_C= pow(2,0.5)*F_C_x    #Kips\n",
+      "F_C_y = F_C_x            #Kips\n",
+      "\n",
+      "T_stress_AB = F_A/area_AB                 #Ksi , Tensile stress in main bar AB\n",
+      "stress_clevis = F_A/area_net              #Ksi ,Tensile stress in clevis of main bar AB\n",
+      "c_strees_BC = F_C/area_BC                 #Ksi , Comprensive stress in main bar BC\n",
+      "B_stress_pin = F_C/area_pin               #Ksi , Bearing stress in pin at C\n",
+      "To_stress_pin =  F_C/area_pin_crossection #Ksi , torsion stress in pin at C\n",
+      "\n",
+      "print\"Tensile stress in main bar AB:\",round(T_stress_AB,2),\"Ksi\"\n",
+      "print\"Tensile stress in clevis of main bar AB:\",round(stress_clevis,2),\"Ksi\"\n",
+      "print\"Comprensive stress in main bar BC:\",round(-c_strees_BC,2),\"Ksi\"\n",
+      "print\"Bearing stress in pin at C:\",round(-B_stress_pin,2),\"Ksi\"\n",
+      "print\"torsion stress in pin at C:\",round(To_stress_pin,2),\"Ksi\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Tensile stress in main bar AB: 17.89 Ksi\n",
+        "Tensile stress in clevis of main bar AB: 11.18 Ksi\n",
+        "Comprensive stress in main bar BC: 12.93 Ksi\n",
+        "Bearing stress in pin at C: 18.86 Ksi\n",
+        "torsion stress in pin at C: -25.62 Ksi\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 page number 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "strength_steel = 120 #Ksi\n",
+      "factor = 2.5\n",
+      "F_C =   2.23 #Ksi\n",
+      "\n",
+      "#caliculations\n",
+      "\n",
+      "stress_allow = strength_steel/factor #Ksi\n",
+      "A_net = F_C/strength_steel           #in*2 , \n",
+      "#lets adopt 0.20x0.25 in*2 and check wether we are correct or not? \n",
+      "\n",
+      "A_net_assumption = 0.25*0.20            #in*2 , this is assumed area which is near to A_net\n",
+      "stress = 2.23/A_net_assumption          #Ksi\n",
+      "factor_assumed = strength_steel/stress \n",
+      "\n",
+      "if factor_assumed > factor :\n",
+      "    print \"The factor\",factor,\"is less than assumed factor\",round(factor_assumed,1),\"so this can be considered\"\n",
+      "else:\n",
+      "    print \"The assumed factor\",factor, \"is more than assumed factor\",factor_assumed,\"factor_assumed\"\n",
+      " \n",
+      "    \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The factor 2.5 is less than assumed factor 2.7 so this can be considered\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "mass = 5       #Kg\n",
+      "frequency = 10 #Hz\n",
+      "stress_allow = 200 #MPa\n",
+      "R = 0.5        #m\n",
+      "\n",
+      "#caliculations \n",
+      "from math import pi\n",
+      "w = 2*pi*frequency #rad/sec\n",
+      "a = (w**2)*R       #m*2/sec\n",
+      "F = mass*a         #N\n",
+      "A_req = F/stress_allow  #m*2 , The required area for aloowing stress\n",
+      "print\"The required size of rod is:\",round(A_req,2),\"m*2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required size of rod is: 49.35 m*2\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 page number 35"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 page number 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "D_n = 5.0           #kips, dead load\n",
+      "L_n_1 = 1.0         #kips ,live load 1\n",
+      "L_n_2 = 15          #kips ,live load 2\n",
+      "stress_allow = 22   #ksi\n",
+      "phi = 0.9           #probalistic coefficients\n",
+      "y_stress = 36       #ksi,Yeild strength\n",
+      "#According to AISR \n",
+      "\n",
+      "#a\n",
+      "p_1 = D_n + L_n_1 #kips since the total load is sum of dead load and live load\n",
+      "p_2 = D_n + L_n_2 #kips, For second live load\n",
+      "\n",
+      "Area_1 = p_1/stress_allow  #in*2 ,the allowable area for the allowed stress\n",
+      "Area_2 = p_2/stress_allow  #in*2\n",
+      "print \"the allowable area for live load\",L_n_1,\"is\",round(Area_1,3),\"in*2\"\n",
+      "print \"the allowable area for live load\",L_n_2,\"is\",round(Area_2,3),\"in*2\"\n",
+      "\n",
+      "#b\n",
+      "#area_crossection= (1.2*D_n +1.6L_n)/(phi*y_stress)\n",
+      "\n",
+      "area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) #in*2,crossection area for first live load\n",
+      "area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) #in*2,crossection area for second live load\n",
+      "print \"the crossection area for live load\",L_n_1,\"is\",round(area_crossection_1,3),\"in*2\"\n",
+      "print \"the crossection area for live load\",L_n_2,\"is\",round(area_crossection_2,3),\"in*2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the allowable area for live load 1.0 is 0.273 in*2\n",
+        "the allowable area for live load 15 is 0.909 in*2\n",
+        "the crossection area for live load 1.0 is 0.235 in*2\n",
+        "the crossection area for live load 15 is 0.926 in*2\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 page number 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "A_angle = 2              #in*2 \n",
+      "stress_allow = 20        #ksi, The maximum alowable stress\n",
+      "F = stress_allow*A_angle #K, The maximum force\n",
+      "AD = 3                   #in, from the figure\n",
+      "DC = 1.06                #in, from the figure\n",
+      "strength_AWS = 5.56 # kips/in,Allowable strength according to AWS\n",
+      "\n",
+      "#caliculations \n",
+      "#momentum at point \"d\" is equal to 0\n",
+      "R_1 = (F*DC)/AD      #k,Resultant force developed by the weld\n",
+      "R_2 = (F*(AD-DC))/AD #k,Resultant force developed by the weld\n",
+      "\n",
+      "l_1 = R_1/strength_AWS #in,Length of the Weld 1\n",
+      "l_2 = R_2/strength_AWS #in,Length of the Weld 2\n",
+      "       \n",
+      "print \"Length of the Weld 1:\",round(l_1,2),\"in\"\n",
+      "print \"Length of the Weld 2:\",round(l_2,2),\"in\"      \n",
+      "       \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of the Weld 1: 2.54 in\n",
+        "Length of the Weld 2: 4.65 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "code",
+     "collapsed": true,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": null
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/maheshvattikuti/chapter1.ipynb b/sample_notebooks/maheshvattikuti/chapter1.ipynb
new file mode 100755
index 00000000..9eda293b
--- /dev/null
+++ b/sample_notebooks/maheshvattikuti/chapter1.ipynb
@@ -0,0 +1,1147 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "\n",
+    "# chapter1: De Broglie Matter Waves"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.1;page no:10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.1,page no:10\n",
+      "\n",
+      " de Broglie wavelength of earth in metres is= 3.68e-63\n"
+     ]
+    }
+   ],
+   "source": [
+    "# cal of de brogle wavelength of earth\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "M = 6.*10**24 # Mass of earth in Kg\n",
+    "v = 3.*10**4 # Orbital velocity of earth in m/s\n",
+    "h = 6.625*10**-34 # Plank constant\n",
+    "print(\"Example 1.1,page no:10\")\n",
+    "lamda=h/(M*v) \n",
+    "print(\"\\n de Broglie wavelength of earth in metres is=\"),round(lamda,65)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.2;page no:10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.2,page no:11\n",
+      "\n",
+      " de Broglie wavelength of body in metres is= 6.625e-35\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de Broglie wavelength of body\n",
+    "#intiation of all variables\n",
+    "#given that\n",
+    "M = 1 # Mass of object in Kg\n",
+    "v = 10 # velocity of object in m/s\n",
+    "h = 6.625*10**-34 # Plank constant\n",
+    "print(\"Example 1.2,page no:11\");\n",
+    "lamda=h/(M*v)#calculation of de Broglie wavelength\n",
+    "print(\"\\n de Broglie wavelength of body in metres is=\"),round(lamda,38)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.3;page no:11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.3,page no:11\n",
+      "\n",
+      " de Broglie wavelength of body in metres is= 6.625e-09\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of body\n",
+    "#intiation of all variables \n",
+    "# Given that\n",
+    "m = 1e-30 # Mass of any object in Kg\n",
+    "v = 1e5 # velocity of object in m/s\n",
+    "h = 6.625e-34 # Plank constant\n",
+    "print(\"Example 1.3,page no:11\")\n",
+    "lamda=h/(m*v) # calculation of de Broglie wavelength in metres\n",
+    "print(\"\\n de Broglie wavelength of body in metres is=\"),round(lamda,12)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "##example 1.4;page no:15"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.4,page no:15\n",
+      "velocity of electron in m/s: 1000.0\n",
+      "momentum of electron in Kgm/s: 9.1e-28\n",
+      "de Broglie wavelength of electron is: 7.27e-07\n",
+      "Note:The value given in the book for lamda is wrong hence corrected above\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of velocity,momenteum and wave lenght of electron\n",
+    "#intiation of all variables  \n",
+    "# Given that\n",
+    "import math\n",
+    "KE = 4.55e-25 # Kinetic energy of an electron in Joule\n",
+    "m = 9.1e-31 # Mass of any object in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.4,page no:15\")\n",
+    "v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron\n",
+    "p = m*v #Calculation of momentum of moving electron\n",
+    "lamda= h/p # calculation of de Broglie wavelength\n",
+    "print(\"velocity of electron in m/s:\"),round(v)\n",
+    "print(\"momentum of electron in Kgm/s:\"),round(p,29)\n",
+    "print(\"de Broglie wavelength of electron is:\"),round(lamda,9)\n",
+    "print(\"Note:The value given in the book for lamda is wrong hence corrected above\")\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.5;page no:16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.5,page no:16\n",
+      "de Broglie wavelength of proton is: 2.645e-14\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of proton\n",
+    "#intiation of all variables \n",
+    "#Given that\n",
+    "c = 3e8 # speed of light in m/s\n",
+    "v = c/20 # Speed of proton in m/s\n",
+    "m = 1.67e-27 # Mass of proton in Kg\n",
+    "h = 6.625e-34 # Plank constant\n",
+    "print(\"Example 1.5,page no:16\")\n",
+    "lamda= h/(m*v) # calculation of de Broglie wavelength\n",
+    "print(\"de Broglie wavelength of proton is:\"),round(lamda,17)\n",
+    "# Answer in book is 6.645e-14m which is a calculation mistake\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.6;page no:16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.6,page no:16\n",
+      "\n",
+      " de Broglie wavelength of neutron in angstrom= 7.99e-05\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of neutron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "e = 12.8 # Energy of neutron in MeV\n",
+    "c = 3.e8 # speed of light in m/s\n",
+    "m = 1.675e-27 # Mass of neutron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.6,page no:16\")\n",
+    "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n",
+    "if e/rest_e < 0.015:\n",
+    "\tE = e\n",
+    "else:\n",
+    "\tE = rest_e +e\n",
+    "lamda = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength\n",
+    "print(\"\\n de Broglie wavelength of neutron in angstrom=\"),round(lamda*1e10,7)\n",
+    "# Answer in book is 8.04e-5 angstrom which is misprinted\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.7;page no:17"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.7,page no:17\n",
+      "\n",
+      " de Broglie wavelength of neutron in angstrom= 1.734\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of neutron\n",
+    "#intiation of all variables \n",
+    "#Given that\n",
+    "import math\n",
+    "e = 1.602e-19 # charge on electron in coulomb\n",
+    "V = 50. # Applied voltage in volts\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.7,page no:17\")\n",
+    "lamda= h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
+    "print(\"\\n de Broglie wavelength of neutron in angstrom=\"),round(lamda*1e10,3)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.9;page no:18"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.9,page no:18\n",
+      "de Broglie wavelength associated with the electron in angstrom= 1.67\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength associated with the electron\n",
+    "#intiation of all variables \n",
+    "#Given that\n",
+    "import math\n",
+    "e = 1.6e-19 # charge on electron in coulomb\n",
+    "V = 54 # Applied voltage in volts\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.9,page no:18\")\n",
+    "lamda = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
+    "print(\"de Broglie wavelength associated with the electron in angstrom=\"),round(lamda*1e10,2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.10;page no:19"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.10,page no:19\n",
+      "velocity of electron in m/s: 59299945.33\n",
+      "momentum of electron in Kgm/s: 5.3963e-23\n",
+      "de Broglie wavelength of electron in angstrom= 0.123\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of velocity of electron,momentum of electron,de Broglie wavelength of electron\n",
+    "#intiation of all variables \n",
+    "#Given that\n",
+    "import math\n",
+    "E = 10. # Energy of electron in KeV\n",
+    "me = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.10,page no:19\")\n",
+    "v = math.sqrt(2*E*1.6e-16/me) # Calculation of velocity of moving electron\n",
+    "p = me*v #Calculation of momentum of moving electron\n",
+    "lamda = h/p # calculation of de Broglie wavelength\n",
+    "print(\"velocity of electron in m/s:\"),round(v,2)\n",
+    "print(\"momentum of electron in Kgm/s:\"),round(p,27)\n",
+    "print(\"de Broglie wavelength of electron in angstrom=\"),round(lamda*1e10,3)\n",
+    "# Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom\n",
+    "# Which is due to wrong calculation"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.11;page no:20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.11,page no:20\n",
+      " velocity of neutron in m/s: 3964.072\n",
+      " Kinetic energy of neutron in eV= 0.082\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of velocity and kinetic energy of neutron\n",
+    "#intiation of all variables \n",
+    "#Given that\n",
+    "lamda= 1 # de Broglie wavelength of neutron in angstrom\n",
+    "m = 1.67e-27 # Mass of electron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.11,page no:20\")\n",
+    "v = h/(m*lamda*1e-10) # Calculation of velocity of moving neutron\n",
+    "print(\" velocity of neutron in m/s:\"),round(v,3)\n",
+    "E = 1./2.*m*v**2 # Calculation of kinetic energy of moving neutron\n",
+    "print(\" Kinetic energy of neutron in eV=\"),round(E/1.6e-19,3)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.12;page no:20"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.12,page no:20\n",
+      "Wavelength of electron in  metres= 2.74e-11\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of electron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "E = 2 # Energy of accelerated electron in KeV\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.12,page no:20\")\n",
+    "lamda = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron\n",
+    "print(\"Wavelength of electron in  metres=\"),round(lamda,13)\n",
+    "# Answer in book is 2.74e-12m\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.13;page no:21"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.13,page no:21\n",
+      "Wavelength of matter wave in  angstrom= 1.48e-05\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of matter wave\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "v = 2e8 # speed of moving proton in m/s\n",
+    "c = 3e8 # speed of light in m/s\n",
+    "m = 1.67e-27 # Mass of proton in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.13,page no:21\")\n",
+    "lamda = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron\n",
+    "print(\"Wavelength of matter wave in  angstrom=\"),round(lamda*1e10,7)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.14;page no:22"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.14,page no:22\n",
+      "Momentum of photon in  Kgm/s while Momentum of electron in Kgm/s which are equal: 6.63e-24 6.63e-24\n",
+      "Total Energy of photon in  joule while Total Energy of electron in  MeV: 1.989e-15 2.42e-17\n",
+      "Ratio of kinetic energies in: 0.0121\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of momentum,total energy and ratio of kinetic energy of photon\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "lamda = 1# wavelength in m/s\n",
+    "m_e = 9.1e-31 # Mass of electron in Kg\n",
+    "m_p = 1.67e-27 # Mass of proton in kg\n",
+    "c = 3e8 # speed of light in m/s\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.14,page no:22\")\n",
+    "p_e = h/(lamda*1e-10) # Momentum of electron\n",
+    "p_p = h/(lamda*1e-10) # Momentum of photon\n",
+    "print(\"Momentum of photon in  Kgm/s while Momentum of electron in Kgm/s which are equal:\"),round(p_p,26),round(p_e,26)\n",
+    "E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron\n",
+    "E_e1=(2.42*10**-17)+(m_e*c**2/1.6*10**-19)\n",
+    "E_p = h*c/(lamda*1e-10) # Total energy of photon\n",
+    "print(\"Total Energy of photon in  joule while Total Energy of electron in  MeV:\"),round(E_p,18),E_e1\n",
+    "K_e = p_e**2/(2*m_e) # Kinetic energy of electron \n",
+    "K_p = h*c/(lamda*1e-10)# Kinetic energy of photon\n",
+    "r_K = K_e/K_p # Ratio of kinetic energies\n",
+    "print(\"Ratio of kinetic energies in:\"),round(r_K,4)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.15;page no:24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.15,page no:24\n",
+      "de Broglie wavelength of neutron in angstrom: 0.0573\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of neutron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "e = 25 # Energy of neutron in eV\n",
+    "c = 3e8 # speed of light in m/s\n",
+    "m = 1.67e-27 # Mass of neutron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.15,page no:24\")\n",
+    "rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV\n",
+    "if e/rest_e < 0.015:\n",
+    "    E = e;\n",
+    "else:\n",
+    "\tE = rest_e +e;\n",
+    "lamda = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength\n",
+    "print(\"de Broglie wavelength of neutron in angstrom:\"),round(lamda*1e10,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.16;page no:24"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.16,page no:24\n",
+      "de Broglie wavelength of neutron in angstrom: 0.00717\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of neutron \n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "e = 2*1.6e-19 # charge on alpha particle in coulomb\n",
+    "V = 200 # Applied voltage in volts\n",
+    "m = 4*1.67e-27 # Mass of alpha particle in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.16,page no:24\")\n",
+    "lamda=h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength\n",
+    "print(\"de Broglie wavelength of neutron in angstrom:\"),round(lamda*1e10,5)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.17;page no:25"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.17,page no:25\n",
+      "de Broglie wavelength of ball in angstrom: 6.62e-26\n",
+      "de Broglie wavelength of electron in angstrom: 7.27\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of ball and electron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "M = 20 # Mass of ball in Kg\n",
+    "V = 5 # velocity of of ball in m/s\n",
+    "m = 9.1e-31 #Mass of electron in Kg\n",
+    "v = 1e6 # velocity of of electron in m/s\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.17,page no:25\")\n",
+    "lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball\n",
+    "lambda_e = h/(m*v) # calculation of de Broglie wavelength electron\n",
+    "print(\"de Broglie wavelength of ball in angstrom:\"),round(lambda_b*1e10,34)\n",
+    "print(\"de Broglie wavelength of electron in angstrom:\"),round(lambda_e*1e10,2)\n",
+    "# answer in book is 6.62e-22 angstrom for ball\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.18;page no:26"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.18,page no:26\n",
+      "Wavelength of neutron in angstrom: 0.286\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of de brogle wavelength of neutron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "E = 1 # Energy of neutron in eV\n",
+    "m = 1.67e-27 # Mass of neutron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "print(\"Example 1.18,page no:26\")\n",
+    "lamda = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron\n",
+    "print(\"Wavelength of neutron in angstrom:\"),round(lamda*1e10,3)\n",
+    "# Answer in book is 6.62e-22 angstrom\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.19;page no:27"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.19,page no:27\n",
+      "Applied voltage on electron in V: 602.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Applied voltage on electron \n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "lamda = 0.5# wavelength of electron in angstrom\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "q = 1.6e-19 # charge on electron in coulomb\n",
+    "print(\"Example 1.19,page no:27\")\n",
+    "V = h**2/(2*m*q*(lamda*1e-10)**2) # Calculation of velocity of moving electron\n",
+    "print(\"Applied voltage on electron in V:\"),round(V,1)\n",
+    "# Answer in book is 601.6 Volt\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.21;page no:29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.21,page no:29\n",
+      "Wavelength of neutron at degree Celsius in angstrom: 1.43\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of  wavelength of neutron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "k = 8.6e-5 # Boltzmann constant\n",
+    "t = 37 # Temperature in degree Celsius\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "m = 1.67e-27 # Mass of neutron\n",
+    "print(\"Example 1.21,page no:29\")\n",
+    "lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n",
+    "print(\"Wavelength of neutron at degree Celsius in angstrom:\"),round(lamda*1e10,2)\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.22;page no:29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.22,page no:29\n",
+      "Wavelength of helium at degree Celsius in angstrom: 0.727\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of wavelength of helium\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "k = 8.6e-5 # Boltzmann constant\n",
+    "t = 27 # Temperature in degree Celsius\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "m = 6.7e-27 # Mass of helium atom\n",
+    "print(\"Example 1.22,page no:29\")\n",
+    "lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength\n",
+    "print(\"Wavelength of helium at degree Celsius in angstrom:\"),round(lamda*1e10,3)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.23;page no:30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.23,page no:30\n",
+      "lamda= 8.67e-11\n",
+      "D/2*x= 0.05\n",
+      "tan(theta)= 0.05\n",
+      "Interatomic spacing of crystal in angstrom: 8.67\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Interatomic spacing of crystal\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "E = 200. # energy of electrons in eV\n",
+    "x = 20. # distance of screen in cm\n",
+    "D = 2. # diameter of ring in cm\n",
+    "h = 6.62e-34 # Plank constant\n",
+    "m = 9.1e-31 # Mass of electron in kg\n",
+    "print(\"Example 1.23,page no:30\")\n",
+    "lamda= h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength\n",
+    "print(\"lamda=\"),round(lamda,13)\n",
+    "print(\"D/2*x=\"),D/(2*x)\n",
+    "p=D/(2*x)\n",
+    "print(\"tan(theta)=\"),p\n",
+    "d = lamda/(2*p)# calculation of interatomic spacing of crystal\n",
+    "print(\"Interatomic spacing of crystal in angstrom:\"),round(d*1e10,2)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.24;page no:31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.24,page no:31\n",
+      "Velocity of electron in ground state in M/s= 2.31\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of velocity of electron \n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "r = 0.5 # Bohr radius of hydrogen in angstrom\n",
+    "m = 9.1e-31 # Mass of neutron in Kg\n",
+    "h = 6.6e-34 # Plank constant\n",
+    "print(\"Example 1.24,page no:31\")\n",
+    "v = h/(2*3.14*r*1e-10*m) # velocity of electron in ground state\n",
+    "print(\"Velocity of electron in ground state in M/s=\"),round(v/10**6,2)\n",
+    "# Answer in book is 2.31e6 m/s\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.25;page no:32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.25,page no:32\n",
+      "Velocity of electron in ground state in m/s: 1237.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Velocity of electron in ground state\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "lamda = 5890 # wavelength of yellow radiation in angstrom\n",
+    "m = 9.1e-31 # Mass of neutron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.25,page no:32\")\n",
+    "v = h/(lamda*1e-10*m) # velocity of electron in ground state\n",
+    "print(\"Velocity of electron in ground state in m/s:\"),round(v,1)\n",
+    "# Answer in book is 1.24e3 m/s\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.26;page no:33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 26,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.26,page no:33\n",
+      "Velocity of neutron in m/s: 1985.0\n",
+      "Kinetic energy of neutron in eV: 0.021\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Velocity and kinetic energy of neutron\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "lamda = 2 # wavelength of neutron in angstrom\n",
+    "m = 1.67e-27 # Mass of neutron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.26,page no:33\")\n",
+    "v = h/(lamda*1e-10*m) # velocity of neutron\n",
+    "k = 0.5*m*v**2 # Kinetic energy of neutron\n",
+    "print(\"Velocity of neutron in m/s:\"),round(v,1)\n",
+    "print(\"Kinetic energy of neutron in eV:\"),round(k/1.6e-19,3)\n",
+    "# Answer in book is 0.021eV\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.29;page no:36"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.29,page no:36\n",
+      "theta 72.6\n",
+      "theta1= 56.84\n",
+      "For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders: 1.91\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of theta and theta1  \n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "v1 = 50 # Previous applied voltage\n",
+    "v2 = 65 # final applied voltage\n",
+    "k = 12.28 \n",
+    "d = 0.91 # Spacing in a crystal in angstrom\n",
+    "print(\"Example 1.29,page no:36\")\n",
+    "lamda = k/math.sqrt(v1)\n",
+    "theta= math.asin(lamda/(2*d))# Angel for initial applied voltage\n",
+    "lamda1 = k/math.sqrt(v2)# wavelength for final applied voltage\n",
+    "theta1 = math.asin(lamda1/(2*d))# Angel for final applied voltage\n",
+    "#print(\"lamda1/1.82=\"),math.asin(lamda1/1.82)\n",
+    "print(\"theta\"),round(theta*180/3.14,1)\n",
+    "print(\"theta1=\"),round(theta1*180/3.14,2)\n",
+    "print(\"For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders:\"),round(2*math.sin(theta),2)\n",
+    "#print(\"Angle of diffraction for first order of beam  is  degree at  Volts:\"),round((math.theta1*180/math.pi),2)\n",
+    "# Answer in book is 57.14 degree"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.30;page no:45"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 28,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.30,page no:45\n",
+      "Group velocity of seawater waves in m/s: 16.29\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Group velocity of seawater waves\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "lamda = 680 # Wavelength in m\n",
+    "g = 9.8 #Acceleration due to gravity\n",
+    "print(\"Example 1.30,page no:45\")\n",
+    "v_g = 0.5*math.sqrt(g*lamda/(2*3.14)) # Calculation of group velocity\n",
+    "print(\"Group velocity of seawater waves in m/s:\"),round(v_g,2)\n",
+    "# Answer in book is 16.29 m/s\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.32;page no:47"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.32,page no:47\n",
+      "Group velocity of de Broglie waves is c  : 0.9966\n",
+      " phase velocity of de Broglie waves is c 1.0034\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of group and phase velocity of de brogle waves \n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "lamda = 2e-13 # de Broglie wavelength of an electron in m\n",
+    "c = 3e8 # Speed of light in m/s\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.32,page no:47\")\n",
+    "E = h*c/(lamda*1.6e-19) \n",
+    "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n",
+    "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n",
+    "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n",
+    "v_p  = c**2/v_g # Phase velocity\n",
+    "print(\"Group velocity of de Broglie waves is c  :\"),round(v_g/c,4)\n",
+    "print(\" phase velocity of de Broglie waves is c\"),round(v_p/c,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## example 1.33;page no:48"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 1.33,page no:48\n",
+      "Kinetic energy of electron in KeV: 293.33\n",
+      "Group velocity of de Broglie waves is c in m/s: 0.7719\n",
+      "phase velocity of de Broglie waves is c in m/s: 1.295\n"
+     ]
+    }
+   ],
+   "source": [
+    "#cal of Kinetic energy of electron,group velocity and phase velocity of de Broglie waves\n",
+    "#intiation of all variables \n",
+    "#given that\n",
+    "import math\n",
+    "lamda = 2.e-12 # de Broglie wavelength of an electron in m\n",
+    "c = 3.e8 # Speed of light in m/s\n",
+    "m = 9.1e-31 # Mass of electron in Kg\n",
+    "h = 6.63e-34 # Plank constant\n",
+    "print(\"Example 1.33,page no:48\")\n",
+    "E = h*c/(lamda*1.6e-19) # Energy due to momentum\n",
+    "E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy\n",
+    "E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV\n",
+    "KE = E_total - E_rest # Kinetic energy\n",
+    "v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity\n",
+    "v_p  = c**2/v_g # Phase velocity\n",
+    "print(\"Kinetic energy of electron in KeV:\"),round(KE/1000,2)\n",
+    "print(\"Group velocity of de Broglie waves is c in m/s:\"),round(v_g/c,4)\n",
+    "print(\"phase velocity of de Broglie waves is c in m/s:\"),round(v_p/c,3)\n",
+    "# Answer in book is v_g = 0.6035c & v_p = 1.657c\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/marupeddisameer chaitanya/Chapter_4_Diffusion_and_Reaction_in_Porous_Catalysts.ipynb b/sample_notebooks/marupeddisameer chaitanya/Chapter_4_Diffusion_and_Reaction_in_Porous_Catalysts.ipynb
new file mode 100755
index 00000000..a01d0a9f
--- /dev/null
+++ b/sample_notebooks/marupeddisameer chaitanya/Chapter_4_Diffusion_and_Reaction_in_Porous_Catalysts.ipynb
@@ -0,0 +1,303 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4 Diffusion and Reaction in Porous Catalysts"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_1 pgno:135"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " OUTPUT Ex4.1.a\n",
+      "\n",
+      "=================================================\n",
+      "\n",
+      "The predicted diffusivity of Chlorine is  cm2/s  0.00217149494706\n",
+      "\n",
+      "\n",
+      " OUTPUT Ex4.1.b\n",
+      "\n",
+      "=================================================\n",
+      "\n",
+      "The tortusity value =  1.25277093159\n",
+      "\n",
+      "\n",
+      " OUTPUT Ex4.1.b\n",
+      "\n",
+      "=================================================\n",
+      "\n",
+      "The Effective diffusivity of Chlorine  K a atm =   cm2/sec  573.0 15.0 1.83302312261e-09\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436.\n",
+    "#Chapter-4 Ex4.1 Pg No. 135\n",
+    "#Title:Diffusivity of Chlorine and tortuosity in catalyst pellet\n",
+    "#===========================================================================================================\n",
+    "# COMMON INPUT \n",
+    "S_g=235.;#Total surface per gram (m2/g)\n",
+    "V_g=0.29E-6;#Pore volume per gram (cm3/g)\n",
+    "rho_p=1.41;#Density of particle (g/cm3)\n",
+    "D_He=0.0065;#Effective diffusivity of He (cm2/sec)\n",
+    "D_AB=0.73;# at 1atm and 298K\n",
+    "M_He=4.;#Molecular weight of He\n",
+    "M_Cl2=70.09;#Molecular weight of Cl2\n",
+    "T_ref=293;#Reference temperature\n",
+    "T_degC=300.;\n",
+    "T_01=T_degC+273;#Reaction temperature(K) (Ex4.1.a)\n",
+    "T_02=298.;#Operating temperature  (Ex4.1.b)\n",
+    "T_03=573.;#operating temperature (Ex4.1.c)\n",
+    "P_ref=1;#Reference pressure\n",
+    "D_Cl2_CH4=0.15;#at 1atm 273K\n",
+    "P=15.;#operating pressure \n",
+    "#tau=1.25;#From value calculated in Ex4.1.b Pg. No. 136\n",
+    "from math import sqrt\n",
+    "\n",
+    "\n",
+    "#CALCULATION (Ex4.1.a)\n",
+    "r_bar=2*V_g/S_g;#Mean Pore radius\n",
+    "D_Cl2_Ex_a=D_He*((M_He/M_Cl2)*(T_01/T_ref))**(0.5);#Assuming Knudsen flow at 573K\n",
+    "\n",
+    "#CALCULATION (Ex4.1.b)\n",
+    "r_bar=2.*V_g*(10**6)/(S_g *(10**4));\n",
+    "D_K=9700.*(r_bar)*(T_ref/M_He)**(0.5);#Knudsen flow\n",
+    "D_AB1=D_AB*(293./298.)**(1.7)# at 1.5 atm and 293K\n",
+    "D_pore=1./((1./D_K)+(1./D_AB1));#pore diffusion\n",
+    "Epsilon=V_g*rho_p*(10**6);\n",
+    "tau=(D_pore*Epsilon)/D_He;#Tortusity\n",
+    "\n",
+    "#CALCULATION (Ex4.1.c)\n",
+    "D_Cl2_CH4_new=D_Cl2_CH4*(P_ref/P)*(T_03/T_ref)**(1.7);\n",
+    "D_K_Cl2=9700*r_bar*sqrt(T_03/M_Cl2);\n",
+    "D_pore=1/((1/D_Cl2_CH4_new)+(1/D_K_Cl2));\n",
+    "Epsilon=V_g*rho_p;\n",
+    "D_Cl2_Ex_c=D_pore*Epsilon/tau;\n",
+    "\n",
+    "\n",
+    "#OUTPUT\n",
+    "print '\\n OUTPUT Ex4.1.a'\n",
+    "print '\\n================================================='\n",
+    "print '\\nThe predicted diffusivity of Chlorine is  cm2/s ',D_Cl2_Ex_a\n",
+    "print '\\n\\n OUTPUT Ex4.1.b'\n",
+    "print '\\n================================================='\n",
+    "print '\\nThe tortusity value = ',tau\n",
+    "print '\\n\\n OUTPUT Ex4.1.b'\n",
+    "print '\\n================================================='\n",
+    "print '\\nThe Effective diffusivity of Chlorine  K a atm =   cm2/sec ',T_03, P, D_Cl2_Ex_c\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_2 pgno:140"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " OUTPUT Ex4.2.a\n",
+      "\n",
+      "=================================================\n",
+      "\n",
+      " The effective diffusivity of O2 in air =  cm2/s 0.0235933499021\n",
+      "\n",
+      "\n",
+      " OUTPUT Ex4.2.b\n",
+      "\n",
+      "=================================================\n",
+      "\n",
+      " The calculated surface mean pore radius = cm 6e-07\n",
+      "\n",
+      " The predicted pore diffusivity =  cm2/sec 0.0218264089105\n",
+      "\n",
+      " The corresponding tortusity =  0.499558598529\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436.\n",
+    "#Chapter-4 Ex4.2 Pg No. 140\n",
+    "#Title:Effective diffusivity of O2 in air\n",
+    "#============================================================================================================\n",
+    "\n",
+    "# COMMON INPUT\n",
+    "S_g=150.;#Total surface per gram (m2/g)\n",
+    "V_g=0.45;#Pore volume per gram (cm3/g)\n",
+    "V_i=0.30;#Micropore volume per gram (cm3/g)\n",
+    "V_a=0.15;# Macropore volume per gram (cm3/g)\n",
+    "rho_P=1.2;#Density of particle (g/cm3)\n",
+    "tau=2.5;# Tortusity\n",
+    "r_bar_i=40*(10**(-8));#Micropore radius\n",
+    "r_bar_a=2000*(10**(-8));#Macropore radius\n",
+    "D_AB=0.49;#For N2O2 at 1 atm (cm2/s)\n",
+    "M_O2=32.;#Molecular weight of O2\n",
+    "T=493.;#Opereating Temperature (K)\n",
+    "from math import sqrt\n",
+    "\n",
+    "\n",
+    "\n",
+    "#CALCULATION (Ex4.2.a)\n",
+    "Epsilon=V_g*rho_P;\n",
+    "D_K_i=9700*(r_bar_i)*sqrt(T/M_O2);#Knudsen flow for micropore\n",
+    "D_Pore_i=1/((1/D_K_i)+(1/D_AB))\n",
+    "D_K_a=9700*(r_bar_a)*sqrt(T/M_O2);\n",
+    "D_Pore_a=1/((1/D_K_a)+(1/D_AB));##Knudsen flow for macropore\n",
+    "D_Pore_Avg=(V_i*D_Pore_i+V_a*D_Pore_a)/(V_i+V_a);\n",
+    "D_e=Epsilon*D_Pore_Avg/tau;\n",
+    "\n",
+    "#CALCULATION (Ex4.2.b)\n",
+    "Epsilon=V_g*rho_P;\n",
+    "r_bar=2*V_g/(S_g*10**4);\n",
+    "D_K=9700*(r_bar)*sqrt(T/M_O2);#Knudsen Flow\n",
+    "D_Pore=1/((1/D_K)+(1/D_AB));\n",
+    "tau=D_Pore*Epsilon/D_e;\n",
+    "\n",
+    "#OUTPUT\n",
+    "print '\\n OUTPUT Ex4.2.a'\n",
+    "print '\\n================================================='\n",
+    "print '\\n The effective diffusivity of O2 in air =  cm2/s',D_e \n",
+    "print '\\n\\n OUTPUT Ex4.2.b'\n",
+    "print '\\n================================================='\n",
+    "print '\\n The calculated surface mean pore radius = cm',r_bar      \n",
+    "print '\\n The predicted pore diffusivity =  cm2/sec',D_Pore          \n",
+    "print '\\n The corresponding tortusity = ',tau\n",
+    "\n",
+    "\n",
+    "\n",
+    "#======================================================END OF PROGRAM========================================\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4_4 pgno:157"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\tBased on average pressures calculated Rate and Effectiveness factor\n",
+      "\n",
+      "\t r :  (mol/s cm3) 1.17056498924e-05\n",
+      "\n",
+      "\t eta_calc :   0.174804371726\n",
+      "\n",
+      " The actual value of Effectiveness factor eta_actual : 0.427402185863\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc. USA,pp 436.\n",
+    "#Chapter-4 Ex4.4 Pg No.157\n",
+    "#Title: Effectiveness factor for solid catalyzed reaction\n",
+    "#======================================================================================================================\n",
+    "\n",
+    "#INPUT\n",
+    "D_e_A=0.02;#(cm2/s)\n",
+    "D_e_B=0.03;#(cm2/s)\n",
+    "D_e_C=0.015;#(cm2/s)\n",
+    "X_f_A=0.3;\n",
+    "X_f_B=(1-X_f_A);\n",
+    "eta_assumed=0.68;#Effectiveness factor from Fig.4.8 for first order reaction\n",
+    "T=150.;#(deg C)\n",
+    "T_K=T+273;#(K)\n",
+    "r=0.3;#(cm)Radius of catalyst sphere\n",
+    "P_opt=4.;#(atm)Operating Pressure \n",
+    "R=82.056;#(cm3 atm/K mol)Gas constant \n",
+    "\n",
+    "\n",
+    "#CALCULATION\n",
+    "#Kinetic equation r= (2.5*10**-5*P_A*P_B)/(1+0.1*P_A+2*P_C)**2\n",
+    "P_A=X_f_A*P_opt;\n",
+    "P_B=X_f_B*P_opt;\n",
+    "r_star=(2.5*10**-5*P_A*P_B)/(1+0.1*P_A)**2;\n",
+    "C_A=P_A/(R*T_K);\n",
+    "k=r_star/C_A;\n",
+    "Phi= r*(k/D_e_A)**(0.5);\n",
+    "P_A_bar=eta_assumed*P_A;\n",
+    "delta_P_A=P_A*(1-eta_assumed);\n",
+    "delta_P_B=delta_P_A*(D_e_A/D_e_B);\n",
+    "P_B_bar=P_B-delta_P_B;\n",
+    "delta_P_C=delta_P_A*(D_e_A/D_e_C);\n",
+    "P_C_bar=delta_P_C;\n",
+    "r_calc=(2.5*10**-5*P_A_bar*P_B_bar)/(1+0.1*P_A_bar+2*P_C_bar)**2\n",
+    "eta_calc=r_calc/r_star;\n",
+    "eta_approx=(eta_calc+eta_assumed)/2;\n",
+    "\n",
+    "#OUTPUT\n",
+    "#Console Output\n",
+    "print'\\tBased on average pressures calculated Rate and Effectiveness factor'\n",
+    "print'\\n\\t r :  (mol/s cm3)',r_calc\n",
+    "print'\\n\\t eta_calc :  ',eta_calc\n",
+    "print'\\n The actual value of Effectiveness factor eta_actual :',eta_approx\n",
+    "\n",
+    "#================================================END OF PROGRAM==================================================================================\n",
+    "\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/mayankagrawal/chapter5.ipynb b/sample_notebooks/mayankagrawal/chapter5.ipynb
new file mode 100755
index 00000000..37fe203a
--- /dev/null
+++ b/sample_notebooks/mayankagrawal/chapter5.ipynb
@@ -0,0 +1,121 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7d02c41b7edfa1d9c3e7a01084e54b2d37ea95fa7e4f6ac43f62128207e0a35d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter5:ULTRASONIC"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Eg1:pg-177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "v=5760   #velocity of propagation of ultrasonic sound wave along X-direction in m/s\n",
+      "t=1.0*10**-3 #thickness of a piezo-electric quartz plate in meter\n",
+      "lamda=2*t #wavelength in meter(since t=lamda/2 corresponding to fundamental frequency)\n",
+      "V=v/lamda #fundamental frequency of the crystal\n",
+      "print\"fundamental frequency of the crystal=\",V/10**6,\"MHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fundamental frequency of the crystal= 2.88 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Eg2:pg-177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "Y=7.9*10**10 #Young's modulus in newton/m**2\n",
+      "p=2650       #density for Quartz in Kg/m**3\n",
+      "t=0.005      #thickness of a Quartz crystal in meter\n",
+      "v=sqrt(Y/p)  #velocity for longitudinal vibrations in m/sec\n",
+      "lamda=2*t #wavelength in meter(since t=lamda/2 corresponding to fundamental frequency)\n",
+      "V=v/lamda #fundamental frequency of the crystal\n",
+      "print\"fundamental frequency of the crystal=\",int(round(V/10**3)),\"KHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fundamental frequency of the crystal= 546 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Eg3:pg-177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "f=200.*10**3 #frequency of ultrasonic sound in Hz\n",
+      "S_a=340      #speed of sound in air in m/s\n",
+      "S_w=1486     #speed of sound in water in m/s\n",
+      "lamda_r=S_a/f#wavelength of reflected sound in metre\n",
+      "print\"Wavelength of reflected sound=\",\"{:.2e}\".format(lamda_r),\"m\"\n",
+      "lamda_t=S_w/f#wavelength of transmitted sound in metre\n",
+      "print\"Wavelength of transmitted sound=\",\"{:.2e}\".format(lamda_t),\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of reflected sound= 1.70e-03 m\n",
+        "Wavelength of transmitted sound= 7.43e-03 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/mokshagunda/Chapter_2_DIFFRACTION.ipynb b/sample_notebooks/mokshagunda/Chapter_2_DIFFRACTION.ipynb
new file mode 100755
index 00000000..9af6a743
--- /dev/null
+++ b/sample_notebooks/mokshagunda/Chapter_2_DIFFRACTION.ipynb
@@ -0,0 +1,365 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 DIFFRACTION"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_1 pg.no:29"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "No of lines per centimeter is 5000\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To calculate the no of lines in one cm of grating surface\n",
+    "from math import pi,sin\n",
+    "k=2.\n",
+    "lamda=5*10**-5 #units in cm\n",
+    "theta=30 # units in degrees\n",
+    "#We have nooflines=1/e=(k∗lamda)/sin(theta)\n",
+    "nooflines=sin(theta*pi/180)/(k*lamda) #units in cm\n",
+    "print \"No of lines per centimeter is %.f\"%nooflines\n",
+    "#In text book the answer is printed wrong as 10ˆ3\n",
+    "#The correct answer is 5∗10ˆ3"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_2 pg.no:30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "For First order spectra theta1=17.5 degrees\n",
+      "For Third order spectra theta3=64.2 degrees\n",
+      "Difference in Angles of deviation in first and third order spectra is theta3−theta1=46.70 degrees\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To Find the difference in angles of deviation in first and third order spectra\n",
+    "from math import pi,asin\n",
+    "lamda=5000. # units in armstrongs\n",
+    "lamda=lamda*10**-8 # units in cm\n",
+    "e=1./6000.\n",
+    "#For first order e∗sin(theta1)=1∗lamda\n",
+    "theta1=asin(lamda/e) # units in radians\n",
+    "theta1=theta1*180./pi # units in degrees\n",
+    "print \"For First order spectra theta1=%.1f degrees\"%theta1\n",
+    "#For third order e∗sin(theta3)=3∗lamda\n",
+    "theta3=asin(3.*lamda/e) # units in radians\n",
+    "theta3=theta3*180/pi # units in degrees\n",
+    "print \"For Third order spectra theta3=%.1f degrees\"%theta3\n",
+    "diffe=theta3-theta1 #units in degrees\n",
+    "print \"Difference in Angles of deviation in first and third order spectra is theta3−theta1=%.2f degrees\"%diffe"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_3 pg.no:30"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false,
+    "scrolled": true
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "No of lines per cm=196.0 \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To calculate minimum no of lines per centimeter\n",
+    "lamda1=5890 # units in armstrongs\n",
+    "lamda2=5896 # units in armstrongs\n",
+    "dlamda=lamda2-lamda1 #units in armstrongs\n",
+    "k=2\n",
+    "n=lamda1/(k*dlamda)\n",
+    "width=2.5 #units in cm\n",
+    "nooflines=n/width\n",
+    "print \"No of lines per cm=%.1f \"%nooflines"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_4 pg.no:31"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "As total no of lines required for resolution in first order is 981 and total no of lines in grating is 850 the lines will not be resolved in first order\n",
+      "As total no of lines required for resolution in first order is 490 and total no of lines in grating is 850 the lines will be resolved in second order\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To examine two spectral lines are clearly resolved in first order and second order\n",
+    "n=425.\n",
+    "tno=2.*n\n",
+    "lamda1=5890 # units in armstrongs\n",
+    "lamda2=5896 # units in armstrongs\n",
+    "dlamda=lamda2 -lamda1\n",
+    "#For first order\n",
+    "n=lamda1/dlamda\n",
+    "print\"As total no of lines required for resolution in first order is %.f and total no of lines in grating is %d the lines will not be resolved in first order\"%(n,tno)\n",
+    "#For second order\n",
+    "n=lamda1/(2*dlamda)\n",
+    "print\"As total no of lines required for resolution in first order is %.f and total no of lines in grating is %d the lines will be resolved in second order\"%(n,tno)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_5 pg.no:32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Angle of separation is 16 minutes\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To find the angle of separation\n",
+    "from math import asin,pi\n",
+    "\n",
+    "lamda1=5016.  # units in armstrongs\n",
+    "lamda2=5048.  # units in armstrongs\n",
+    "lamda1=lamda1*10**-8   # units in cm\n",
+    "lamda2=lamda2*10**-8   # units in cm\n",
+    "k=2.\n",
+    "n=15000\n",
+    "e=2.54/n     # units in cm \n",
+    "theta1=asin((2*lamda1)/e)*(180/pi)    # units in in degrees\n",
+    "theta2=asin((2*lamda2)/e)*(180/pi)  # units in in degrees\n",
+    "diffe=theta2-theta1 # units in in degrees\n",
+    "diffe=diffe*60 # units in minutes\n",
+    "print \"Angle of separation is %.f minutes\"%diffe"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_6 pg.no:32"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The dispersive power of the grating is 15000\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To Calculate the dispersive power of the grating\n",
+    "from math import pi,asin,cos\n",
+    "n=4000.\n",
+    "e=1/n #units in cm\n",
+    "k=3.\n",
+    "lamda=5000 # units in armstrongs\n",
+    "lamda=lamda*10**-8 # units in cm\n",
+    "theta=asin((k*lamda)/e)*(180/pi) # units in degrees\n",
+    "costheta=cos(theta*pi/180)\n",
+    "disppower=(k*n)/costheta\n",
+    "print \"The dispersive power of the grating is %.f\"%disppower"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_7 pg.no:33"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The highest order spectrum Seen with monochromatic light is 3.33\n"
+     ]
+    }
+   ],
+   "source": [
+    "#To Calculate highest power of spectrum seen with mono chromaic light\n",
+    "lamda=6000.    # units in armstrongs\n",
+    "lamda=lamda*10**-8   #units in cm\n",
+    "n=5000.\n",
+    "e=1/n #units in cm\n",
+    "k=e/lamda\n",
+    "print \"The highest order spectrum Seen with monochromatic light is %.2f\"%k"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_8 pg.no:35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Wavelength of the lines is 0.0000606 cms\n",
+      "Minimum grating width required is 4.2 cm \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To calculate the wavelength\n",
+    "from math import pi,sin\n",
+    "k=2.\n",
+    "theta1=10.\n",
+    "dtheta=3.\n",
+    "dlamda=5*10**-9\n",
+    "lamda=(sin((theta1*pi)/180)*dlamda*60*60)/(cos((theta1*pi)/180)*dtheta*(pi/180)) # units in cm\n",
+    "print \"Wavelength of the lines is %.7f cms\"%lamda\n",
+    "lamda_dlamda=lamda+dlamda # units in cm\n",
+    "N=6063\n",
+    "Ne=(N*k*lamda)/sin((theta1*pi)/180) # units in cm\n",
+    "print \"Minimum grating width required is %.1f cm \"%Ne"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_9 pg.no:35"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Resolving power is 10000 \n"
+     ]
+    }
+   ],
+   "source": [
+    "#To calculate resolving power in second order\n",
+    "#We have e∗sin(theta)=k∗lamda\n",
+    "#We have e∗0.2=k∗lamda −>1\n",
+    "#And e∗0.3=(k+1)∗lamda −>2\n",
+    "#Subtracting one and two 3∗0.1=lamda\n",
+    "lamda=5000. # units in armstrongs\n",
+    "lamda=lamda*10**-8 # units in cm\n",
+    "e=lamda/0.1 # units in cm\n",
+    "width=2.5 #units in cm\n",
+    "N=width/e\n",
+    "respower=2*N\n",
+    "print \"Resolving power is %.f \"%respower"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/nishumittal/chapter2_1.ipynb b/sample_notebooks/nishumittal/chapter2_1.ipynb
new file mode 100755
index 00000000..cacd8a20
--- /dev/null
+++ b/sample_notebooks/nishumittal/chapter2_1.ipynb
@@ -0,0 +1,306 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d284e33c3a44645a717587479459459fbad97c0b83247c27d3e8959966515278"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 Transport Phenomena in semiconductors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page no 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "n=10**20\n",
+      "q=1.6*10**-19\n",
+      "u=800\n",
+      "e=1\n",
+      "\n",
+      "#Calculation\n",
+      "J=n*q*u*e\n",
+      "\n",
+      "#Result\n",
+      "print\"Electron current density is \",J*10**-4,\"*10**4 A/cm**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Electron current density is  1.28 *10**4 A/cm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page no 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Nd=10**12                        #/cm**3\n",
+      "ni=10**10                        #/cm**3\n",
+      "Nd1=10**18\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "n=Nd+math.sqrt(Nd+4*(ni**2))/2.0\n",
+      "n1=Nd1+math.sqrt(Nd1+4*(ni**2))/2.0\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) free electron at 10**12 is \", round(n*10**-12,2),\"*10**12 /cm**3\",\"and hole density is  p=9.999*10**7 /cm**3\"\n",
+      "print\"(b) free electron at 10**18 is \",round(n1*10**-18,2),\"*10**18 /cm**3\",\"and hole density is p=10**2 /cm**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) free electron at 10**12 is  1.01 *10**12 /cm**3 and hole density is  p=9.999*10**7 /cm**3\n",
+        "(b) free electron at 10**18 is  1.0 *10**18 /cm**3 and hole density is p=10**2 /cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page no 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "K=6.02*10**23                     #atoms/mole\n",
+      "b=72.6                            #g\n",
+      "c=5.32                            #g/cm**3\n",
+      "n=2.5*10**13                      #cm**3\n",
+      "q=1.60*10**-19                     #C\n",
+      "un=3800                            #cm**2/V-s\n",
+      "up=1800                            #cm**2/V-s\n",
+      "ni=2.5*10**13\n",
+      "Nd=4.41*10**14\n",
+      "x=1.60*10**-19\n",
+      "n1=4.41*10**22                     #electron/cm**3\n",
+      "\n",
+      "#Calculation\n",
+      "C=K*(1/b)*c\n",
+      "A=n*q*(un+up)\n",
+      "R=1/A\n",
+      "p=ni**2/Nd\n",
+      "A1=Nd*x*un\n",
+      "R1=1/A1\n",
+      "A2=n1*x*un\n",
+      "X=A2/A1\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Concentration of atoms in germanium is \", round(C*10**-22,2) *10**22,\"atoms/cm**3\"\n",
+      "print\"(b) resistivity of intrinsic germanium at 300 degree K is \", round(R,1),\"ohm-cm\"\n",
+      "print\"(c) Resistivity is \", round(R1,2),\"ohm-cm\"\n",
+      "print\"(d) Ratio of conductivity is \",X"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Concentration of atoms in germanium is  4.41e+22 atoms/cm**3\n",
+        "(b) resistivity of intrinsic germanium at 300 degree K  44.6 ohm-cm\n",
+        "(c) Resistivity is  3.73 ohm-cm\n",
+        "(d) Ratio of conductivity is  100000000.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 53
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page no 35 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "g=5*10**21\n",
+      "t=2*10**-6                  #/cm**3\n",
+      "\n",
+      "#Calculation\n",
+      "p=g*t\n",
+      "\n",
+      "#Result\n",
+      "print\"Hole density in the semiconductor is \", p,\"/cm**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hole density in the semiconductor is  1e+16 /cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 56
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page no 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "un=1200                             #cm**2/V-s\n",
+      "n0=10.0**18                         #/cm**3\n",
+      "ni=10**10                           #/cm**3\n",
+      "up=500                              #cm**2/V-s\n",
+      "t=2*10**-6                          #S\n",
+      "K=5*10**15                          #/cm**3\n",
+      "K1=8.620*10**-5                     #eV/degree K\n",
+      "q=1.602*10**-19                     #C\n",
+      "T=50\n",
+      "Lp=51.0*10**-4\n",
+      "p0=100\n",
+      "Dn=31.2\n",
+      "x=0\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "p0=ni**2/n0\n",
+      "Dp=(((K1*T)/q)*up)*10**-18\n",
+      "Jp=((q*Dp)/Lp)*(K-p0)*math.exp(x/Lp)\n",
+      "Lp=math.sqrt(Dp*t)\n",
+      "Jn=-((Dn/Dp)*Jp)*math.exp(x/Lp)\n",
+      "Jn1=((Dn/Dp-1)*Jp)*math.exp(x/Lp)\n",
+      "Jp1=((K*up)/(n0*un))*(Dn/Dp-1)*Jp\n",
+      "\n",
+      "#Result\n",
+      "print\"The drift current density for holes is \", round(Jp1,4),\"exp(-x/Lp)\",\"A/cm**2\"\n",
+      "print\"The diffusion current density for holes is \",round(Jp,2),\"exp(-x/Lp)\",\"A/cm**2\"\n",
+      "print\"The diffusion current density for electrons is \",round(Jn,2),\"exp(-x/Lp)\",\"A/cm**2\"\n",
+      "print\"The drift current density for electrons is \",round(Jn1,2),\"exp(-x/Lp)\",\"A/cm**2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The drift current density for holes is  0.0058 exp(-x/Lp) A/cm**2\n",
+        "The diffusion current density for holes is  2.11 exp(-x/Lp) A/cm**2\n",
+        "The diffusion current density for electrons is  -4.9 exp(-x/Lp) A/cm**2\n",
+        "The drift current density for electrons is  2.79 exp(-x/Lp) A/cm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page no 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "ni=1.5*10**10                       #/cm**3\n",
+      "Nd=10**18                           #/cm**3\n",
+      "Na=10**14\n",
+      "Vt=2.4*10**18\n",
+      "Na1=10**15\n",
+      "Na2=10**16\n",
+      "Na3=10**17\n",
+      "Na4=10**18\n",
+      "Na5=10**19\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "V0=Vt*(math.log(Na*Nd)/ni**2)\n",
+      "V01=Vt*(math.log(Na1*Nd)/ni**2)\n",
+      "V02=Vt*(math.log(Na2*Nd)/ni**2)\n",
+      "V03=Vt*(math.log(Na3*Nd)/ni**2)\n",
+      "V04=Vt*(math.log(Na4*Nd)/ni**2)\n",
+      "V05=Vt*(math.log(Na5*Nd)/ni**2)\n",
+      "\n",
+      "#Result\n",
+      "print\"Contact potential across the junction is \"\n",
+      "print round(V0,2),\"\\n\", round(V01,2),\"\\n\" ,round(V02,2),\"\\n\" ,round(V03,2),\"\\n\",round(V04,1),\"\\n\",round(V05,0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Contact potential across the junction is \n",
+        "0.79 \n",
+        "0.81 \n",
+        "0.84 \n",
+        "0.86 \n",
+        "0.9 \n",
+        "1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/pramodkumardesu/Chapter_2_Transmission_Lines.ipynb b/sample_notebooks/pramodkumardesu/Chapter_2_Transmission_Lines.ipynb
new file mode 100755
index 00000000..b232b9ae
--- /dev/null
+++ b/sample_notebooks/pramodkumardesu/Chapter_2_Transmission_Lines.ipynb
@@ -0,0 +1,263 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2 Transmission Lines"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_1 pgno:65"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Maximum field =  V/m per volt 42064315640.1\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Example 1, page 65\n",
+    "#Calculate the maximum field at the sphere surface\n",
+    "\n",
+    "#Calulating Field at surface E based on figure 2.31 and table 2.3\n",
+    "from math import pi\n",
+    "Q1 = 0.25\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))\n",
+    "RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)\n",
+    "RV= RV1+RV2\n",
+    "E = (Q1*RV)/(4*pi*e0)\n",
+    "print\"Maximum field =  V/m per volt\",E\n",
+    "\n",
+    "#Answers vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_2 pgno:66"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Part a\t\n",
+      "Equivalent radius =  m \t0.0887411967465\n",
+      "Charge per bundle =  uC/m \t4.88704086264e-06\n",
+      "Charge per sunconducter =  uC/m \t2.44352043132e-06\n",
+      "\tPart b\n",
+      "\tSub part 1\t\n",
+      "Maximum feild =  kV/m \t2607466.95017\n",
+      "Maximum feild =  kV/m \t2412255.52075\n",
+      "Maximum feild =  kV/m \t2509861.23546\n",
+      "\tSub part 2\t\n",
+      "EO1 =  kV/m \t2597956.83558\n",
+      "EO2 =  kV/m \t2597429.47744\n",
+      "EI1 =  kV/m \t2402709.21273\n",
+      "EI2 =  kV/m \t2402258.0563\n",
+      "\tPart c\t\n",
+      "The average of the maximum gradient =  kV/m \t2597693.15651\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 2, page 66\n",
+    "\n",
+    "\n",
+    "#calculation based on figure 2.32\n",
+    "from math import sqrt,pi,log\n",
+    "\n",
+    "#(a)Charge on each bundle\n",
+    "print\"Part a\\t\"\n",
+    "req = sqrt(0.0175*0.45)\n",
+    "print\"Equivalent radius =  m \\t\", req\n",
+    "V = 400*10**3 #Voltage\n",
+    "H = 12. #bundle height in m\n",
+    "d = 9. #pole to pole spacing in m\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "Hd = sqrt((2*H)**2+d**2)#2*H**2 + d**2\n",
+    "Q = V*2*pi*e0/(log((2*H/req))-log((Hd/d)))\n",
+    "q = Q/2\n",
+    "print\"Charge per bundle =  uC/m \\t\",Q #micro C/m\n",
+    "print\"Charge per sunconducter =  uC/m \\t\",q #micro C/m\n",
+    "\n",
+    "#(b part i)Maximim & average surface feild\n",
+    "print\"\\tPart b\"\n",
+    "print\"\\tSub part 1\\t\"\n",
+    "r = 0.0175 #subconductor radius\n",
+    "R = 0.45 #conductor to subconductor spacing\n",
+    "MF = (q/(2*pi*e0))*((1/r)+(1/R)) # maximum feild\n",
+    "print\"Maximum feild =  kV/m \\t\",MF\n",
+    "MSF = (q/(2*pi*e0))*((1/r)-(1/R)) # maximum surface feild\n",
+    "print\"Maximum feild =  kV/m \\t\",MSF\n",
+    "ASF = (q/(2*pi*e0))*(1/r) # Average surface feild\n",
+    "print\"Maximum feild =  kV/m \\t\",ASF\n",
+    "\n",
+    "#(b part ii) Considering the two sunconductors on the left\n",
+    "print\"\\tSub part 2\\t\"\n",
+    "#field at the outer point of subconductor #1 \n",
+    "drO1 = 1/(d+r)\n",
+    "dRrO1 = 1/(d+R+r)\n",
+    "EO1 =  MF -((q/(2*pi*e0))*(drO1+dRrO1))\n",
+    "print\"EO1 =  kV/m \\t\",EO1\n",
+    "#field at the outer point of subconductor #2 \n",
+    "drO2 = 1/(d-r)\n",
+    "dRrO2 = 1/(d-R-r)\n",
+    "EO2 =  MF -((q/(2*pi*e0))*(dRrO2+drO2))\n",
+    "print\"EO2 =  kV/m \\t\",EO2\n",
+    "\n",
+    "#field at the inner point of subconductor #1 \n",
+    "drI1 = 1/(d-r)\n",
+    "dRrI1 = 1/(d+R-r)\n",
+    "EI1 =  MSF -((q/(2*pi*e0))*(drI1+dRrI1))\n",
+    "print\"EI1 =  kV/m \\t\",EI1\n",
+    "#field at the inner point of subconductor #2 \n",
+    "drI2 = 1/(d+r)\n",
+    "dRrI2 = 1/(d-R+r)\n",
+    "EI2 =  MSF -((q/(2*pi*e0))*(dRrI2+drI2)) \n",
+    "print\"EI2 =  kV/m \\t\",EI2\n",
+    "\n",
+    "#(part c)Average of the maximim gradient\n",
+    "print\"\\tPart c\\t\"\n",
+    "Eavg = (EO1+EO2)/2\n",
+    "print\"The average of the maximum gradient =  kV/m \\t\",Eavg\n",
+    "\n",
+    "#Answers might vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_3 pgno:69"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Electric Feild =  V/m \t35950238891.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 3, page 69\n",
+    "#Electric feild induced at x\n",
+    "from math import pi\n",
+    "e0 = 8.85418*10**-12 #Epselon nought\n",
+    "q = 1 # C/m\n",
+    "C = (q/(2*pi*e0))\n",
+    "#Based on figure 2.33\n",
+    "E = C-(C*(1/3+1/7))+(C*(1+1/5+1/9))+(C*(1/5+1/9))-(C*(1/3+1/7))\n",
+    "print\"Electric Feild =  V/m \\t\",E\n",
+    "\n",
+    "#Answers might vary due to round off error\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2_4 pgno:70"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\tThickness of graded design=  cm \t4.24264068712\n",
+      "Curve =  cm**2 \t62.4264068712\n",
+      "V1 =  cm**3 \t47402.906725\n",
+      "Thickness of regular design =  cm \t14.684289433\n",
+      "V2 =  cm**3 \t861.944682812\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Chapter 2, Exmaple 4, page 70\n",
+    "#Calculate the volume of the insulator\n",
+    "from math import sqrt,pi,e\n",
+    "#Thinkness of graded design\n",
+    "V = 150*sqrt(2)\n",
+    "Ebd = 50\n",
+    "T = V/Ebd\n",
+    "print\"\\tThickness of graded design=  cm \\t\",T\n",
+    "#Based on figure 2.24\n",
+    "r = 2 # radius of the conductor\n",
+    "l = 10 #length of graded cylinder; The textbook uses 10 instead of 20\n",
+    "zr = l*(T+r)\n",
+    "print\"Curve =  cm**2 \\t\",zr\n",
+    "#Volume of graded design V1\n",
+    "V1 = 4*pi*zr*(zr-r)\n",
+    "print\"V1 =  cm**3 \\t\",V1 #Unit is wrong in the textbook\n",
+    "#Thickness of regular design as obtained form Eq.2.77\n",
+    "pow = V/(2*Ebd)\n",
+    "t = 2*(e**pow-1)\n",
+    "print\"Thickness of regular design =  cm \\t\",t\n",
+    "#Volume of regular design V2\n",
+    "V2 = pi*((2+t)**2-4)\n",
+    "print\"V2 =  cm**3 \\t\",V2#unit not mentioned in textbook\n",
+    " \n",
+    "#Answers may vary due to round off error\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/prashanthkumar/Chapter1.ipynb b/sample_notebooks/prashanthkumar/Chapter1.ipynb
new file mode 100755
index 00000000..7cc7b973
--- /dev/null
+++ b/sample_notebooks/prashanthkumar/Chapter1.ipynb
@@ -0,0 +1,561 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:04b8728a2a112ab03efb0f888eacfe0c6847d92d7043fbf038fc0d7708b1bf9e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-Introduction of Electronic Device"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.1\n",
+      "#calcualte fusing current for given values\n",
+      "import math\n",
+      "print(\"I = K(d^1.5)\") ##formula used for fusing current\n",
+      "d=0.0031\n",
+      "print\"%s %.3f %s\"%(\"d = \",d,\"inches\") ##initializing values of diameter\n",
+      "I1=10244*(d**1.5);\n",
+      "I2=7585*(d**1.5);\n",
+      "I3=5320*(d**1.5); \n",
+      "I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d^1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d^1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d^1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d^1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d^1.5) = \",I5,\"Amp.\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "I = K(d^1.5)\n",
+        "d =  0.003 inches\n",
+        "for Copper, I = 10244*(d^1.5) =  1.77 Amp.\n",
+        "for Aluminum, I = 7585*(d^1.5) =  1.31 Amp.\n",
+        "for Silver, I = 5320*(d^1.5) =  0.92 Amp.\n",
+        "for Iron, I = 3148*(d^1.5) =  0.54 Amp.\n",
+        "for Tin, I = 1642*(d^1.5) =  0.28 Amp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.2\n",
+      "#calculate fusing current for given values\n",
+      "print(\"fusing current, I = K(d**1.5) Amp.\")##formula used for fusing current\n",
+      "d=0.0201\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "## note : calculation for fusing current of Iron is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fusing current, I = K(d**1.5) Amp.\n",
+        "d =  0.02 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  29.19 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  21.61 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  15.16 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  8.97 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  4.68 Amp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.3\n",
+      "#calculate for fusing current in all four cases\n",
+      "import math\n",
+      "print(\"fusing current, I = K(d**1.5) Amp.\") ##formula used for fusing current\n",
+      "print(\"(a)\") \n",
+      "d=0.0159\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "print(\"(b)\")\n",
+      "d=0.0063\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "print(\"(c)\")\n",
+      "d=0.0403\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "print(\"(d)\")\n",
+      "d=0.0452\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "print(\"(e)\")\n",
+      "d=0.0508\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "print(\"(f)\")\n",
+      "d=0.162\n",
+      "print\"%s %.2f %s\"%(\"d = \",d,\"inches\") ##initializing value of diameter\n",
+      "I1=10244*(d**1.5);I2=7585*(d**1.5); I3=5320*(d**1.5); I4=3148*(d**1.5); I5=1642*(d**1.5)  ##calculation for fusing current\n",
+      "print\"%s %.2f %s\"%(\"for Copper, I = 10244*(d**1.5) = \",I1,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Aluminum, I = 7585*(d**1.5) = \",I2,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Silver, I = 5320*(d**1.5) = \",I3,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Iron, I = 3148*(d**1.5) = \",I4,\"Amp.\")\n",
+      "print\"%s %.2f %s\"%(\"for Tin, I = 1642*(d**1.5) = \",I5,\"Amp.\")\n",
+      "\n",
+      "\n",
+      "\n",
+      "## note : in part (e) ... calculation for fusing current of silver is wrong.\n",
+      "## note : in part (f) ... calculation for fusing current of Iron is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fusing current, I = K(d**1.5) Amp.\n",
+        "(a)\n",
+        "d =  0.02 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  20.54 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  15.21 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  10.67 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  6.31 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  3.29 Amp.\n",
+        "(b)\n",
+        "d =  0.01 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  5.12 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  3.79 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  2.66 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  1.57 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  0.82 Amp.\n",
+        "(c)\n",
+        "d =  0.04 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  82.88 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  61.36 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  43.04 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  25.47 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  13.28 Amp.\n",
+        "(d)\n",
+        "d =  0.05 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  98.44 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  72.89 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  51.12 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  30.25 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  15.78 Amp.\n",
+        "(e)\n",
+        "d =  0.05 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  117.29 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  86.85 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  60.91 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  36.04 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  18.80 Amp.\n",
+        "(f)\n",
+        "d =  0.16 inches\n",
+        "for Copper, I = 10244*(d**1.5) =  667.95 Amp.\n",
+        "for Aluminum, I = 7585*(d**1.5) =  494.57 Amp.\n",
+        "for Silver, I = 5320*(d**1.5) =  346.88 Amp.\n",
+        "for Iron, I = 3148*(d**1.5) =  205.26 Amp.\n",
+        "for Tin, I = 1642*(d**1.5) =  107.06 Amp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.4\n",
+      "#calculate resistance for given resistivity\n",
+      "import math\n",
+      "A=0.5189*10**-6##wire cross sectional area\n",
+      "rho=1.725*10**-8##resistivity\n",
+      "l=100 ##wire length\n",
+      "print\"%s %.3e %s\"%(\"A =\",A,\"merer square\") \n",
+      "print\"%s %.2e %s\"%(\"rho =\",rho,\"ohm-m\")\n",
+      "print\"%s %.2f %s\"%(\"l =\",l,\"m\")\n",
+      "print\"%s %.2f %s\"%(\"R = rho*l/A = \",rho*l/A,\"ohm\") ##resistance\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A = 5.189e-07 merer square\n",
+        "rho = 1.73e-08 ohm-m\n",
+        "l = 100.00 m\n",
+        "R = rho*l/A =  3.32 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.5\n",
+      "#calculate resistance wire\n",
+      "import  math\n",
+      "A=0.2588*10**-6##wire cross-sectional area\n",
+      "rho=1.725*10**-8##resistivity\n",
+      "l=100 ##wire length\n",
+      "print\"%s %.2e %s\"%(\"A =\",A,\"merer square\")\n",
+      "print\"%s %.2e %s\"%(\"rho =\",rho,\"ohm-m\")\n",
+      "print\"%s %.2f %s\"%(\"l =\",l,\"m\")\n",
+      "print\"%s %.2f %s\"%(\"R = rho*l/A = \",rho*l/A,\"ohm\") ##resistance of wire\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A = 2.59e-07 merer square\n",
+        "rho = 1.73e-08 ohm-m\n",
+        "l = 100.00 m\n",
+        "R = rho*l/A =  6.67 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6-pg39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.6\n",
+      "#calculate resistance at temperature at T2\n",
+      "R1 = 14##resistance at temperature T1 \n",
+      "alpha=0.005\n",
+      "T1=20;##initial temperature\n",
+      "T2=120 ##final temperature\n",
+      "print\"%s %.2f %s %.2f %s %.2f %s%.2f %s \"%(\"R1 = \",R1, \"ohm\"and\" alpha = \",alpha,\"\"and \" T1 = \",T1,\"degreeC\"and \"T2 = \",T2,\"degreeC\")\n",
+      "print\"%s %.2f %s\"%(\"R2 = R1(1+(alpha*(T1-T2))) = \",R1*(1+(alpha*(T2-T1))),\"ohm\") ##resistance at temperature T2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R1 =  14.00  alpha =  0.01  20.00 T2 = 120.00 degreeC \n",
+        "R2 = R1(1+(alpha*(T1-T2))) =  21.00 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7-pg40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##EX1.7\n",
+      "#calculate force of electron charge\n",
+      "import math\n",
+      "Ex=3;Ey=4;Ez=2##electric field\n",
+      "e=1.6*10**-19 ##electorn charge\n",
+      "print(\"E = 3ax + 4ay + 2az k V/m\")\n",
+      "print(\"e = 1.6*10**-19 C\")\n",
+      "print\"%s %.2e %s %.2e %s %.2e %s \"%(\" F=eE = \",Ex*e*1000,\"ax + \",Ey*e*1000,\"ay + \",Ez*e*1000,\"az N\") ##force\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E = 3ax + 4ay + 2az k V/m\n",
+        "e = 1.6*10**-19 C\n",
+        " F=eE =  4.80e-16 ax +  6.40e-16 ay +  3.20e-16 az N \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8-pg40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.8\n",
+      "#calculate elctric field\n",
+      "import math\n",
+      "F=0.1*10**-12##force applied\n",
+      "e = 1.6*10**-19##electron charge\n",
+      "print\"%s %.2e %s  %.2e %s \"%(\"F= \",F,\"N \"and \" e = \",e,\"C\")\n",
+      "print\"%s %.2f %s\"%(\"E = F/e =\",F/e,\"V/m\")##electric field\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "F=  1.00e-13  e =   1.60e-19 C \n",
+        "E = F/e = 625000.00 V/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9-pg41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.9\n",
+      "#calculate charge of electron\n",
+      "import math\n",
+      "F = 3*(10**-12) ##force applied\n",
+      "E = 5*(10**-6)  ##electric field\n",
+      "print\"%s %.2e %s\"%(\"F = \",F,\"N\")\n",
+      "print\"%s %.2e %s\"%(\"E = \",E,\"V/m\")\n",
+      "print\"%s %.2e %s\"%(\"Q= F/E = \",F/E,\"C\") ##chage\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "F =  3.00e-12 N\n",
+        "E =  5.00e-06 V/m\n",
+        "Q= F/E =  6.00e-07 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10-pg44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.10\n",
+      "#calculate force \n",
+      "import math\n",
+      "B = 2*10**-6 ##magnetic flux density\n",
+      "V = 4*10**6  ##electron velocity\n",
+      "e= 1.6*10**-19##elcetron charge\n",
+      "print\"%s %.2e %s\"%(\"B =\",B,\"ax wb/m.sq\")\n",
+      "print\"%s %.2f %s\"%(\"V =\",V,\"az m/s\")\n",
+      "print\"%s %.3e %s\"%(\"e = \",e, \"C\")\n",
+      "print\"%s %.2e %s\"%(\"F = e[VxB] =\",e*V*B,\"ay N\")##force\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B = 2.00e-06 ax wb/m.sq\n",
+        "V = 4000000.00 az m/s\n",
+        "e =  1.600e-19 C\n",
+        "F = e[VxB] = 1.28e-18 ay N\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Ex11-pg44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##Ex1.11\n",
+      "#calculate force on electron due to field\n",
+      "import math\n",
+      "Hx = 1*10**-3  ##magnetic field in x-axis\n",
+      "Hy = 2*10**-3  ##magnetic field in y-axis\n",
+      "V = (4*10**6)  ##electron velocity\n",
+      "micro_not=(4*math.pi*(10**-7)) ##permitivity in vaccum\n",
+      "e=1.6*10**-19  ##charge of electorn\n",
+      "print\"%s %.2e %s %.2e %s \"%(\" H = \",Hx,\"ax  + \",Hy,\"ay A/m\")\n",
+      "print\"%s %.2f %s\"%(\"V = \",V,\"ay m/s\")\n",
+      "Bx = micro_not*Hx; By = micro_not*Hy ##magnetic flux density\n",
+      "print\"%s %.2e %s %.2e %s \"%(\"B = micro_not*H = \",Bx,\"ax + \",By,\"ay wb/m.sq\")\n",
+      "print\"%s %.2e %s \"%(\"F = e[VxB] = \",e*V*Bx,\"az N\") ##force on electron due to field\n",
+      "\n",
+      "\n",
+      "## note : there is a misprint in the textbook for the above problem\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " H =  1.00e-03 ax  +  2.00e-03 ay A/m \n",
+        "V =  4000000.00 ay m/s\n",
+        "B = micro_not*H =  1.26e-09 ax +  2.51e-09 ay wb/m.sq \n",
+        "F = e[VxB] =  8.04e-22 az N \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/pratiksonone/Chapter1.ipynb b/sample_notebooks/pratiksonone/Chapter1.ipynb
new file mode 100755
index 00000000..9ad0850a
--- /dev/null
+++ b/sample_notebooks/pratiksonone/Chapter1.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2328a7dc3cc0ee8a847caf61a092e7220be4bed1f89464a7bc727661ff6069c6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##calculate the The ultimate strength  and the yield strength\n",
+      "## initialization of variables\n",
+      "import math\n",
+      "## part (a)\n",
+      "a=700. ## M Pa from figure 1.8\n",
+      "b=100. ## M Pafrom figure 1.8\n",
+      "m=1/6. ## from figure 1.8\n",
+      "Y=450. ## M Pa from figure 1.9\n",
+      "##calculations\n",
+      "sigma_u=a+m*b\n",
+      "## results\n",
+      "print('\\n part (a) \\n')\n",
+      "print\"%s %.2f %s\"%(' The ultimate strength is sigma = ',sigma_u,' M Pa')\n",
+      "print\"%s %.2f %s\"%('\\n and the yield strength is Y = ',Y,'M Pa')\n",
+      "\n",
+      "## part (b)\n",
+      "c1=62. ## from figure 1.8\n",
+      "d1=0.025 ## from figure 1.8\n",
+      "c2=27. ## from figure 1.10a\n",
+      "d2=0.04 ## from figure 1.10a\n",
+      "## calculations\n",
+      "U_f1=c1*b*d1*10**6\n",
+      "U_f2=c2*b*d2*10**6\n",
+      "## results\n",
+      "print('\\n part (b)')\n",
+      "print\"%s %.2e %s\"%('\\n The modulus of toughness for alloy steel is Uf = ',U_f1,' N/m^2')\n",
+      "print\"%s %.2e %s\"%('\\n and structural steel is Uf = ',U_f2,' N/m^2')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " part (a) \n",
+        "\n",
+        " The ultimate strength is sigma =  716.67  M Pa\n",
+        "\n",
+        " and the yield strength is Y =  450.00 M Pa\n",
+        "\n",
+        " part (b)\n",
+        "\n",
+        " The modulus of toughness for alloy steel is Uf =  1.55e+08  N/m^2\n",
+        "\n",
+        " and structural steel is Uf =  1.08e+08  N/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##calculate the permanet strain\n",
+      "## initialization of variables\n",
+      "import math\n",
+      "sigma=500. ## Stress M Pa\n",
+      "eps=0.0073 ## Strain\n",
+      "sigma_A=343. ## M Pa from figure 1.9\n",
+      "eps_A=0.00172 ## from figure 1.9\n",
+      "## part (a)\n",
+      "E=sigma_A/eps_A\n",
+      "\n",
+      "## part (B)\n",
+      "eps_e=sigma/E\n",
+      "eps_p=eps-eps_e\n",
+      "## results\n",
+      "print(' part (a) \\n')\n",
+      "print\"%s %.2f %s\"%(' The modulus of elasticity of the rod is E = ',E/1000,' G Pa')\n",
+      "print('\\n part (b)')\n",
+      "print\"%s %.4f %s\"%('\\n the permanent strain is = ',eps_p,'')\n",
+      "print\"%s %.4f %s\"%('\\n and the strain recovered is =',eps_e,'')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " part (a) \n",
+        "\n",
+        " The modulus of elasticity of the rod is E =  199.42  G Pa\n",
+        "\n",
+        " part (b)\n",
+        "\n",
+        " the permanent strain is =  0.0048 \n",
+        "\n",
+        " and the strain recovered is = 0.0025 \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "##calculate the diameter\n",
+      "## initialization of variables\n",
+      "import math\n",
+      "D=25. ## kN\n",
+      "L=60. ## kN\n",
+      "W=30. ##kN\n",
+      "Y=250. ## M Pa\n",
+      "safety=5./3. ## AISC, 1989\n",
+      "## calculations\n",
+      "Q=(D+L+W)*10**3. ## converted to N\n",
+      "A=safety*Q/Y\n",
+      "r=math.sqrt(A/math.pi)+0.5 ## additional 0.5 mm is for extra safety\n",
+      "d=1.8*r ## diameter\n",
+      "## results\n",
+      "print('Part (a) \\n ')\n",
+      "print\"%s %.2f %s %.2f %s \"%('A rod of ',d,' mm'and ' in diameter, with a cross sectional area of ',math.pi*(d**2./4.),' mm^2, is adequate')\n",
+      "## The diameter is correct as given in the textbook. Area doesn't match due to rounding off error and partly because it's a design problem.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a) \n",
+        " \n",
+        "A rod of  29.02  in diameter, with a cross sectional area of  661.39  mm^2, is adequate \n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/rohitmahadev/Chapter1_1.ipynb b/sample_notebooks/rohitmahadev/Chapter1_1.ipynb
new file mode 100755
index 00000000..16205a41
--- /dev/null
+++ b/sample_notebooks/rohitmahadev/Chapter1_1.ipynb
@@ -0,0 +1,651 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d669fd33d214dfe64d2894959328f05abdede2e1f58e2a68176378d9cf3931d3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter1-circuit analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "r1=4.;##resistance between point A and B in ohms which is in series with 10 volts d.c. supply.\n",
+      "r2=3.;##resistance between points C and D in ohms which is in series with a d.c. supply of 8 volts.\n",
+      "r3=5.;##resistance betwwen points F and G in ohms\n",
+      "##arms AB.CD,FG are in parallel with each other.\n",
+      "v1=10.;##d.c. supply voltage in the arm AB in volts\n",
+      "v2=8.;##d.c. supply voltage in the arm CD in volts\n",
+      "\n",
+      "##calculations\n",
+      "##using SUPER POSITION THEOREM\n",
+      "##voltage source of 8 volts is neglected and supply is 10 volts d.c\n",
+      "R1=r1+((r2*r3)/(r2+r3));## total resistance in ohms\n",
+      "bIa1=v1/R1;##current in arm AB in amperes \n",
+      "cId1=v1*(r3/(R1*(r2+r3)));##current in arm CD in amperes\n",
+      "dIc1= -cId1;\n",
+      "fIg1=(v1/R1)-cId1;##current in arm FG in amperes\n",
+      "##voltage source of 10 volts is neglected and supply is 8 volts d.c\n",
+      "R2=r2+((r1*r3)/(r1+r3));##total resistance in ohms\n",
+      "dIc2=v2/R2;##current in arm CD in amperes\n",
+      "aIb2=v2*(r3/(R2*(r3+r1)));##current in arm AB in amperes\n",
+      "bIa2= -aIb2;\n",
+      "fIg2=(v2/R2)-aIb2;##current in arm FG in amperes\n",
+      "I1=bIa1+bIa2;##current in 10 V source in amperes\n",
+      "I2=dIc1+dIc2;##current in 8V source in amperes\n",
+      "I3=fIg1+fIg2;##current in arm FG in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s %.2f %s '%('the currents in the circuit are ',I1,'A' and '',I2,'A' and  '',I3,'A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the currents in the circuit are  0.85  0.47  1.32 A \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v1=10.;## d.c. voltage source in volts present in arm 1 in series with a 2 ohm resistor\n",
+      "v2=15.;##d.c. voltage source in volts present in arm 2 in series with a 3 ohm resistor\n",
+      "r1=2.; ##resistance in arm 1 in ohms\n",
+      "r2=3.;## resistance in arm 2 in ohms\n",
+      "r3=1.8;##resistance between node formed by arm 1 and 2 and point A\n",
+      "R=3.;##load resistance in ohms placed in arm AB\n",
+      "## point A and B are in open condition and arm 1 and 2 are in parallel\n",
+      "\n",
+      "##calculations\n",
+      "##thevenin equivalent circuit method\n",
+      "i1=(v2-v1)/(r1+r2);## current in the parallel circuit in amperes\n",
+      "e=v2-(i1*r2);## open cicuit e.m.f in volts i.e. thevenin's voltage\n",
+      "r=r3+((r1*r2)/(r1+r3));## resistance to be considered between AandB in ohms i.e. thevenin's resistance\n",
+      "I=e/(r+R);##current through the load resistance in amperes\n",
+      "\n",
+      "##output \n",
+      "print'%s %.2f %s %.2f %s %.2f %s '%(' the thevenin equivalent generator will have a constant e.m.f. of ',e,' V' and 'internal resistance of ',r,' ohm.' and '\\n the current in 3 ohm resistor is ',I,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " the thevenin equivalent generator will have a constant e.m.f. of  12.00 internal resistance of  3.38 \n",
+        " the current in 3 ohm resistor is  1.88  A \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "r1=0.2;##resistance in arm 1 in ohms which is in series with 10 volts d.c. supply.\n",
+      "r2=0.2;##resistance in arm 2 in ohms which is in series with a d.c. supply of 12 volts.\n",
+      "r3=0.4;##resistance in arm 3 in ohms whichis in series with 15 volts d.c. supply .\n",
+      "##arms 1,2 and 3 are in parallel with each other and are parallel with the arm AB.\n",
+      "v1=10;##d.c. supply voltage in the arm 1 in volts\n",
+      "v2=12;##d.c. supply voltage in the arm 2 in volts\n",
+      "v3=15;##d.c. supply voltage in the arm 3 in volts\n",
+      "R1=2.28;## resistance in arm AB in ohms in one case\n",
+      "R2=5.82;## resistance in arm AB in ohms in another\n",
+      "\n",
+      "##calculations\n",
+      "##thevenin equivalent circuit method\n",
+      "e=((v3/r3)+(v2/r2)+(v1/r1))/((1/r1)+(1/r2)+(1/r3));## thevenin's voltage in volts\n",
+      "r=1/((1/r1)+(1/r2)+(1/r3));##thevenin's resistance in ohms\n",
+      "I1=e/(r+R1);## current when resistance in AB arm is 2.28 ohms\n",
+      "I2=e/(r+R2);## current when resistance in AB arm is 5.82 ohms\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s  %.2f %s %.2f %s '%('the equivalent generator has a constant voltage of ',e,' V 'and 'an internal resistance of ',r,' ohms' '\\n the load currents are ',I1,' A' and '',I2,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the equivalent generator has a constant voltage of  11.80 an internal resistance of  0.08  ohms\n",
+        " the load currents are   5.00  2.00  A \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##AB,BC,CD,DA are arms of a wheatstone bridge\n",
+      "r1=4.;##resistance in arm AB in ohms\n",
+      "r2=6.;##resistance in arm BC in ohms\n",
+      "r3=5.;##resistance in arm CD in ohms\n",
+      "r4=3.;##resistance in arm DA in ohms\n",
+      "v=4.;##d.c. supply given between points A and C in volt\n",
+      "R=10.;##resistance of the detector placed between the points B and D in ohms\n",
+      "\n",
+      "##calculations\n",
+      "aIb=v/(r1+r2);##current in arm AB in amperes\n",
+      "aId=v/(r3+r4);##current in arm DA in amperes\n",
+      "aVb=aIb*r1;##voltage drop along arm AB in volts\n",
+      "aVd=aId*r4;##voltage drop across arm AD in volts\n",
+      "dVb=aVb-aVd;##since D is positive with respect to B\n",
+      "e=dVb;## open circuit voltage in volts\n",
+      "r0=((r1*r2)/(r1+r2))+((r3*r4)/(r3+r4));##equivalent resistance in ohms when the supply neglected\n",
+      "I=e/(r0+R);##current through the 10 ohms resistance in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s'%('the current through the detector will be ',I,' A in the direction from D to B')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the current through the detector will be  0.01  A in the direction from D to B\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg19"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v1=21.;##voltage of first battery in arm 1 in volts\n",
+      "v2=16.;##voltage of second battery in arm 2 in volts\n",
+      "r1=3.;##internal resistance of first battery in ohms\n",
+      "r2=2.;##internal resistance of second battery in ohms\n",
+      "R=6.;##resistance going to be introduced in arm AB in ohms\n",
+      "##arms 1,2 and AB are in parallel\n",
+      "##arm AB is a short circuit path\n",
+      "\n",
+      "##calculations\n",
+      "##norton's equivalent circuit method\n",
+      "Isc=(v1/r1)+(v2/r2);##current through short circuit path in amperes\n",
+      "aRb=(r1*r2)/(r1+r2);##equivalent resistance in ohms\n",
+      "##now 6ohm resistor is placed in arm AB\n",
+      "aIb=Isc*((aRb)/(aRb+R));##current through 6 ohm resistor in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s %.2f %s '%('the constants for norton equivalent generator are ',Isc,' A' and '',aRb,'ohm' '\\n the current through the 6 ohm resistor is ',aIb,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the constants for norton equivalent generator are  15.00  1.20 ohm\n",
+        " the current through the 6 ohm resistor is  2.50  A \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6-pg20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v1=5.;##voltage of battery in arm 1 in volts\n",
+      "v2=10.;##voltage of battery in arm 2 in volts\n",
+      "v3=20.;##voltage of battery in arm 3 in volts\n",
+      "r1=3.;##internal resistance of battery in arm 1 in ohms\n",
+      "r2=8.;##internal resistance of battery in arm 2 in ohms\n",
+      "r3=24.;##internal resistance of battery in arm 3 in ohms\n",
+      "##arms 1,2,3 and AB are in parallel with each other and AB are in open condition\n",
+      "r4=6.;##resistance between node formed by arms 1,2 and 3 and point A in ohms\n",
+      "R0=7.;##load  resistance to be connected in arm AB in ohms\n",
+      "##calculations\n",
+      "##norton's equivalent method\n",
+      "##batteries are neglected. so, only internal resistances remain in the arms\n",
+      "R=1./((1./r1)+(1./r2)+(1./r3));##equivalent resistance in ohms\n",
+      "aRb=R+r4;## total resistance when looked into the circuit from arm AB in ohm\n",
+      "##applying superposition principle to determine the short circuit current\n",
+      "##battery in arm 1 alone is considered\n",
+      "R1=r1+(1./((1./r2)+(1./r3)+(1./r4)));##effective resistance in ohms\n",
+      "I1=v1/R1;##current in amperes\n",
+      "pd=I1*r1;##potential drop across the parallel combination in volts\n",
+      "aIb1=pd/r4;##current in amperes\n",
+      "##battery in the arm 2 alone is considered\n",
+      "R2=r2+(1./((1./r1)+(1./r3)+(1./r4)));## effective resistance in ohms\n",
+      "I2=v2/R2;##current in amperes\n",
+      "V1=I2/((1./r1)+(1./r3)+(1./r4));##voltage in volts\n",
+      "aIb2=V1/r4;##current in amperes\n",
+      "##battery in the arm 3 alone is considered\n",
+      "R3=r3+(1./((1./r1)+(1./r2)+(1./r4)));##effective resistance in ohms\n",
+      "I3=v3/R3;##current in amperes\n",
+      "V2=I3/((1./r1)+(1./r2)+(1./r4));##voltage in volts\n",
+      "aIb3=V2/r4;##current in amperes\n",
+      "Isc=aIb1+aIb2+aIb3;##short circuit current in amperes\n",
+      "I=Isc*(aRb/(aRb+R0));##current through load resistor in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s %.2f %s'%('Nortons equivalent generator will produce a constant current of ',Isc,' A' and 'has a shunt resistance of ',r2,' ohms' '\\n the current through the external resistor will be ',I,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Nortons equivalent generator will produce a constant current of  0.94 has a shunt resistance of  8.00  ohms\n",
+        " the current through the external resistor will be  0.50  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7-pg20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##AB,BC,CD,DA are arms of a wheatstone bridge\n",
+      "r1=4.;##resistance in arm AB in ohms\n",
+      "r2=6.;##resistance in arm BC in ohms\n",
+      "r3=5.;##resistance in arm CD in ohms\n",
+      "r4=3.;##resistance in arm DA in ohms\n",
+      "v=4.;##d.c. supply given between points A and C in volt\n",
+      "R0=10.;##resistance of the detector placed between the points B and D in ohms\n",
+      "##a detector is placed between the point B and D\n",
+      "\n",
+      "##calculations\n",
+      "## noerton's equivalent circuit method\n",
+      "R1=((r1*r2)/(r1+r2))+((r3*r4)/(r3+r4));## equivalent resistance assuming short circuit between poin A and C in ohms\n",
+      "R2=((r1*r4)/(r1+r4))+((r2*r3)/(r2+r3));##equivalent resistance assuming short circuit between points B and D in ohms\n",
+      "I1=v/R2;## total current in amperes\n",
+      "aIb=v*(r4/(R2*(r4+r1)));##current in arm AB in amperes\n",
+      "aVDb=v*aIb;##voltage drop in arm AB\n",
+      "bVDc=v-aVDb;##voltage drop in arm DC\n",
+      "bIc=bVDc/r2;##currrent in arm BC in amperes\n",
+      "dIb=bIc-aIb;##current in arm DB in amperes\n",
+      "Isc=dIb;##short circuit current in amperes\n",
+      "I=Isc*(R1/(R1+R0));##current through the detector in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.4f %s %.2f %s %.2f %s'%('Nortons equivalent generator will produce a constant current of ',Isc,' A' and 'has a shunt resistance of ',R1,' ohms' '\\n the current through the external resistor will be ',I,' A')\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Nortons equivalent generator will produce a constant current of  0.0234 has a shunt resistance of  4.28  ohms\n",
+        " the current through the external resistor will be  0.01  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8-pg21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##arma AB,BC and CA forms delta connection\n",
+      "r1=2.;##resistance in arm AB in ohms\n",
+      "r2=3.;##resistance in arm BC in ohms\n",
+      "r3=5.;##resistance in arm CA in ohms\n",
+      "\n",
+      "##calculations\n",
+      "##conversion of given delta into star connection\n",
+      "##let N be the star point\n",
+      "R1=(r1*r2)/(r1+r2+r3);##resistance in arm AN in ohms\n",
+      "R2=(r2*r3)/(r1+r2+r3);##resistance in arm BN in ohms\n",
+      "R3=(r1*r3)/(r1+r2+r3);##resistance in arm CN in ohms\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s %.2f %s'%('the respective star connected resistances are ',R1,' ohm'and '',R2,' ohm' and '',R3,'ohm')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the respective star connected resistances are  0.60  1.50  1.00 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex9-pg22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##AB,BC,CD,DA are arms of a wheatstone bridge\n",
+      "r1=5.;##resistance in arm AB in ohms\n",
+      "r2=20.;##resistance in arm BC in ohms\n",
+      "r3=15.;##resistance in arm CD in ohms\n",
+      "r4=4.;##resistance in arm DA in ohms\n",
+      "v=4.;##d.c. supply given between points A and C in volt\n",
+      "r0=0.5;## internal resistances pf the d.c. supply in ohms\n",
+      "r5=15.;##resistance in arm BD in ohms\n",
+      "\n",
+      "##calculations\n",
+      "##BCD is replaced by equivalent star connection\n",
+      "##assume N as star piont after conversion\n",
+      "bRn=(r2*r3)/(r3+r2+r5);##resistance in arm BN in ohms\n",
+      "cRn=(r2*r5)/(r3+r2+r5);##resistance in arm CN in ohms\n",
+      "dRn=(r5*r3)/(r3+r2+r5);##resistance in arm DN in ohms\n",
+      "R=r0+cRn+(((r1+bRn)*(r4+dRn))/(r1+bRn+r4+dRn));##total resistance in ohms after conversion\n",
+      "I=v/R;##totalcurrent supply in amperes\n",
+      "I1=(v/R)*((r4+dRn)/(r1+bRn+r4+dRn));##current between points A and B in amperes\n",
+      "I2=I-I1;##current between points A and D in amperes\n",
+      "V1=I1*r1;##voltage drop across r1 in volts\n",
+      "V2=I2*r4;##voltage drop across r4 in volts\n",
+      "V3=V2-V1;##voltage drop across r5 in volts and B is positive to D\n",
+      "I3=V3/r5;##current between points B and D in amperes\n",
+      "I4=I1-I3;##current between points B and C in amperes\n",
+      "I5=I2+I3;##current between points D and C in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s %.2f %s  %.2f %s %.2f %s %.2f %s '%('the currents in each part of the circuit are \\n It= ',I,' A' and  '\\n aIb= ',I1,' A' and  '\\n aId= ',I2,' A 'and '\\n bId= ',I3,' A' and '\\n bIc= ',I4,' A' and '\\n dIc= ',I5,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the currents in each part of the circuit are \n",
+        " It=  0.35 \n",
+        " aIb=  0.15 \n",
+        " aId=  0.20 \n",
+        " bId=   0.00 \n",
+        " bIc=  0.15 \n",
+        " dIc=  0.20  A \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##AN,BN,CN are connected in star fashion where N is the nuetral point\n",
+      "r1=5.;##resistance in arm AN in ohms\n",
+      "r2=20.;##resistance in arm BN in ohms\n",
+      "r3=10.;##resistance in arm CN in ohms\n",
+      "\n",
+      "##calculations\n",
+      "##star to delta conversion\n",
+      "Y1=1./r1;##conductance of arm AN in seimens\n",
+      "Y2=1./r2;##conductance of arm BN in seimens\n",
+      "Y3=1./r3;##conductance of arm CN in seimens\n",
+      "R1=1./((Y1*Y2)/(Y1+Y2+Y3));##resistance of arm AB in ohms\n",
+      "R2=1./((Y2*Y3)/(Y1+Y2+Y3));##resistance of arm BC in ohms\n",
+      "R3=1./((Y1*Y3)/(Y1+Y2+Y3));##resistance of arm CA in ohms\n",
+      "\n",
+      "##ouput\n",
+      "print'%s %.2f %s %.2f %s %.2f %s '%('the equivalent resistances values for delta circuit are ',R1,' ohms,'and '',R2,' ohms' and '',R3,' ohms')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the equivalent resistances values for delta circuit are  35.00  70.00  17.50  ohms \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "##AB,BC,CD,DA forms an unbalanced wheatstone's bridge\n",
+      "r1=2.;##resistance in arm AB in ohms\n",
+      "r2=5.;##resistance in arm BC in ohms\n",
+      "r3=6.;##resistance in arm CD in ohms\n",
+      "r4=2.;##resistance in arm DA in ohms\n",
+      "r5=10.;##resistance of detector placed between the points B and D\n",
+      "v=4.;##batterry supplying d.c. voltage in volts which is placed between points A and C\n",
+      "r0=0.2;## internal resistance of the battery in ohms\n",
+      "\n",
+      "##calculations\n",
+      "##AB,BC and BD are cosidered to be in star connection with B as star point\n",
+      "Y1=1./r1;##conductacne of r1 in seimens\n",
+      "Y2=1./r2;##conductance of r2 in seimens\n",
+      "Y3=1./r5;##conductance of r5 in seimens\n",
+      "##after delta conversion\n",
+      "R1=1./((Y1*Y2)/(Y1+Y2+Y3));##resistance between points A and B in ohms\n",
+      "R2=1./((Y2*Y3)/(Y1+Y2+Y3));##resistance between points C and D in ohms\n",
+      "R3=1./((Y1*Y3)/(Y1+Y2+Y3));##resistance between points D and A in ohms\n",
+      "Rad=(r4*R3)/(r4+R3);##effective resistance of arm AD in ohms\n",
+      "Rdc=(r3*R2)/(r3+R2);##effective resistance of arm DC in ohms\n",
+      "Radc=(Rad+Rdc);##effective resistance if arms AD and DC in ohms\n",
+      "R=r0+((R1*Radc)/(R1+Radc));## total resistance of hte circuit in ohms\n",
+      "I=v/R;##total current in the circuit in amperes\n",
+      "I1=I*(R1/(R1+Radc));##current in arm AD in amperes\n",
+      "I2=I-I1;##current in arm AB in amperes\n",
+      "V1=I1*r4;##voltage across arm AD in volts\n",
+      "V2=I2*r1;##voltage across arm AB in volts\n",
+      "V3=V1-V2;##voltage across arm BD in volts and B is positive to D\n",
+      "I3=V3/r5;##current in arm BD in amperes\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s'%('the current in the detector is ',I3,' A')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the current in the detector is  0.01  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12-pg24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "## a battery consists of 10cells connected in series\n",
+      "v=1.5;##e.m.f. of each cell in volts\n",
+      "r=0.2;## internal resistance of each cell in ohms\n",
+      "n=10.;##number of cells in the battery\n",
+      "\n",
+      "##calculations\n",
+      "##for maximum power load resistance=internal resistance\n",
+      "R=n*r;##total internal resistance of hte battery in ohms\n",
+      "Rl=R;##load resistance in ohms\n",
+      "e=n*v;##total e.m.f. of battery in volts\n",
+      "I=e/(R+Rl);##current from battery in amperes\n",
+      "P=(I**2)*R;##heating loss in the battery in watts\n",
+      "V=e-(I*R);##terminal voltage in volts\n",
+      "\n",
+      "##output\n",
+      "print'%s %.2f %s %.2f %s '%('The maximum value of power which the battery may transfer is ',P,' W' and 'an equal quantity of power is dissipated in the battery. \\n under these conditions the terminal p.d. is',V,'')\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum value of power which the battery may transfer is  28.12 an equal quantity of power is dissipated in the battery. \n",
+        " under these conditions the terminal p.d. is 7.50  \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/saikiran/Chapter3.ipynb b/sample_notebooks/saikiran/Chapter3.ipynb
new file mode 100755
index 00000000..9b328651
--- /dev/null
+++ b/sample_notebooks/saikiran/Chapter3.ipynb
@@ -0,0 +1,240 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e5ce604b49bc138fd65826c495598b13c742523cc595beeb292a2a1aa22f7c49"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter3-Engine Thrust and Performance Parameters"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate The ram drag for given engine in kN\n",
+      "M0=0.85 ##Mach no.\n",
+      "a0=300. ##speed of sound in m/s\n",
+      "m=50. ##Air mass flow rate in kg/s\n",
+      "##Calculations\n",
+      "V0=M0*a0 ##Flight speed\n",
+      "Dr=m*V0 ##Ram drag\n",
+      "Dk=Dr/1000. ##in kN\n",
+      "print'%s %.2f %s'%(\"The ram drag for given engine in kN:\",Dk,\"\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ram drag for given engine in kN: 12.75 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate gross thurst of the core and fan nozzles\n",
+      "\n",
+      "Cv=450. ##exhaust velocity at core in m/s\n",
+      "Nv=350. ##exhaust velocity at nozzle in m/s\n",
+      "Cm=50. ##Mass flow rate through core in kg/s\n",
+      "Nm=350. ##Mass flow rate through nozzle in kg/s\n",
+      "##Calculations:\n",
+      "##Newton's second law\n",
+      "Fgc=Cm*Cv ##gross thrust of the core\n",
+      "Fgf=Nm*Nv ##gross thrust of the nozzle fan\n",
+      "print'%s %.f %s'%(\"Gross thrust of the core in\",Fgc,\"N\")\n",
+      "print'%s %.f %s'%(\"Gross thrust of the fan nozzles in\",Fgf,\"N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gross thrust of the core in 22500 N\n",
+        "Gross thrust of the fan nozzles in 122500 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calculate The rocket gross thrust and pressure thurst\n",
+      "V9=4000 ##in m/s\n",
+      "p9=200*10**3 ##in Pa\n",
+      "p0=100*10**3 ## in Pa\n",
+      "D=2. ##in meter\n",
+      "m=200.+50. ## in kg/s\n",
+      "A=math.pi*(D**2)/4. ##nozzle exit area\n",
+      "##let p=(p9-p0)*A i.e. pressure thrust\n",
+      "p=(p9-p0)*A\n",
+      "mt=m*V9 ##momentum thrust.\n",
+      "t=p+mt ##rocket gross thrust\n",
+      "print'%s %.2f %s'%(\"The pressure thrust in\",p,\"N\")\n",
+      "print'%s %.1f %s'%(\"The rocket gross thrust in\",t,\"N\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure thrust in 314159.27 N\n",
+        "The rocket gross thrust in 1314159.3 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#calcualte engine thurst takeoff\n",
+      "m0=100. ##air flow rate in kg/s\n",
+      "V0=0. ##takeoff assumptions in m/s\n",
+      "mf=2. ##2% of fuel-to-air ratio\n",
+      "Qr=43000. ##Heating value of typical hydrocarbon fuel in kJ/kg\n",
+      "V9=900. ##high speed exhaust jet (in m/s)\n",
+      "e=((m0+mf)*(V9)**2.)/(2.*(mf)*(Qr)*1000.)\n",
+      "m9=m0+mf\n",
+      "t=m9*V9 ## the engine thrust at takeoff.\n",
+      "print'%s %.f %s'%(\"The engine thrust at takeoff in SI units\",t,\"N\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The engine thrust at takeoff in SI units 91800 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the  Engine propulsive efficiency\n",
+      "V9=900. ## in m/s\n",
+      "V0=200. ## in m/s\n",
+      "e=2./(1.+(V9/V0))\n",
+      "print'%s %.7f %s'%(\"Engine propulsive efficiency\",e,\"\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Engine propulsive efficiency 0.3636364 \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex6-pg118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#estimate Propulsive efficiency\n",
+      "import math\n",
+      "V9=250. ##in m/s\n",
+      "V0=200. ##in m/s\n",
+      "##Calculations:\n",
+      "e=2./(1.+(V9/V0))\n",
+      "print'%s %.3f %s'%(\"Propulsive efficiency:\",e,\"\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Propulsive efficiency: 0.889 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/srikanthgugloth/Chapter2.ipynb b/sample_notebooks/srikanthgugloth/Chapter2.ipynb
new file mode 100755
index 00000000..d4e7a3ad
--- /dev/null
+++ b/sample_notebooks/srikanthgugloth/Chapter2.ipynb
@@ -0,0 +1,1124 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4425378c6999e9724676588b0097c2038f3833cd503390f3d7cf7bb3508521f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2-Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input from given graph\n",
+      "##calculation of initial accleration\n",
+      "ia=18/4.\n",
+      "## calculation of final accleration\n",
+      "fa=-18/10.\n",
+      "decel=-(fa)##calculation of deceleration\n",
+      "##calculation of total distance covered\n",
+      "d=0.5*(4.*18.)+(8.*18.)+0.5*(10.*18.)##area under velocity time graph\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"\\n the initial acceleration is \",ia,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the final acceleration is \",decel,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the distance covered is is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " the initial acceleration is  4.50  m/s^2\n",
+        "\n",
+        " the final acceleration is  1.80  m/s^2\n",
+        "\n",
+        " the distance covered is is  270.00  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v=0. ##car stops => final velocity=0\n",
+      "u=29. ##initial velocity\n",
+      "t=11. ##time \n",
+      "##calculation of acceleration\n",
+      "a=(v-u)/t##eqn of uniformly accelerated body\n",
+      "##calculating distance travelled during this period\n",
+      "d=(v+u)*t*0.5##eqn of uniformly accelerated body\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the accleration is \",a,\" ms^-2 \")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe distance travelled is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the accleration is  -2.64  ms^-2 \n",
+        "\n",
+        "the distance travelled is  159.50  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "v=120. ##velocity\n",
+      "a=75 ##accleration\n",
+      "##ca.lculation of time\n",
+      "t=2.*v/(a*math.cos(45/57.3))##eqn of uniformly accelerated body\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the time taken is \",t,\" s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the time taken is  4.53  s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex4-pg28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "f1=50.\n",
+      "f2=50.\n",
+      "##calculation of net force\n",
+      "f=f1+f2 ## the two forces act in same direction\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the resultant force is \",f,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the resultant force is  100.00  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex5-pg29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "##input\n",
+      "vc=25. ##velocity of car\n",
+      "va=10. ##velocity of wind\n",
+      "va1=15. ##velocity of wind westward\n",
+      "##calculation\n",
+      "v1=vc+va##resultant velocity for a tail of wind\n",
+      "v2=vc-va##when wind blows westward at 10 m/s^resultant velocity \n",
+      "v3=vc-va1##resultant velocity when wind blows westward at 15m/s^2\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"1. the resultant velocity of wind is \",v1,\" ms^-1 eastwards \")\n",
+      "print\"%s %.2f %s\"%(\"\\n2. the resultant velocity of wind is \",v2,\" ms^-1 westwards \")\n",
+      "print\"%s %.2f %s\"%(\"\\n3. the resultant velocity of wind is \",v3,\"  ms^-1westwards \")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. the resultant velocity of wind is  35.00  ms^-1 eastwards \n",
+        "\n",
+        "2. the resultant velocity of wind is  15.00  ms^-1 westwards \n",
+        "\n",
+        "3. the resultant velocity of wind is  10.00   ms^-1westwards \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex7-pg31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=30. ##velocity of speedboat\n",
+      "vw=40. ##velocity of wind\n",
+      "##calculation\n",
+      "x=(30./40.)##angle between original velocity of boat and resultant velocity\n",
+      "y=math.atan(x)*(57.3)##applying trigonometry\n",
+      "b=90.+y##bearing of boat\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the bearing of speedboat is \",b,\" deg\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the bearing of speedboat is  126.87  deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex8-pg32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#input\n",
+      "f1=6. ##tension on string AB\n",
+      "f2=6. ##tension on string BC\n",
+      "##calculation of tension\n",
+      "t=2.*f1*math.sin(55/57.3)## the resultant tension is the diagonal of rhombus formed\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"/n the resultant tension is \",t,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "/n the resultant tension is  9.83  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex10-pg33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input magnitude of forces\n",
+      "f1=40.\n",
+      "f2=50.\n",
+      "##calculation\n",
+      "d=50**2+40**2+2.*50.*40.*math.cos(50./57.3)##finding the diagonal\n",
+      "r=50**2+40**2+2.*50.*(40.)*math.cos(130./57.3)##reversing the side and finding diagonlprint\"%s %.2f %s\"%(\"the resultant is %3.3f\",d1)\n",
+      "r1=math.sqrt(r)##resultant sum\n",
+      "d1=math.sqrt(d)## resultant when smaller force is subtracted from larger\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"1. the resultant sum is \",d1,\" N\")\n",
+      "print\"%s %.2f %s\"%(\"\\n 2. the resultant when smaller force is subtracted from larger is \",r1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. the resultant sum is  81.68  N\n",
+        "\n",
+        " 2. the resultant when smaller force is subtracted from larger is  39.11  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex11-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=380.##velocity\n",
+      "##calculation\n",
+      "vh=v*math.cos(60./57.3)##horizontal component\n",
+      "vv=v*math.sin(60./57.3)##vertical component\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the horizontal component is \",vh,\" ms**-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe vertical component is \",vv,\" ms**-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the horizontal component is  190.03  ms**-1\n",
+        "\n",
+        "the vertical component is  329.07  ms**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex12-pg34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "fc=50.##force applied by magnet\n",
+      "x=90.-20. ##angle of force\n",
+      "##calculation\n",
+      "fb=fc*math.sin(70./57.3)##force due to b\n",
+      "fa=fc*math.cos(70./57.3)##force due to a\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the force due to b is \",fb,\" N\")\n",
+      "print\"%s %.2f %s\"%(\"\\nthe force due to b is \",fa,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the force due to b is  46.98  N\n",
+        "\n",
+        "the force due to b is  17.11  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex13-pg35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m1=1.\n",
+      "v1=25.\n",
+      "m2=2.\n",
+      "v2=0.\n",
+      "##calculation\n",
+      "v=(m1*v1)+(m2*v2)##applying princilpe of conservation of linear momentum\n",
+      "v4=v/(m1+m2)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the velocity with which both will move is \",v4,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the velocity with which both will move is  8.33  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex14-pg35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m1=1.##mass of object 1\n",
+      "v1=25.##velocity of object 1\n",
+      "m2=2.##mass of object 2\n",
+      "v2=0.##body at rest,velocity =0\n",
+      "v3=10.\n",
+      "##caclulation\n",
+      "u=((m1*v1)+(m2*v2)-(m2*v3))/2.##applying princilpe of conservation of linear momentum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"\\n the value of u is \",-u,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " the value of u is  -2.50  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex15-pg36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=2.##mass\n",
+      "r=4.##radius\n",
+      "v=6.##uniform speed\n",
+      "##calculation\n",
+      "w=v/r##angular velocity\n",
+      "f=m*r*w*w##centripetal force\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the angular velocity is \",w,\" rads^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the centripetal force is \",f,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the angular velocity is  1.50  rads^-1\n",
+        "\n",
+        " the centripetal force is  18.00  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex16-pg37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=140.##mass\n",
+      "v=8.##speed\n",
+      "r=5.##radius\n",
+      "g=9.8##acceleration due to gravity\n",
+      "##calculation\n",
+      "t=((m*v**2/5.)**2)+(140.*9.8)**2 ##applying parallelogram of vectors\n",
+      "t1=math.sqrt(t)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the tension in arm is \",t1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the tension in arm is  2256.91  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex17-pg38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=15.##velocity\n",
+      "m=70.##mass\n",
+      "r=50.##radius\n",
+      "##calculation\n",
+      "x=v*v/(r*10.)##applying parallelogram of vectors,then for equilibrium\n",
+      "y=math.atan(x)*57.3\n",
+      "r1=(m*10.)/math.cos(y/57.3)\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the inclination is \",y,\" deg\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the reaction is \",r1,\" N\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the inclination is  24.23  deg\n",
+        "\n",
+        " the reaction is  767.61  N\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex18-pg39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "r=5500.##radius\n",
+      "g1=6.7*10**-11\n",
+      "g=7##acceleration due to gravity\n",
+      "##calculation of mean density\n",
+      "p=3.*g/(4.*math.pi*r*10**3*g1)##mean density\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the mean density of planet is \",p,\" kgm^-3\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean density of planet is  4534.94  kgm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex19-pg40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=5.*10**24##mass of earth\n",
+      "g1=6.7*10**-11\n",
+      "##calculation\n",
+      "r=((6.7*10**-11.*5.*10**24*(3600.*24.)**2)/(4.*math.pi**2))**(1./3.)##orbit radius\n",
+      "v=(2.*math.pi*r)/(3600.*24.)##linear velocity\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the orbit radius is \",r,\"\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the linear velocity is \",v,\" ms^-1\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the orbit radius is  39863080.05 \n",
+        "\n",
+        " the linear velocity is  2898.92  ms^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex20-pg41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=3.*10**5##orbit speed\n",
+      "r=4.6*10**20##distance\n",
+      "g1=6.7*10**-11\n",
+      "##calculation of mass\n",
+      "m=v*v*r/g1 ##Newtons law\n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the mass is \",m,\" kg\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mass is  6.18e+41  kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex21-pg42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "v=0.6##speed\n",
+      "m=0.3##mass\n",
+      "##calculation\n",
+      "e=0.75*m*v*v##total kinetic energy is kinetic energy+moment of inertia\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the total kinetic energy is \",e,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the total kinetic energy is  0.08  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex22-pg43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "t1=34.\n",
+      "u=0.##starts from rest\n",
+      "x=3.##distance to move\n",
+      "##calculation\n",
+      "t=(3.*3./(10.*math.sin(t1)))**0.5##from law of conservation of energy\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the time taken is \",t,\" s\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the time taken is  1.30  s\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex23-pg43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "i1=53.##inertia when it spins with panels carrying solar cells\n",
+      "i2=37.##inertia about axis of rotation\n",
+      "##calculation\n",
+      "r=i1/i2##law of conservation of angular momentum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the ratio of angular velocities is\",r,\"\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of angular velocities is 1.43 \n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex25-pg45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "f=9.##frequency\n",
+      "x=0.##at midpoint of stroke x=0\n",
+      "##calculation\n",
+      "t=1./f\n",
+      "a=-4.*math.pi**2*f**2*x##acceleration for shm\n",
+      "v=2.*math.pi*f*0.05##velocity for shm\n",
+      "a1=-4.*math.pi**2*9**2*0.05##acceleration at amplitude\n",
+      "v1=0.##velocity at amplitude is 0\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the period of oscillation is \",t,\" s^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the velocity  at midpoint of stroke is \",v,\" ms^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the acceleration at midpoint of stroke is \",a,\" ms^-2\")\n",
+      "\n",
+      "print\"%s %.2f %s\"%(\"\\n the velocity at amplitude is \",v1,\"  ms^-1\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the acceleration at amplitude is \",a1,\" ms^-2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the period of oscillation is  0.11  s^-1\n",
+        "\n",
+        " the velocity  at midpoint of stroke is  2.83  ms^-1\n",
+        "\n",
+        " the acceleration at midpoint of stroke is  -0.00  ms^-2\n",
+        "\n",
+        " the velocity at amplitude is  0.00   ms^-1\n",
+        "\n",
+        " the acceleration at amplitude is  -159.89  ms^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex26-pg47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "g=10.\n",
+      "t=0.3##period of shm\n",
+      "##calculation\n",
+      "x=g*t**2/(4.*math.pi**2)##for shm maximum amplitude\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the maximum amplitude for bead to be in contact is \",x,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum amplitude for bead to be in contact is  0.02  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex27-pg48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "p1=2.3##period of pendulum\n",
+      "p2=3.1##period when pendulum is lengthened\n",
+      "##calculation\n",
+      "g=4.*math.pi**2/(p2**2-p1**2)##acceleration of free fall\n",
+      "l=p1**2*g/(4.*math.pi**2)##length of pendulum\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the acceleration of free fall is \",g,\" m/s^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the length of pendulum is \",l,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the acceleration of free fall is  9.14  m/s^2\n",
+        "\n",
+        " the length of pendulum is  1.22  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex28-pg49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##INPUT DATA\n",
+      "f=55. ##frequency\n",
+      "a=7.*10**-3 ##amplitude\n",
+      "\n",
+      "\n",
+      "##calculation\n",
+      "a=(-2.*math.pi*f)**2*a\n",
+      "\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the acceleration of the body when it is at its maximum displacement from its zero position is \",a,\" ms^-2\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the acceleration of the body when it is at its maximum displacement from its zero position is  835.96  ms^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex29-pg50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "f=55.##frequency\n",
+      "amp=7.*10**-3##amplitude\n",
+      "m=1.2##mass\n",
+      "##calculation\n",
+      "e=0.5*m*4.*math.pi**2*f**2*amp**2##maximum pe occurs at zero position\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the maximum pe is \",e,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum pe is  3.51  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex30-pg51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "l=6.5##length\n",
+      "m=0.06##mass of wire\n",
+      "m1=10##mass attached\n",
+      "g=9.8##acceleration due to gravity\n",
+      "e=2.1*10**11##youngs modulus\n",
+      "ro=8.*10**3##density of steel\n",
+      "##calculation\n",
+      "e1=m1*g*ro*l*l/(e*m)##extension caused \n",
+      "pe=0.5*g*m1*e1##potential energy \n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the extension caused is \",e1,\" m\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the potential energy is \",pe,\" J\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the extension caused is  2.63e-03  m\n",
+        "\n",
+        " the potential energy is  0.13  J\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex31-pg52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "w=250.*10**3\n",
+      "s=0.00003##strain\n",
+      "a=0.04##area\n",
+      "w1=320.*10**3\n",
+      "##calculation\n",
+      "e=w/(a*s)##youngs module\n",
+      "st=w1/a##stress\n",
+      "##output\n",
+      "print\"%s %.2f %s\"%(\"the youngs modulus is \",e,\" N/m^2\")\n",
+      "print\"%s %.2f %s\"%(\"\\n the stress is \",st,\" N/m^2\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the youngs modulus is  208333333333.33  N/m^2\n",
+        "\n",
+        " the stress is  8000000.00  N/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex32-pg53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "##input\n",
+      "m=40.##mass\n",
+      "g=9.8##acceleration due to gravity\n",
+      "E=2*10**11##youngs modulus\n",
+      "##calculation\n",
+      "t1=m*g/5.##principle of momentum\n",
+      "t2=4*m*g/5. ##principle of momentum\n",
+      "d=4.*(t2-t1)/(4.*math.pi*10**-6*E)##difference in length\n",
+      "##output\n",
+      "print\"%s %.2e %s\"%(\"the difference is \",d,\" m\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the difference is  3.74e-04  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/testingtesting/ajinkya.ipynb b/sample_notebooks/testingtesting/ajinkya.ipynb
new file mode 100755
index 00000000..2e81c43d
--- /dev/null
+++ b/sample_notebooks/testingtesting/ajinkya.ipynb
@@ -0,0 +1,483 @@
+{
+ "metadata": {
+  "celltoolbar": "Raw Cell Format",
+  "name": "",
+  "signature": "sha256:8ade9358728d8873e3787f09037539bd355e8453094fed25ae256529fe52d8d4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: Water Treatment"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 1,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import  math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 95    #mass of MgCl2\n",
+      "\n",
+      "m4 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 7.5     #amount of Mg(HCO3)2 in mg/l\n",
+      "\n",
+      "amnt_2 = 16     #amount of Ca(HCO3)2 in mg/l\n",
+      "\n",
+      "amnt_3 = 9     #amount of MgCl2 in mg/l\n",
+      "\n",
+      "amnt_4 = 13.6     #amount of CaSO4 in mg/l\n",
+      "\n",
+      "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n",
+      "\n",
+      "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)\n",
+      "\n",
+      "total= temp_hard +perm_hard\n",
+      "\n",
+      "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the total  hardness is =\",total,\"mg/l\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 15.0135295112 mg/l\n",
+        "the total  hardness is = 34.4872137218 mg/l\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 2,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1= 136   # mass of FeSO4\n",
+      "\n",
+      "m2 = 100  #mass of CaCO3\n",
+      "\n",
+      "#for 100 ppm hardness FeSO4 required per 10**6 parts of water is 136 parts\n",
+      "#for 200 ppm hardness\n",
+      "\n",
+      "amt= m1*200./m2\n",
+      "\n",
+      "print\"the amount of FeSO4 required is =\",amt,\"mg/l\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the amount of FeSO4 required is = 272 mg/l\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 3,Page number 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "conc = 15.6 *10**-6    #concentration of (CO3)-2\n",
+      "\n",
+      "m = 60                 #mass of CO3\n",
+      "\n",
+      "Molarity= conc*100./m\n",
+      "\n",
+      "print\"the molarity of (CO3)-2  is =\",Molarity,\"M\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the molarity of (CO3)-2  is = 2.6e-05 M\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 4,Page number 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 111    #mass of CaCl2\n",
+      "\n",
+      "m4 = 120    #mass of MgSO4\n",
+      "\n",
+      "m5 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 12.5     #amount of Mg(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_2 = 10.5     #amount of Ca(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_3 = 8.2     #amount of CaCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 2.6     #amount of MgSO4 in ppm\n",
+      "\n",
+      "amnt_5 = 7.5     #amount of CaSO4 in ppm\n",
+      "\n",
+      "temp_hard= (amnt_1*100./m1)+(amnt_2*100./m2)\n",
+      "\n",
+      "perm_hard= (amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5)\n",
+      "\n",
+      "total= temp_hard +perm_hard\n",
+      "\n",
+      "print\"the temporary hardness is =\",temp_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the permanent hardness is =\",perm_hard,\"mg/l\"\n",
+      "\n",
+      "print\"the total  hardness is =\",total,\"mg/l\"\n",
+      "\n",
+      "v= 100    #volume of sample\n",
+      "\n",
+      "v_EDTA = total*v/1000   #volume of EDTA \n",
+      "\n",
+      "print\"the volume of M/100 EDTA required is =\",v_EDTA,\"ml\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 15.0431253171 mg/l\n",
+        " the permanent hardness is = 15.0687599364 mg/l\n",
+        " the total  hardness is = 30.1118852535 mg/l\n",
+        " the volume of M/100 EDTA required is = 3.01118852535 ml\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 5,Page number 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 50000  #volume of water\n",
+      "\n",
+      "m1 = 84    #mass of MgCO3\n",
+      "\n",
+      "m2 = 100    #mass of CaCO3\n",
+      "\n",
+      "m3 = 95    #mass of MgCl2\n",
+      "\n",
+      "m4 = 111    #mass of CaCl2\n",
+      "\n",
+      "amnt_1 = 144     #amount of MgCO3 in ppm\n",
+      "\n",
+      "amnt_2 = 25     #amount of CaCO3 in ppm\n",
+      "\n",
+      "amnt_3 = 95     #amount of MgCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 111     #amount of CaCl2 in ppm\n",
+      "\n",
+      "lime = (74./100)*(2*(amnt_1*100./m1)+(amnt_2*100./m2)+(amnt_3*100./m3))*v\n",
+      "\n",
+      "print\"the lime required is =\",lime,\"mg\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lime required is = 17310714.2857 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 6,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 10**6  #volume of water\n",
+      "\n",
+      "m1 = 40.0  #mass of Ca+2\n",
+      "\n",
+      "m2 = 24.0    #mass of Mg+2\n",
+      "\n",
+      "m3 = 44.0    #mass of CO2\n",
+      "\n",
+      "m4 = 122.0    #mass of HCO3-\n",
+      "\n",
+      "amnt_1 = 20.0     #amount of Ca+2 in ppm\n",
+      "\n",
+      "amnt_2 = 25.0     #amount of Mg+2 in ppm\n",
+      "\n",
+      "amnt_3 = 30.0     #amount of CO2 in ppm \n",
+      "\n",
+      "amnt_4 = 150.0     #amount of HCO3- in ppm\n",
+      "\n",
+      "lime_1 = (74./100)*((amnt_2*100./m2)+(amnt_3*100./m3)+(amnt_4*100./m4))*v\n",
+      "\n",
+      "soda = (106./100)*((amnt_1*100./m1)+(amnt_2*100./m2)-(amnt_4*100./m4))*v\n",
+      "\n",
+      "print\"the lime required is =\",lime_1,\"mg\"\n",
+      "\n",
+      "print\"the soda required is =\",soda,\"mg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the lime required is = 218521485.345 mg\n",
+        "the soda required is = 33088797.8142 mg\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 7,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "v= 150        #volume of NaCl \n",
+      "\n",
+      "conc = 150.   #concentration of NaCl\n",
+      "\n",
+      "amnt =v*conc *100./117    #amnt of NaCl\n",
+      "\n",
+      "hard = 600.   #hardness of water\n",
+      "\n",
+      "vol= amnt*1000./hard\n",
+      "\n",
+      "print\"the volume of water is =\",round(vol,4),\"litres\"\n",
+      "\n",
+      "#calculation mistake in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the volume of water is = 32051.2821 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 8,Page number 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "strength = 10*0.85/9    #strength of EDTA\n",
+      "\n",
+      "#1000 ml EDTA solution == 1 g CaCO3\n",
+      "\n",
+      "#for 20 ml EDTA solution\n",
+      "\n",
+      "amnt= 20.*strength/1000\n",
+      "\n",
+      "#50 ml smple of water contains amnt CaCO3\n",
+      "\n",
+      "hard= amnt*10**6/50    #hardness of water \n",
+      "\n",
+      "print\"the hardness of water is =\",round(hard,3),\"ppm\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the hardness of water is = 377.778 ppm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Problem 9,Page number 372"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#Given Data\n",
+      "\n",
+      "m1 = 146    #mass of Mg(HCO3)2\n",
+      "\n",
+      "m2 = 162    #mass of Ca(HCO3)2\n",
+      "\n",
+      "m3 = 111    #mass of CaCl2\n",
+      "\n",
+      "m4 = 120    #mass of MgSO4\n",
+      "\n",
+      "m5 = 136    #mass of CaSO4\n",
+      "\n",
+      "amnt_1 = 12.5     #amount of Mg(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_2 = 10.5     #amount of Ca(HCO3)2 in ppm\n",
+      "\n",
+      "amnt_3 = 8.2     #amount of CaCl2 in ppm \n",
+      "\n",
+      "amnt_4 = 2.6     #amount of MgSO4 in ppm\n",
+      "\n",
+      "amnt_5 = 7.5     #amount of CaSO4 in ppm\n",
+      "\n",
+      "temp_hard= ((amnt_1*100./m1)+(amnt_2*100./m2))*0.1\n",
+      "\n",
+      "perm_hard= ((amnt_3*100./m3)+(amnt_4*100./m4)+(amnt_5*100./m5))*0.1\n",
+      "\n",
+      "print\"the temporary hardness is =\",round(temp_hard,3),\"degree Fr\"\n",
+      "\n",
+      "print\"the permanent hardness is =\",round(perm_hard,3),\"degree Fr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temporary hardness is = 1.504 degree Fr\n",
+        "the permanent hardness is = 1.507 degree Fr\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/sample_notebooks/vijayadurga/Chapter_5_Force_Torque_and_Shaft_power_Measurement.ipynb b/sample_notebooks/vijayadurga/Chapter_5_Force_Torque_and_Shaft_power_Measurement.ipynb
new file mode 100755
index 00000000..379da0ca
--- /dev/null
+++ b/sample_notebooks/vijayadurga/Chapter_5_Force_Torque_and_Shaft_power_Measurement.ipynb
@@ -0,0 +1,265 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 Force Torque and Shaft power Measurement"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_1 pgno:204"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "x=(Sg*sig_f*(1+v))/(2*E)\n",
+      "('a voltmeter with a maximum range of  mV is suitable for measurement', 94.9385766342288)\n",
+      "Round it off to get the suitable range voltmeter\n"
+     ]
+    }
+   ],
+   "source": [
+    "#CHAPTER 5_ Force,Torque and Shaft Power Measurement\n",
+    "#Caption : Load cell\n",
+    "# Example 1# Page 294\n",
+    "from math import sqrt\n",
+    "\n",
+    "Sg=2.;   # Strain gage factor\n",
+    "Rg=120.;    # Gage resistance\n",
+    "v=0.3    # poissons ratio\n",
+    "E=210*10**9;    # for steel\n",
+    "Pd=1.   #('enter the power dissipation capacity=:')\n",
+    "# Looking for a suitable voltage measuring system\n",
+    "sig_f=700*10**6   #('enter the fatigue strength=:')\n",
+    "P_max=10000.   #('enter the maximum load=:')\n",
+    "# For a load cell of square cross-section d,\n",
+    "d=sqrt(P_max/sig_f);\n",
+    "Ei=sqrt(4*Rg*Pd)   #input excitation to the bridge circuit\n",
+    "x=(Sg*sig_f*(1+v))/(2*E);\n",
+    "dEo_max=x*Ei*10**3;\n",
+    "print (\"x=(Sg*sig_f*(1+v))/(2*E)\")\n",
+    "print ('a voltmeter with a maximum range of  mV is suitable for measurement',dEo_max)\n",
+    "print (\"Round it off to get the suitable range voltmeter\")"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_2 pgno:295"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('(dE/V)_max= d\\n ', 4285714.285714285)\n",
+      "Sensitivity of this load cell is  nV/N/per unit excitation 42.8571428571\n"
+     ]
+    }
+   ],
+   "source": [
+    "#CHAPTER 5_ Force,Torque and Shaft Power Measurement\n",
+    "#Caption : Load cell\n",
+    "# Example 2# Page 295\n",
+    "\n",
+    "b=.2   #('enter the width of load cell=:')\n",
+    "h=.05   #('enter the thickness of load cell=:')\n",
+    "Sg=2.;\n",
+    "Rg=120.;\n",
+    "sig_f=150*10**6   #('enter the fatigue strength=:')\n",
+    "E=70.;      #(in GPa) for aluminium\n",
+    "v=0.33;           #poissons ratio\n",
+    "# Let dE/V_max be represented by W\n",
+    "W=Sg*sig_f/E;\n",
+    "print('(dE/V)_max= d\\n ',W)\n",
+    "P_max=100000.   #('enter the value of maximum load=:')\n",
+    "l=sig_f*b*h**2/(6*P_max);\n",
+    "\n",
+    "S=(6*Sg*l)/(E*b*h**2);\n",
+    "print'Sensitivity of this load cell is  nV/N/per unit excitation',S\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_3 pgno:296"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Sensitivity of this load cell is  micro V/N\n",
+      "0.13\n",
+      "The maximum density that can be achieved without endangering the strain gage sensors is micro V/N\n",
+      "0.284815729903\n",
+      "The voltage ratio is mV/V 3.9\n"
+     ]
+    }
+   ],
+   "source": [
+    "#CHAPTER 5_ Force,Torque and Shaft Power Measurement\n",
+    "#Caption : Load cell\n",
+    "# Example 3# Page 296\n",
+    "from math import sqrt\n",
+    "Sg=2;\n",
+    "v=0.3;    #poissons ratio\n",
+    "Ei=10.   #('enter the excitation voltage=:')\n",
+    "A=5*10**-4  #('enter the area of load cell=:')\n",
+    "E=200.;   #(in Gpa) Youngs modulus\n",
+    "# Let sensitivity Eo/P be represented by Se\n",
+    "Se=Sg*(1+v)*Ei/(2*A*E)*.001;\n",
+    "print'Sensitivity of this load cell is  micro V/N\\n',Se\n",
+    "Rg=120.   #given\n",
+    "Pd=1. #('enter the power dissipated in each gage=:')\n",
+    "Ei_max=sqrt(4*Rg*Pd)\n",
+    "Se_max=Sg*(1+v)*Ei_max/(2*A*E)*.001\n",
+    "print'The maximum density that can be achieved without endangering the strain gage sensors is micro V/N\\n',Se_max\n",
+    "# Let (Eo/Ei)_max be represented by Em\n",
+    "sig_f=600*10**6  #('enter the fatigue strength=:')\n",
+    "Em=Sg*sig_f*(1+v)/(2*E)*10**-6\n",
+    "print'The voltage ratio is mV/V',Em\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_4 pgno:302"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Relative displacement is d', 1.9999999999999997e-08)\n",
+      "wnc**2 is approx. 10**9. So,\n",
+      "Z is approx. 20nm(rms)\n",
+      "Actual force transmitted to the plate is d N 18.0260791198\n"
+     ]
+    }
+   ],
+   "source": [
+    "#CHAPTER 5_ Force,Torque and Shaft Power Measurement\n",
+    "#Caption : Piezoelectric Transducers\n",
+    "# Example 4# Page 302\n",
+    "from math import sqrt,pi\n",
+    "mc=0.04   #('enter the connector mass=:')\n",
+    "m=0.01   #('enter the seismic mass=:')\n",
+    "k=10**9   #('enter the stiffness of the sensing element=:')\n",
+    "Sf=.005   #('enter the sensitivity of the transducer=:')\n",
+    "Xi=100*10**-6 #  ('enter the displacement amplitude of the shaker vibration=:')\n",
+    "Eo=.1   #('enter the reading of voltage recorder connected to the transducer=:')\n",
+    "wnc=sqrt(k/(m+mc));\n",
+    "R=20;    #20N (rms)\n",
+    "Z=(1/(m+mc))*(1/wnc**2)*R;\n",
+    "print('Relative displacement is d',Z)\n",
+    "print(\"wnc**2 is approx. 10**9. So,\")\n",
+    "print(\"Z is approx. 20nm(rms)\")\n",
+    "f=100.;   # given\n",
+    "\n",
+    "F=R-((2*pi*f)**2*(m+mc)*Xi);\n",
+    "print'Actual force transmitted to the plate is d N',F\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_5 pgno:308"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The load torque is d N-m 1636.24617374\n"
+     ]
+    }
+   ],
+   "source": [
+    "#CHAPTER 5_ Force,Torque and Shaft Power Measurement\n",
+    "#Caption : Torque measurement on rotating shaft\n",
+    "# Example 5# Page 308\n",
+    "Sg=2.;\n",
+    "Rg=120.;\n",
+    "G=80*10**9   #('enter the sheer modulus of elasticity=:')\n",
+    "D=0.05   #('enter the shaft diameter=:')\n",
+    "dR=0.1   # given\n",
+    "# we have to find the load torque\n",
+    "from math import pi\n",
+    "\n",
+    "y=2*dR/(Rg*Sg);\n",
+    "tou_xy=y*G;\n",
+    "j=pi*D**4;\n",
+    "T=tou_xy*2*j/(D*32);\n",
+    "print'The load torque is d N-m',T"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/yashwanth kumarmada/Chapter_5_Laser.ipynb b/sample_notebooks/yashwanth kumarmada/Chapter_5_Laser.ipynb
new file mode 100755
index 00000000..e5562a27
--- /dev/null
+++ b/sample_notebooks/yashwanth kumarmada/Chapter_5_Laser.ipynb
@@ -0,0 +1,272 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 Laser"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_1 pgno:242"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 1 # \n",
+      "\n",
+      "\n",
+      " Number of oscillation corresponding to coherent length is  \n",
+      " Coherent time is sec. 50000.0 9.81666666667e-11\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Given that\n",
+    "l = 2.945e-2 # coherent length of sodium light\n",
+    "lamda = 5890 # wavelength of light used in angstrom\n",
+    "c = 3e8 # speed of light\n",
+    "# Sample Problem 1 on page no. 242\n",
+    "print(\"\\n # PROBLEM 1 # \\n\")\n",
+    "n = l/(lamda*1e-10) # number of oscillation corresponding to coherent length\n",
+    "t = l/c # coherent time\n",
+    "print\"\\n Number of oscillation corresponding to coherent length is  \\n Coherent time is sec.\",n,t\n",
+    "\n",
+    "\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_2 pgno:242"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 2 # \n",
+      "\n",
+      "\n",
+      " Angular  spread  is  rad. \n",
+      " Areal spread is m^2. 0.00016 4096000000.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "# Given that\n",
+    "l = 4e5 # Distance of moon in km\n",
+    "lamda = 8e-7 # wavelength of light used\n",
+    "a = 5e-3 # Aperture of laser\n",
+    "c = 3e8 # speed of light\n",
+    "# Sample Problem 2 on page no. 242\n",
+    "print\"\\n # PROBLEM 2 # \\n\"\n",
+    "theta = lamda/a  # Angular of spread \n",
+    "Areal_spread = (l*1000*theta)**2 # Areal spread\n",
+    "print\"\\n Angular  spread  is  rad. \\n Areal spread is m^2.\",theta,Areal_spread\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_3 pgno:242"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 3 # \n",
+      "\n",
+      "\n",
+      " Number of oscillation corresponding to coherent length is  \n",
+      " Coherent time is sec. 50000.0 9.81666666667e-11\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "# Given that\n",
+    "l = 2.945e-2 # coherent length of sodium light\n",
+    "lamda = 5890 # wavelength of light used\n",
+    "c = 3e8 # speed of light\n",
+    "# Sample Problem 3 on page no. 242\n",
+    "print\"\\n # PROBLEM 3 # \\n\"\n",
+    "n = l/(lamda *1e-10) # number of oscillation corresponding to coherent length\n",
+    "t = l/c # coherent time\n",
+    "print\"\\n Number of oscillation corresponding to coherent length is  \\n Coherent time is sec.\",n,t\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_4 pgno:243"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 4 # \n",
+      "\n",
+      "\n",
+      " Energy difference is eV. 0.365641494412\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "# Given that\n",
+    "k = 12400 # constant\n",
+    "lamda = 3.3913 # wavelength IR radiation\n",
+    "\n",
+    "# Sample Problem 4 on page no. 243\n",
+    "print\"\\n # PROBLEM 4 # \\n\"\n",
+    "E = k/(lamda*1e4) # Energy difference\n",
+    "print\"\\n Energy difference is eV.\",E\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_5 pgno:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 5 # \n",
+      "\n",
+      "\n",
+      " Energy of one photon  is eV. \n",
+      " Total energy is J 1 4.8\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "k = 12400 # constant\n",
+    "lamda = 6943 # wavelength of radiation in angstrom\n",
+    "n = 3e19 # Total number of ions\n",
+    "# Sample Problem 5 on page no. 243\n",
+    "print\"\\n # PROBLEM 5 # \\n\"\n",
+    "E = k/(lamda) # Energy difference\n",
+    "E_total = E*n*1.6e-19 # Total Energy emitted \n",
+    "print\"\\n Energy of one photon  is eV. \\n Total energy is J\",E,E_total\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5_6 pgno:244"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "\n",
+      " # PROBLEM 6 # \n",
+      "\n",
+      "\n",
+      " Required length of cavity is cm. 10.010896\n"
+     ]
+    }
+   ],
+   "source": [
+    "\n",
+    "# Given that\n",
+    "h_w = 2e-3 # half width of gain profile of laser in nm\n",
+    "mu = 1 # refractive index\n",
+    "lamda = 6328 # wavelength of light used in angstrom\n",
+    "# Sample Problem 6 on page no. 244\n",
+    "print\"\\n # PROBLEM 6 # \\n\"\n",
+    "L = (lamda*1e-10)**2/(2*mu*h_w*1e-9) # Length of cavity \n",
+    "print\"\\n Required length of cavity is cm.\",L*100\n",
+    "\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.9"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
author	FOSSEE SysAds	2015-12-08 15:04:13 +0600
committer	FOSSEE SysAds	2015-12-08 15:04:13 +0600
commit	af2e802cfaf390cbc43b476ec9f0c73c7e651d9f (patch)
tree	a024377882455868ac98f790805df5ee7cdf6f68 /sample_notebooks
parent	2ea1980b3f812234480f978b7a0aff0ffca0a0eb (diff)
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