diff options
Diffstat (limited to '1316')
55 files changed, 650 insertions, 0 deletions
diff --git a/1316/CH1/EX1.1/example1_1.sce b/1316/CH1/EX1.1/example1_1.sce new file mode 100644 index 000000000..3605ee8f7 --- /dev/null +++ b/1316/CH1/EX1.1/example1_1.sce @@ -0,0 +1,14 @@ +//Chapter 1
+//Example 1.1
+//Page 3
+
+clear;
+clc;
+
+K=0.222;
+Qout=2;
+
+
+//Calculation of value of h at which level stabilizes
+
+printf("The level will stabilize from self regulation when Qout=Qin.Thus the value of h is %.0f ft",((Qout/K)^2)/27);
diff --git a/1316/CH1/EX1.1/resultexample1_1.txt b/1316/CH1/EX1.1/resultexample1_1.txt new file mode 100644 index 000000000..4f5b054e6 --- /dev/null +++ b/1316/CH1/EX1.1/resultexample1_1.txt @@ -0,0 +1 @@ +The level will stabilize from self regulation when Qout=Qin.Thus the value of h is 3 ft
\ No newline at end of file diff --git a/1316/CH1/EX1.10/example1_10.sce b/1316/CH1/EX1.10/example1_10.sce new file mode 100644 index 000000000..c0d72bf92 --- /dev/null +++ b/1316/CH1/EX1.10/example1_10.sce @@ -0,0 +1,15 @@ +//Chapter 1
+//Example 1.10
+//Page 28
+
+clear;
+clc;
+
+Read=27.5;
+T_min=4.95;
+T_max=5.05;
+
+
+//Calculation of possible temperature values
+printf("The range of Transfer function is 4.95 to 5.05 mV per degree celcius \n The possible temperature values that could be inferred from reading of 27.5 degree celcius are %.2f and %.2f",Read/T_min,Read/T_max);
+
diff --git a/1316/CH1/EX1.10/resultexample1_10.txt b/1316/CH1/EX1.10/resultexample1_10.txt new file mode 100644 index 000000000..b874b4b40 --- /dev/null +++ b/1316/CH1/EX1.10/resultexample1_10.txt @@ -0,0 +1,2 @@ + The range of Transfer function is 4.95 to 5.05 mV per degree celcius
+ The possible temperature values that could be inferred from reading of 27.5 degree celcius are 5.56 and 5.45
\ No newline at end of file diff --git a/1316/CH1/EX1.11/example1_11.sce b/1316/CH1/EX1.11/example1_11.sce new file mode 100644 index 000000000..9a7bfb34f --- /dev/null +++ b/1316/CH1/EX1.11/example1_11.sce @@ -0,0 +1,15 @@ +//Chapter 1
+//Example 1.11
+//Page 29
+
+clear;
+clc;
+
+Delta_K=1.5;
+Delta_G=0.05;
+K=100;
+G=20;
+
+//Calculation of system accuracy of flow process
+printf("Here we have direct application of Delta V/V = %.2f \n",(Delta_K/K)+(Delta_G/G));
+printf("If we use more statistically correct rms approach,the system accuracy would be = %.4f",sqrt((Delta_K/K)^2+(Delta_G/G)^2));
diff --git a/1316/CH1/EX1.11/resultexample1_11.txt b/1316/CH1/EX1.11/resultexample1_11.txt new file mode 100644 index 000000000..22fb6898c --- /dev/null +++ b/1316/CH1/EX1.11/resultexample1_11.txt @@ -0,0 +1,2 @@ + Here we have direct application of Delta V/V = 0.02
+If we use more statistically correct rms approach,the system accuracy would be = 0.0152
\ No newline at end of file diff --git a/1316/CH1/EX1.12/example1_12.sce b/1316/CH1/EX1.12/example1_12.sce new file mode 100644 index 000000000..1c22b80cc --- /dev/null +++ b/1316/CH1/EX1.12/example1_12.sce @@ -0,0 +1,13 @@ +//Chapter 1
+//Example 1.12
+//Page 31
+
+clear;
+clc;
+
+F=150;
+Res_FS=0.1;
+
+
+//Calculation of smallest change in force that can be measured
+printf("Hence the smallest measurable change in force which cam be measured is = %.2f N",(Res_FS/100)*F);
diff --git a/1316/CH1/EX1.12/resultexample1_12.txt b/1316/CH1/EX1.12/resultexample1_12.txt new file mode 100644 index 000000000..087472fd1 --- /dev/null +++ b/1316/CH1/EX1.12/resultexample1_12.txt @@ -0,0 +1 @@ + Hence the smallest measurable change in force which cam be measured is = 0.15 N
\ No newline at end of file diff --git a/1316/CH1/EX1.13/example1_13.sce b/1316/CH1/EX1.13/example1_13.sce new file mode 100644 index 000000000..46d36c187 --- /dev/null +++ b/1316/CH1/EX1.13/example1_13.sce @@ -0,0 +1,12 @@ +//Chapter 1
+//Example 1.13
+//Page 31
+
+clear;
+clc;
+
+TF=5;
+Temp_Res=0.2;
+
+//Calculation of required voltage resolution
+printf("The temperature change of 0.2 degree celcius will result in a voltage change of = %.1f mV",TF*Temp_Res);
diff --git a/1316/CH1/EX1.13/resultexample1_13.txt b/1316/CH1/EX1.13/resultexample1_13.txt new file mode 100644 index 000000000..56136bd2a --- /dev/null +++ b/1316/CH1/EX1.13/resultexample1_13.txt @@ -0,0 +1 @@ + The temperature change of 0.2 degree celcius will result in a voltage change of = 1.0 mV
\ No newline at end of file diff --git a/1316/CH1/EX1.14/example1_14.sce b/1316/CH1/EX1.14/example1_14.sce new file mode 100644 index 000000000..4ee6adef0 --- /dev/null +++ b/1316/CH1/EX1.14/example1_14.sce @@ -0,0 +1,18 @@ +//Chapter 1
+//Example 1.14
+//Page 32
+
+clear;
+clc;
+
+R_min=100;
+R_max=180;
+T_min=20;
+T_max=120;
+m=(R_max-R_min)/(T_max-T_min);
+R0=R_min-(T_min*m);
+
+//Finding a linear euation relating resistance and temperature
+
+printf("The linear equation would be of the form R=mT+R0 \n");
+printf("Hence the equation relating temperature and resistance is R = %.1f T + %.f",m,R0)
diff --git a/1316/CH1/EX1.14/resultexample1_14.txt b/1316/CH1/EX1.14/resultexample1_14.txt new file mode 100644 index 000000000..a4cf3ef5f --- /dev/null +++ b/1316/CH1/EX1.14/resultexample1_14.txt @@ -0,0 +1,2 @@ + The linear equation would be of the form R=mT+R0
+Hence the equation relating temperature and resistance is R = 0.8 T + 84
\ No newline at end of file diff --git a/1316/CH1/EX1.15/example1_15.sce b/1316/CH1/EX1.15/example1_15.sce new file mode 100644 index 000000000..52791bc04 --- /dev/null +++ b/1316/CH1/EX1.15/example1_15.sce @@ -0,0 +1,20 @@ +//Chapter 1
+//Example 1.15
+//Page 38
+
+clear;
+clc;
+
+STF = 33;
+t1 = 1.5;
+t2 = 0.75;
+T1 = 20;
+T2 = 41;
+bi = STF*T1;
+bf = STF*T2;
+e = 2.718
+//Finding the error in temperature represents
+
+printf("The value of b(0.75) is %.1f mV \n",bi+(bf-bi)*[1-e^(-t2/t1)]);
+printf("This corresponds to an indicated temperature of %.1f degree celcius\n",(bi+(bf-bi)*[1-e^(-t2/t1)])/STF);
+printf("So the error is %.1f degree celcius because the actual temperature is 41 degree celcius \n",T2-((bi+(bf-bi)*[1-e^(-t2/t1)])/STF));
diff --git a/1316/CH1/EX1.15/resultexample1_15.txt b/1316/CH1/EX1.15/resultexample1_15.txt new file mode 100644 index 000000000..30922a2df --- /dev/null +++ b/1316/CH1/EX1.15/resultexample1_15.txt @@ -0,0 +1,4 @@ + The value of b(0.75) is 932.7 mV
+This corresponds to an indicated temperature of 28.3 degree celcius
+So the error is 12.7 degree celcius because the actual temperature is 41 degree celcius
+
\ No newline at end of file diff --git a/1316/CH1/EX1.16/example1_16.sce b/1316/CH1/EX1.16/example1_16.sce new file mode 100644 index 000000000..c2a991dbc --- /dev/null +++ b/1316/CH1/EX1.16/example1_16.sce @@ -0,0 +1,15 @@ +//Chapter 1
+//Example 1.16
+//Page 40
+
+clear;
+clc;
+
+R = 12.5;
+I = 2.21;
+Scale = 10;
+Acc = 0.2;
+
+//Finding the voltage across resistor
+printf("The voltage across resistor is %.1f V",I*R);
+
diff --git a/1316/CH1/EX1.16/resultexample1_16.txt b/1316/CH1/EX1.16/resultexample1_16.txt new file mode 100644 index 000000000..59b00c45c --- /dev/null +++ b/1316/CH1/EX1.16/resultexample1_16.txt @@ -0,0 +1 @@ +The voltage across resistor is 27.6 V
\ No newline at end of file diff --git a/1316/CH1/EX1.17/example1_17.sce b/1316/CH1/EX1.17/example1_17.sce new file mode 100644 index 000000000..a8b4c7f96 --- /dev/null +++ b/1316/CH1/EX1.17/example1_17.sce @@ -0,0 +1,13 @@ +//Chapter 1
+//Example 1.17
+//Page 40
+
+clear;
+clc;
+
+TF=22.4;
+V=412;
+
+//Finding the value of temperature
+printf("The value of temperature is equal to %.6f degree celcius \n",V/TF);
+printf("Our result can be significant to 3 places = %.1f degree celcius",V/TF);
diff --git a/1316/CH1/EX1.17/resultexample1_17.txt b/1316/CH1/EX1.17/resultexample1_17.txt new file mode 100644 index 000000000..2635145b0 --- /dev/null +++ b/1316/CH1/EX1.17/resultexample1_17.txt @@ -0,0 +1,2 @@ +The value of temperature is equal to 18.392857 degree celcius
+Our result can be significant to 3 places = 18.4 degree celcius
\ No newline at end of file diff --git a/1316/CH1/EX1.18/example1_18.sce b/1316/CH1/EX1.18/example1_18.sce new file mode 100644 index 000000000..da47b38df --- /dev/null +++ b/1316/CH1/EX1.18/example1_18.sce @@ -0,0 +1,20 @@ +//Chapter 1
+//Example 1.18
+//Page 42
+
+clear;
+clc;
+
+t1=21.2;
+t2=25;
+t3=18.5;
+t4=22.1;
+t5=19.7;
+t6=27.1;
+t7=19;
+t8=20;
+AVE=(t1+t2+t3+t4+t5+t6+t7+t8)/8;
+
+//Finding the arithmetic mean of the temperature
+printf("The arithmetic mean of the temperatue is %f degree celcius \n",AVE);
+printf("The standard deviation is %.2f degree celcius",sqrt(((t1-AVE)^2)+((t2-AVE)^2)+((t3-AVE)^2)+((t4-AVE)^2)+((t5-AVE)^2)+((t6-AVE)^2)+((t7-AVE)^2)+((t8-AVE)^2))/sqrt(8-1));
diff --git a/1316/CH1/EX1.18/resultexample1_18.txt b/1316/CH1/EX1.18/resultexample1_18.txt new file mode 100644 index 000000000..a25086f90 --- /dev/null +++ b/1316/CH1/EX1.18/resultexample1_18.txt @@ -0,0 +1,2 @@ +The arithmetic mean of the temperatue is 21.575000 degree celcius
+The standard deviation is 3.04 degree celcius
\ No newline at end of file diff --git a/1316/CH1/EX1.19/example1_19.sce b/1316/CH1/EX1.19/example1_19.sce new file mode 100644 index 000000000..30cfc0a8f --- /dev/null +++ b/1316/CH1/EX1.19/example1_19.sce @@ -0,0 +1,46 @@ +//Chapter 1
+//Example 1.18
+//Page 42
+
+clear;
+clc;
+
+b1=201;
+b2=205;
+b3=197;
+b4=185;
+b5=202;
+b6=207;
+b7=215;
+b8=220;
+b9=179;
+b10=201;
+b11=197;
+b12=221;
+b13=202;
+b14=200;
+b15=195;
+a1=197;
+a2=202
+a3=193
+a4=210
+a5=207
+a6=195;
+a7=199;
+a8=202;
+a9=193;
+a10=195;
+a11=201;
+a12=201;
+a13=200;
+a14=189;
+a15=197;
+AVE1=(b1+b2+b3+b4+b5+b6+b7+b8+b9+b10+b11+b12+b13+b14+b15)/15;
+AVE2=(a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14+a15)/15;
+
+//Finding the arithmetic mean of the temperature
+printf("The arithmetic mean samples before is %f g \n",AVE1);
+printf("The stbndbrd devibtion of sbmples before is %f g \n",sqrt(((b1-AVE1)^2)+((b2-AVE1)^2)+((b3-AVE1)^2)+((b4-AVE1)^2)+((b5-AVE1)^2)+((b6-AVE1)^2)+((b7-AVE1)^2)+((b8-AVE1)^2)+((b9-AVE1)^2)+((b10-AVE1)^2)+((b11-AVE1)^2)+((b12-AVE1)^2)+((b13-AVE1)^2)+((b14-AVE1)^2)+((b15-AVE1)^2))/sqrt(15-1));
+printf("The arithmetic mean samples before is %f g \n",AVE2);
+printf("The standard deviation of samples before is %f g \n",sqrt(((a1-AVE2)^2)+((a2-AVE2)^2)+((a3-AVE2)^2)+((a4-AVE2)^2)+((a5-AVE2)^2)+((a6-AVE2)^2)+((a7-AVE2)^2)+((a8-AVE2)^2)+((a9-AVE2)^2)+((a10-AVE2)^2)+((a11-AVE2)^2)+((a12-AVE2)^2)+((a13-AVE2)^2)+((a14-AVE2)^2)+((a15-AVE2)^2))/sqrt(15-1));
+
diff --git a/1316/CH1/EX1.19/resultexample1_19.txt b/1316/CH1/EX1.19/resultexample1_19.txt new file mode 100644 index 000000000..1c0c4f333 --- /dev/null +++ b/1316/CH1/EX1.19/resultexample1_19.txt @@ -0,0 +1,5 @@ + The arithmetic mean samples before is 201.800000 g
+The stbndbrd devibtion of sbmples before is 11.371644 g
+The arithmetic mean samples before is 198.733333 g
+The standard deviation of samples before is 5.496319 g
+
\ No newline at end of file diff --git a/1316/CH1/EX1.2/example1_2.sce b/1316/CH1/EX1.2/example1_2.sce new file mode 100644 index 000000000..17860f9e9 --- /dev/null +++ b/1316/CH1/EX1.2/example1_2.sce @@ -0,0 +1,12 @@ +//Chapter 1
+//Example 1.2
+//Page 23
+
+clear;
+clc;
+
+p=2.1*10^3;
+
+//Calculation of pressure in pascals
+
+printf("As we know that 10^2 cm= 1m and 10^5 dyne=1 Newton \n Thus the value of pressure is %.f Pascals",(p*10000)/(100000));
diff --git a/1316/CH1/EX1.2/resultexample1_2.txt b/1316/CH1/EX1.2/resultexample1_2.txt new file mode 100644 index 000000000..8714ad5bd --- /dev/null +++ b/1316/CH1/EX1.2/resultexample1_2.txt @@ -0,0 +1,2 @@ + As we know that 10^2 cm= 1m and 10^5 dyne=1 Newton
+ Thus the value of pressure is 210 Pascals
\ No newline at end of file diff --git a/1316/CH1/EX1.3/example1_3.sce b/1316/CH1/EX1.3/example1_3.sce new file mode 100644 index 000000000..5faf08465 --- /dev/null +++ b/1316/CH1/EX1.3/example1_3.sce @@ -0,0 +1,12 @@ +//Chapter 1
+//Example 1.3
+//Page 23
+
+clear;
+clc;
+
+K=5.7;
+
+//Calculation of number of feets in 5.7m
+
+printf("As 1m=39.37 in. Thus number of feets in 5.7m is %.1f ft",K*39.37/12);
diff --git a/1316/CH1/EX1.3/resultexample1_3.txt b/1316/CH1/EX1.3/resultexample1_3.txt new file mode 100644 index 000000000..3a51abd96 --- /dev/null +++ b/1316/CH1/EX1.3/resultexample1_3.txt @@ -0,0 +1 @@ + As 1m=39.37 in. Thus number of feets in 5.7m is 18.7 ft
\ No newline at end of file diff --git a/1316/CH1/EX1.4/example1_4.sce b/1316/CH1/EX1.4/example1_4.sce new file mode 100644 index 000000000..2fc625d72 --- /dev/null +++ b/1316/CH1/EX1.4/example1_4.sce @@ -0,0 +1,11 @@ +//Chapter 1
+//Example 1.4
+//Page 23
+
+clear;
+clc;
+
+K=6;
+
+//Calculation of feets in meter
+printf("As 1m=39.37 in and 1m=12ft Thus number of meter in 6ft is %.2f m",(K*12)/39.37);
diff --git a/1316/CH1/EX1.4/resultexample1_4.txt b/1316/CH1/EX1.4/resultexample1_4.txt new file mode 100644 index 000000000..0d860bb20 --- /dev/null +++ b/1316/CH1/EX1.4/resultexample1_4.txt @@ -0,0 +1 @@ + As 1m=39.37 in and 1m=12ft Thus number of meter in 6ft is 1.83 m
\ No newline at end of file diff --git a/1316/CH1/EX1.5/example1_5.sce b/1316/CH1/EX1.5/example1_5.sce new file mode 100644 index 000000000..801829d5e --- /dev/null +++ b/1316/CH1/EX1.5/example1_5.sce @@ -0,0 +1,11 @@ +//Chapter 1
+//Example 1.5
+//Page 23
+
+clear;
+clc;
+
+K=2;
+
+//Calculation of mass in kg of 2 lb object
+printf("As 1 lb=0.454 kg Therefore we have \n m=%.3f kg",K*0.454);
diff --git a/1316/CH1/EX1.5/resultexample1_5.txt b/1316/CH1/EX1.5/resultexample1_5.txt new file mode 100644 index 000000000..a9d3910b8 --- /dev/null +++ b/1316/CH1/EX1.5/resultexample1_5.txt @@ -0,0 +1,2 @@ + As 1 lb=0.454 kg Therefore we have
+ m=0.908 kg
\ No newline at end of file diff --git a/1316/CH1/EX1.7/example1_7.sce b/1316/CH1/EX1.7/example1_7.sce new file mode 100644 index 000000000..18ce19f4f --- /dev/null +++ b/1316/CH1/EX1.7/example1_7.sce @@ -0,0 +1,27 @@ +//Chapter 1
+//Example 1.7
+//Page 25
+
+clear;
+clc;
+
+T_min=20;
+T_max=120;
+i_min=4;
+i_max=20;
+T1=66;
+I1=6.5;
+m=(i_max-i_min)/(T_max-T_min);
+I0=i_min-(T_min*m);
+
+
+//Calculation of current represented by 66 degree celcius
+printf("Here the value of current for 66 degree celcius = %.2f mA \n",(m*T1)+I0);
+printf("Here the value of temperature for 6.5mA current = %.2f degree celcius",(I1-I0)/m);
+
+
+
+
+
+
+
diff --git a/1316/CH1/EX1.7/resultexample1_7.txt b/1316/CH1/EX1.7/resultexample1_7.txt new file mode 100644 index 000000000..155bcb17a --- /dev/null +++ b/1316/CH1/EX1.7/resultexample1_7.txt @@ -0,0 +1,2 @@ + Here the value of current for 66 degree celcius = 11.36 mA
+Here the value of temperature for 6.5mA current = 35.63 degree celcius
\ No newline at end of file diff --git a/1316/CH1/EX1.8/example1_8.sce b/1316/CH1/EX1.8/example1_8.sce new file mode 100644 index 000000000..781ec4c0a --- /dev/null +++ b/1316/CH1/EX1.8/example1_8.sce @@ -0,0 +1,27 @@ +//Chapter 1
+//Example 1.8
+//Page 27
+
+clear;
+clc;
+
+T_min=20;
+T_max=250;
+T=55;
+A1=0.5;
+A2=0.75;
+A3=0.8;
+
+
+//Calculation of Errors for each case
+printf("error for accuracy of 0.5 percent of full scale value is = %.2f degree celcius \n",(A1/100)*T_max);
+printf("Thus the actual temperature is in range of = %.2f degree celcius to =%.2f degree celcius \n",(T-(A1/100)*T_max),((A1/100)*T_max)+T);
+printf("error for accuracy of 0.75 percent of span is = %.3f degree celcius \n",(A2/100)*(T_max-T_min));
+printf("Thus the actual temperature is in range of = %.3f degree celcius to =%.3f degree celcius \n",(T-((A2/100)*(T_max-T_min))),(A2/100)*(T_max-T_min)+T);
+printf("error for accuracy of 0.8 percent of reading is = %.2f degree celcius \n",(A3/100)*(T));
+printf("Thus the actual temperature is in range of = %.2f degree celcius to = %.2f degree celcius \n",T-((A3/100)*(T)),((A3/100)*(T))+T);
+
+
+
+
+
diff --git a/1316/CH1/EX1.8/resultexample1_8.txt b/1316/CH1/EX1.8/resultexample1_8.txt new file mode 100644 index 000000000..00e5c3232 --- /dev/null +++ b/1316/CH1/EX1.8/resultexample1_8.txt @@ -0,0 +1,7 @@ + error for accuracy of 0.5 percent of full scale value is = 1.25 degree celcius
+Thus the actual temperature is in range of = 53.75 degree celcius to =56.25 degree celcius
+error for accuracy of 0.75 percent of span is = 1.725 degree celcius
+Thus the actual temperature is in range of = 53.275 degree celcius to =56.725 degree celcius
+error for accuracy of 0.8 percent of reading is = 0.44 degree celcius
+Thus the actual temperature is in range of = 54.56 degree celcius to = 55.44 degree celcius
+
\ No newline at end of file diff --git a/1316/CH1/EX1.9/example1_9.sce b/1316/CH1/EX1.9/example1_9.sce new file mode 100644 index 000000000..ebcc9ba14 --- /dev/null +++ b/1316/CH1/EX1.9/example1_9.sce @@ -0,0 +1,14 @@ +//Chapter 1
+//Example 1.9
+//Page 28
+
+clear;
+clc;
+
+TF=5;
+A=1;
+
+
+//Calculation of possible range of transfer function
+printf("The transfer function will be = %.2f mV per degree celcius \n",((A/100)*(TF)));
+printf("Thus the range is %.2f mV per degree celcius to %.2f mV per degree celcius \n",(TF)-((A/100)*(TF)),(TF)+((A/100)*(TF)));
diff --git a/1316/CH1/EX1.9/resultexample1_9.txt b/1316/CH1/EX1.9/resultexample1_9.txt new file mode 100644 index 000000000..14ed0e92b --- /dev/null +++ b/1316/CH1/EX1.9/resultexample1_9.txt @@ -0,0 +1,2 @@ + The transfer function will be = 0.05 mV per degree celcius
+Thus the range is 4.95 mV per degree celcius to 5.05 mV per degree celcius
diff --git a/1316/CH1/EX2.6/example1_6.sce b/1316/CH1/EX2.6/example1_6.sce new file mode 100644 index 000000000..c18f4a835 --- /dev/null +++ b/1316/CH1/EX2.6/example1_6.sce @@ -0,0 +1,13 @@ +//Chapter 1
+//Example 1.6
+//Page 24
+
+clear;
+clc;
+
+K1=0.0000215;
+K2=3781000000;
+
+//expressing numbers in decimal prefix
+printf("0.0000215 micro second = %.1f micro second \n",K1*10^6);
+printf("3781000000 W=%.3f GW",K2*10^-9);
diff --git a/1316/CH1/EX2.6/resultexample1_6.txt b/1316/CH1/EX2.6/resultexample1_6.txt new file mode 100644 index 000000000..c8a4c1ebc --- /dev/null +++ b/1316/CH1/EX2.6/resultexample1_6.txt @@ -0,0 +1,2 @@ + 0.0000215 micro second = 21.5 micro second
+3781000000 W=3.781 GW
\ No newline at end of file diff --git a/1316/CH2/EX2.1/example2_1.sce b/1316/CH2/EX2.1/example2_1.sce new file mode 100644 index 000000000..a7f5357e7 --- /dev/null +++ b/1316/CH2/EX2.1/example2_1.sce @@ -0,0 +1,29 @@ +//Chapter 2
+//Example 2.1
+//Page 55
+
+clear;
+clc;
+
+Ri = 10;
+TF = 0.020;
+Ro = 5;
+T = 50;
+
+printf("The unloaded output of sensor is simply ")
+//Calculation of Vt
+x = TF*T;
+printf("Vt = %.1f \n",x)
+printf("Since the amplifier has a gain of 10,the output of the amplifier appears to be ")
+//Calculation of Vout
+y = Ri*x;
+printf("Vout = %.0f V \n",y)
+printf("But this is wrong because of loading !\n ")
+//Correct Analysis
+printf("Here we see that there will be a voltage dropped across the \n output resistance of the sensor.The actual amplifier input voltage will be given by ")
+//Calculation of Vin
+a = x*(1-((Ro)/(Ro+Ri)));
+printf("Vin = %.2f V.\n",a)
+printf("Thus the output of amplifier is actually Vout = %.1f V \n",Ri*a)
+
+
diff --git a/1316/CH2/EX2.1/resultexample2_1.txt b/1316/CH2/EX2.1/resultexample2_1.txt new file mode 100644 index 000000000..a2a032eb1 --- /dev/null +++ b/1316/CH2/EX2.1/resultexample2_1.txt @@ -0,0 +1,7 @@ + The unloaded output of sensor is simply Vt = 1.0
+Since the amplifier has a gain of 10,the output of the amplifier appears to be Vout = 10 V
+But this is wrong because of loading !
+ Here we see that there will be a voltage dropped across the
+ output resistance of the sensor.The actual amplifier input voltage will be given by Vin = 0.67 V.
+Thus the output of amplifier is actually Vout = 6.7 V
+
\ No newline at end of file diff --git a/1316/CH2/EX2.10/example2_10.sce b/1316/CH2/EX2.10/example2_10.sce new file mode 100644 index 000000000..7417d62b6 --- /dev/null +++ b/1316/CH2/EX2.10/example2_10.sce @@ -0,0 +1,25 @@ +//Chapter 2
+//Example 2.10
+//Page 68
+
+clear;
+clc;
+
+R1 = 1000;
+R2 = 2000;
+R3 = 1000;
+C1 = 1;
+
+printf("Because the bridge is at null , we have \n ")
+printf("Z2*Z3 = Z1*Zx \n")
+printf("R2(R3-j/wC)=R1(Rx-j/wCx) \n")
+printf("The real and imaginary parts must be indpendently equal,so that \n")
+printf("Rx-(R2*R3/R1)=0 \n")
+//Calculation of value of Rx
+x=(R2*R3/R1)/1000
+printf("Rx = %.0f kilo ohm \n",x)
+//Calculation of value of Cx
+y=(C1*(R1/R2))
+printf("Cx=C(R1/R2) \n Cx = %.1f uF",y)
+
+
diff --git a/1316/CH2/EX2.2/example2_2.sce b/1316/CH2/EX2.2/example2_2.sce new file mode 100644 index 000000000..239adea00 --- /dev/null +++ b/1316/CH2/EX2.2/example2_2.sce @@ -0,0 +1,36 @@ +//Chapter 2
+//Example 2.2
+//Page 58
+
+clear;
+clc;
+
+R1 = 10;
+Vs = 5;
+R_low = 4;
+R_high = 12;
+
+
+printf("a. The solution is given by VsR2/R1+R2.For R2=4k.ohm,we have\n")
+//Calculation of Vd
+x = (Vs*R_low)/(R1+R_low);
+printf("VD = %.2f V\n",x)
+printf("For R2=12k.ohm,we have\n")
+//Calculation of Vd
+y = (Vs*R_high)/(R1+R_high);
+printf("VD = %.2f V\n",y)
+printf("b. Thus the voltage varies from %.2f to %.2f V",x,y)
+printf("c. The range of output impedance is found from the parallel \n combination of R1 and R2 for the minimum and maximum of R2.\n Simple parallel resistance computation shows that this will be from ")
+//Calculation of parallel resistances
+a = (R1*R_low)/(R1+R_low);
+b = (R1*R_high)/(R1+R_high);
+printf("%.2f to %.2f k.ohm",a,b)
+printf("d. The power dissipated by the sensor can be determined most easily from V^2/R2,as the voltage across R2 has been calculated.The power dissipated varies from ")
+c = (Vs^2)/R_high;
+d = (Vs^2)/R_low;
+printf("%.4f to %.4f W",c,d)
+
+
+
+
+
diff --git a/1316/CH2/EX2.2/resultexample2_2.txt b/1316/CH2/EX2.2/resultexample2_2.txt new file mode 100644 index 000000000..a4bdafdfc --- /dev/null +++ b/1316/CH2/EX2.2/resultexample2_2.txt @@ -0,0 +1,7 @@ +a. The solution is given by VsR2/R1+R2.For R2=4k.ohm,we have
+VD = 1.43 V
+For R2=12k.ohm,we have
+VD = 2.73 V
+b. Thus the voltage varies from 1.43 to 2.73 Vc. The range of output impedance is found from the parallel
+ combination of R1 and R2 for the minimum and maximum of R2.
+ Simple parallel resistance computation shows that this will be from 2.86 to 5.45 k.ohmd. The power dissipated by the sensor can be determined most easily from V^2/R2,as the voltage across R2 has been calculated.The power dissipated varies from 2.0833 to 6.2500 W
\ No newline at end of file diff --git a/1316/CH2/EX2.3/example2_3.sce b/1316/CH2/EX2.3/example2_3.sce new file mode 100644 index 000000000..edf7d6df7 --- /dev/null +++ b/1316/CH2/EX2.3/example2_3.sce @@ -0,0 +1,17 @@ +//Chapter 2
+//Example 2.3
+//Page 60
+
+clear;
+clc;
+
+R1 = 1000;
+R2 = 842;
+R3 = 500;
+
+printf("Because bridge is nulled we find R4 using equation:\n R1R4=R3R2 \n")
+//Calculation of R4
+R4=R3*R2/R1;
+printf("%.0f ohm",R4)
+
+
diff --git a/1316/CH2/EX2.3/resultexample2_3.txt b/1316/CH2/EX2.3/resultexample2_3.txt new file mode 100644 index 000000000..82b35d588 --- /dev/null +++ b/1316/CH2/EX2.3/resultexample2_3.txt @@ -0,0 +1,3 @@ + Because bridge is nulled we find R4 using equation:
+ R1R4=R3R2
+421 ohm
\ No newline at end of file diff --git a/1316/CH2/EX2.4/example2_4.sce b/1316/CH2/EX2.4/example2_4.sce new file mode 100644 index 000000000..f1b08d7a8 --- /dev/null +++ b/1316/CH2/EX2.4/example2_4.sce @@ -0,0 +1,18 @@ +//Chapter 2
+//Example 2.4
+//Page 60
+
+clear;
+clc;
+
+R1 = 120;
+R2 = 120;
+R3 = 120;
+R4 = 121;
+V=10;
+printf("Assuming detector impedance to be very high , we find the offset as \n")
+//Calculation of Delta Vd
+Del_Vd = V*(((R3*R2)-(R1*R4))/((R1+R3)*(R2+R4)))
+printf("Delta Vd = %f V",Del_Vd)
+
+
diff --git a/1316/CH2/EX2.4/resultexample2_4.txt b/1316/CH2/EX2.4/resultexample2_4.txt new file mode 100644 index 000000000..c9b08d046 --- /dev/null +++ b/1316/CH2/EX2.4/resultexample2_4.txt @@ -0,0 +1,2 @@ + Assuming detector impedance to be very high , we find the offset as
+Delta Vd = -0.020747 V
diff --git a/1316/CH2/EX2.5/example2_5.sce b/1316/CH2/EX2.5/example2_5.sce new file mode 100644 index 000000000..7d3a3d660 --- /dev/null +++ b/1316/CH2/EX2.5/example2_5.sce @@ -0,0 +1,31 @@ +//Chapter 2
+//Example 2.5
+//Page 61
+
+clear;
+clc;
+
+R1 = 2000.00;
+R2 = 2000.00;
+R3 = 2000.00;
+R4 = 2050.00;
+V = 5.00;
+Rg = 50.0;
+
+//Calculation of Offset Current
+printf("From equation the offset voltage is Vth. \n")
+
+//Calculation of Vth
+x = V *(((R3*R2)-(R1*R4))/((R1+R3)*(R2+R4)))
+printf("Vth = %f V \n",x)
+
+//Calculation of Thevenin Resistance
+printf("We next find the bridge Thevenin Resistance form equation as : ")
+y = ((R1*R2)/(R1+R2))+((R3*R4)/(R3+R4))
+printf("Rth = %f ohm \n",y)
+
+//Calculation of Current
+printf("Finally the current is given by the equation ")
+z = x/(y + Rg)
+printf("Ig = %f Ampere \n",z)
+printf("The negative sign on current simply means that current flows fromright to left.")
diff --git a/1316/CH2/EX2.5/resultexample2_5.txt b/1316/CH2/EX2.5/resultexample2_5.txt new file mode 100644 index 000000000..d8e003697 --- /dev/null +++ b/1316/CH2/EX2.5/resultexample2_5.txt @@ -0,0 +1,5 @@ + From equation the offset voltage is Vth.
+Vth = -0.030864 V
+We next find the bridge Thevenin Resistance form equation as : Rth = 2012.345679 ohm
+Finally the current is given by the equation Ig = -0.000015 Ampere
+The negative sign on current simply means that current flows fromright to left.
\ No newline at end of file diff --git a/1316/CH2/EX2.7/example2_7.sce b/1316/CH2/EX2.7/example2_7.sce new file mode 100644 index 000000000..4f13b9be6 --- /dev/null +++ b/1316/CH2/EX2.7/example2_7.sce @@ -0,0 +1,34 @@ +//Chapter 2
+//Example 2.7
+//Page 65
+
+clear;
+clc;
+
+R1 = 10000.00;
+R2 = 10000.00;
+R3 = 1000;
+R4 = 950.00;
+R5 = 50.00;
+Del_R3 = 1;
+V =10.00;
+
+//Calculation of Current required
+printf("First,for the nominal resistance values given, \n the bridge is set at anull with I=0,because\n")
+
+//Calculation of Va
+x = (V * R3)/(R1 + R3)
+printf("Va = %f V \n",x)
+
+//Calculation of Vb
+printf("With I=0 we get \n")
+y = ((V) * (R4+R5))/(R2 + R4 + R5)
+printf("Vb = %f V \n",y)
+
+//Calculation of Va for change in R3
+printf("When R3 increases by 1 ohm to 1001 ohm, Va becomes")
+z = (V * (R3+1000))/(R1 + (R3+1000))
+printf("Va = %f V \n",z)
+printf("which shows that the voltage at b must increase by 0.0008 V to renull the bridge.\nThis can be proved by current,from equation and Delta V=0,found from \n")
+I = 0.0008/50;
+printf("I = %f A",I)
diff --git a/1316/CH2/EX2.7/resultexample2_7.txt b/1316/CH2/EX2.7/resultexample2_7.txt new file mode 100644 index 000000000..041ea5956 --- /dev/null +++ b/1316/CH2/EX2.7/resultexample2_7.txt @@ -0,0 +1,9 @@ + First,for the nominal resistance values given,
+ the bridge is set at anull with I=0,because
+Va = 0.909091 V
+With I=0 we get
+Vb = 0.909091 V
+When R3 increases by 1 ohm to 1001 ohm, Va becomesVa = 1.666667 V
+which shows that the voltage at b must increase by 0.0008 V to renull the bridge.
+This can be proved by current,from equation and Delta V=0,found from
+I = 0.000016 A
\ No newline at end of file diff --git a/1316/CH2/EX2.8/example2_8.sce b/1316/CH2/EX2.8/example2_8.sce new file mode 100644 index 000000000..38bc4c129 --- /dev/null +++ b/1316/CH2/EX2.8/example2_8.sce @@ -0,0 +1,22 @@ +//Chapter 2
+//Example 2.8
+//Page 66
+
+clear;
+clc;
+
+R1 = 1000.00;
+R2 = 1000.00;
+R3 = 605.00;
+R4 = 500.00;
+V =10.00;
+
+//Finding Unknown Potential
+printf("Here simply we use the equation to solve for Vx \n")
+
+//Calculation of Vx
+x = -[(V*R3)/(R3+R1)]+[(V*R4)/(R4+R2)]
+printf("Vx = %f V \n",x)
+
+
+printf("The negative sign tells us the polarity of the voltage , Vx.Since Vx numerically \n substracts from Va,we see that its positive terminal must be connected to pointa in figure \n")
diff --git a/1316/CH2/EX2.8/resultexample2_8.txt b/1316/CH2/EX2.8/resultexample2_8.txt new file mode 100644 index 000000000..02116a092 --- /dev/null +++ b/1316/CH2/EX2.8/resultexample2_8.txt @@ -0,0 +1,4 @@ + Here simply we use the equation to solve for Vx
+Vx = -0.436137 V
+The negative sign tells us the polarity of the voltage , Vx.Since Vx numerically
+ substracts from Va,we see that its positive terminal must be connected to pointa in figure
\ No newline at end of file diff --git a/1316/CH2/EX2.9/example2_9.sce b/1316/CH2/EX2.9/example2_9.sce new file mode 100644 index 000000000..b540a958a --- /dev/null +++ b/1316/CH2/EX2.9/example2_9.sce @@ -0,0 +1,24 @@ +//Chapter 2
+//Example 2.9
+//Page 67
+
+clear;
+clc;
+
+R1 = 5000;
+R2 = 5000;
+R3 = 1000;
+R4 = 990;
+R5 = 10;
+Vx =10;
+Pot = 0.012;
+
+//Finding Current necessary to null the bridge
+printf("First an examination of the resistancs show \n that the bridge is nulled when I = 0 and Vx = 0\n from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0")
+x=(R3*Vx)/(R1+R3);
+printf("(R3*V)/(R1+R3))=%.3f V\n",x)
+y=(Vx*(R4+R5)/(R2+R4+R5))
+printf("(V*(R4+R5)/(R2+R4+R5))=%.3f V",y)
+z = Pot/R5*1000;
+printf("\n Thus,we can use equation Vx-IR5=0: \n 12mV-10I=0 \n Thus I = %0.1f mA",z)
+
diff --git a/1316/CH2/EX2.9/resultexample2_9.txt b/1316/CH2/EX2.9/resultexample2_9.txt new file mode 100644 index 000000000..99fb723ab --- /dev/null +++ b/1316/CH2/EX2.9/resultexample2_9.txt @@ -0,0 +1,7 @@ + First an examination of the resistancs show
+ that the bridge is nulled when I = 0 and Vx = 0
+ from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0(R3*V)/(R1+R3))=1.667 V
+(V*(R4+R5)/(R2+R4+R5))=1.667 V
+ Thus,we can use equation Vx-IR5=0:
+ 12mV-10I=0
+ Thus I = 1.2 mA
\ No newline at end of file |