Added new codeHEADmaster

90 files changed, 2050 insertions, 0 deletions

diff --git a/3830/CH1/EX1.1/Ex1_1.sce b/3830/CH1/EX1.1/Ex1_1.sce
new file mode 100644
index 000000000..e88d81754
--- /dev/null
+++ b/3830/CH1/EX1.1/Ex1_1.sce
@@ -0,0 +1,24 @@
+// Exa 1.1

+

+clc;

+clear;

+

+// Given

+

+I = 10; // Current in Amp

+dI = 0.1; // Probability of error in I (Amp)

+R = 100; // Resistor value in Ohms

+dR = 2; // Probability of error in R (Ohms)

+

+

+// Solution

+

+printf('The power dissipated P = I^2*R \n');

+

+printf(' The probable error can be determined with the help of Erss(Root Sum Square) Formula, i.e Error = sqrt((dR*I^2)^2 + (2*I*R*dI)^2)  \n');

+

+PE = sqrt((dR*I^2)^2 + (dI*2*I*R)^2); // Probable error

+P = I^2 * R;

+

+printf(' Error = %d W \n',round(PE));

+printf(' Power dissipated P = %d kW \n',P*10^-3);

diff --git a/3830/CH1/EX1.10/Ex1_10.sce b/3830/CH1/EX1.10/Ex1_10.sce
new file mode 100644
index 000000000..fb8484166
--- /dev/null
+++ b/3830/CH1/EX1.10/Ex1_10.sce
@@ -0,0 +1,23 @@
+// Exa 1.10

+

+clc;

+clear;

+

+// Given

+

+// Referring Fig. 1.28 and 1.29

+Rm = 78; // Meter resistance in Ohms

+Ra = 1000; // in Ohms

+Rb = 1000; // in Ohms

+Rc = 1000; // in Ohms

+

+// Solution

+

+Rth = Rc + Ra*Rb/(Ra+Rb);

+

+// Let Im/Ie be 'x' where Ie = expected value and Im be shown value

+x = Rth/(Rth+Rm);

+

+Error = (1-x)*100;

+

+printf('The meter reading shown is %d percent of the expected value. Therefore, Error = %d percent \n',100-round(Error),round(Error));

diff --git a/3830/CH1/EX1.11/Ex1_11.sce b/3830/CH1/EX1.11/Ex1_11.sce
new file mode 100644
index 000000000..e4b04e3cd
--- /dev/null
+++ b/3830/CH1/EX1.11/Ex1_11.sce
@@ -0,0 +1,30 @@
+// Exa 1.11

+

+clc;

+clear;

+

+// Given

+

+// An Ohm meter

+Ifs = 0.001; // Current in Amp

+V =5; // Supply voltage in Volts

+Rm = 200; // Meter resistance in Ohms

+

+// Solution

+

+// Ifs = V/(Rm+Rp);

+

+Rp =V/Ifs - Rm;

+

+printf('Value of Rx with 20 percent full-scale deflection :- \n');

+P1 = 0.2;

+Rx1 = (Rp+Rm)/P1 - (Rp+Rm);

+printf('The resistor value for 20 percent of full scale deflection = %d k Ohms \n',Rx1/1000);

+printf(' Value of Rx with 40 percent full-scale deflection :- \n');

+P2 = 0.4;

+Rx2 = (Rp+Rm)/P2 - (Rp+Rm);

+printf('The resistor value for 40 percent of full scale deflection = %.1f k Ohms \n',Rx2/1000);

+printf(' Value of Rx with 50 percent full-scale deflection :- \n');

+P3 = 0.5;

+Rx3 = (Rp+Rm)/P3 - (Rp+Rm);

+printf('The resistor value for 50 percent of full scale deflection = %d k Ohms \n',Rx3/1000);

diff --git a/3830/CH1/EX1.12/Ex1_12.sce b/3830/CH1/EX1.12/Ex1_12.sce
new file mode 100644
index 000000000..b968d1b88
--- /dev/null
+++ b/3830/CH1/EX1.12/Ex1_12.sce
@@ -0,0 +1,18 @@
+// Exa 1.12

+

+clc;

+clear;

+

+// Given

+

+// Referring Fig. 1.49

+Ifs = 10; // Full scale deflection current in Amp

+Im = 0.015; // Meter resistance in Amp

+R = 5; // Resistance in Ohms

+

+// Solution

+

+Isg = Ifs-Im;

+Rsg = Im*R/Isg;

+

+printf('The value of shunt resistance for the milli Ammeter = %.2f * 10^-3  Ohms \n',Rsg*1000);

diff --git a/3830/CH1/EX1.13/Ex1_13.sce b/3830/CH1/EX1.13/Ex1_13.sce
new file mode 100644
index 000000000..54a876b6b
--- /dev/null
+++ b/3830/CH1/EX1.13/Ex1_13.sce
@@ -0,0 +1,24 @@
+// Exa 1.13

+

+clc;

+clear;

+

+// Given

+

+Rm = 1; // Meter resistance in ohms

+Vmax = 250; // Max voltage(V)

+R1 = 4.999; // Series resistance in Ohms

+R2 = 1/499; //Shunt resistance in Ohms

+V = 500; // Volage to be measured(V)

+// Solution

+

+Imeter = V/(R1+Rm);

+printf('Meter current Imeter = %.2f A \n ',Imeter);

+printf('The current through the meter should not exceed %.1f A \n Hence, Voltage drop = %.2f V \n',Imeter,Imeter*Rm);

+Vdrop = Imeter*Rm;

+// When Shunt resistance is connected accross meter

+CurrentRange = Vdrop/R2 + Imeter;

+Rs =(V-Imeter)/0.05;

+printf(' Series resistance Rs = %d Ohms \n',Rs);

+

+//The answer provided in the textbook are incorrect

diff --git a/3830/CH1/EX1.14/Ex1_14.sce b/3830/CH1/EX1.14/Ex1_14.sce
new file mode 100644
index 000000000..6eb3a70a0
--- /dev/null
+++ b/3830/CH1/EX1.14/Ex1_14.sce
@@ -0,0 +1,29 @@
+// Formulae's from Example 1.6 are used here

+

+// Exa 1.14

+

+clc;

+clear;

+

+// Given

+

+// Referring Fig. 1.50

+Im = 100*10^-6; // Meter resistance in Amp

+Rm = 1000; // Meter resistance in Ohms

+I1 = 1; // in Amp

+I2 = 0.1; // in Amp

+I3 = 0.01; // in Amp

+

+// Solution

+

+n = I3/Im;

+Rsh = Rm/(n-1);

+printf('  Ra + Rb + Rc = Rsh; \n ');

+printf(' The value of Rsh by calculations = %.4f Ohms \n',Rsh);

+

+Rc = Im*(Rsh+Rm)/I1;

+Rb = (Im/I2)*(Rsh+Rm) - Rc;

+Ra = Rsh-(Rb+Rc);

+printf('  Ra = %.3f Ohms, Rb = %.3f Ohms and Rc = %.3f Ohms \n',Ra,Rb,Rc);

+

+//The answer provided in the textbook is wrong for Ra calculation

diff --git a/3830/CH1/EX1.15/Ex1_15.sce b/3830/CH1/EX1.15/Ex1_15.sce
new file mode 100644
index 000000000..febb9fc98
--- /dev/null
+++ b/3830/CH1/EX1.15/Ex1_15.sce
@@ -0,0 +1,29 @@
+// Exa 1.15

+

+clc;

+clear;

+

+// Given

+

+// Refer fig. 1.73 

+Rtotal = 11*10^6; // Total resistance in Ohms

+Vg = 1; // Voltage at gate of balancing circuit

+Vmax = 1000; // Max voltage in Volts

+// Solution

+

+printf('When the selector switch is at 1000 V position, and the mass input is 1000 V, the drop across R7 should be 1 value \n');

+R7 = (Vg/Vmax)*Rtotal;

+printf('R7 = %d k Ohms \n',R7/1000);

+printf('Similarly, when the selector switch is at the 300 V position, the drop across (R6+R7) should be 1 V \n');

+R6 = (Vg/300)*Rtotal-R7;

+printf('R6 = %.2f k Ohms \n',R6/1000);

+R5 = (Vg/100)*Rtotal-(R7+R6);

+printf('R5 = %.2f k Ohms \n',R5/1000);

+R4 = (Vg/30)*Rtotal-(R7+R6+R5);

+printf('R4 = %.2f k Ohms \n',R4/1000);

+R3 = (Vg/10)*Rtotal-(R7+R6+R5+R4);

+printf('R3 = %.2f k Ohms \n',R3/1000);

+R2 = (Vg/3)*Rtotal-(R7+R6+R5+R4+R3);

+printf('R2 = %.2f k Ohms \n',R2/1000);

+R1 = Rtotal - (R2+R3+R4+R5+R6+R7);

+printf('R1 = %.2f k Ohms \n',R1/1000);

diff --git a/3830/CH1/EX1.16/Ex1_16.sce b/3830/CH1/EX1.16/Ex1_16.sce
new file mode 100644
index 000000000..e11e5ebcd
--- /dev/null
+++ b/3830/CH1/EX1.16/Ex1_16.sce
@@ -0,0 +1,32 @@
+// Exa 1.16

+

+clc;

+clear;

+

+// Given

+

+// Refer Fig.1.74

+Rmv = 200; // Voltmeter resistance in Ohms

+Vt = 3; // Terminal voltage(V)

+S = 1; // Sensitivity in mm/microV

+Rmi = 100; // Galvanometer resistance in Ohms

+Deflection = 250; // in mm

+S1 = 5; // sesitivity of second galvanometer in mm/micro Amp

+Ri = 1000; // Internal resistance of 2nd galvanometer

+// Solution

+

+Ig = Deflection/S;

+printf(' Current through the galvanometer = %d micro Amp \n',Ig);

+Rtotal = Rmv+Rmi;

+Vrtotal = Rtotal*Ig*10^-6; // in Volts

+Ek = Vt - Vrtotal;

+printf('The emf of the unknown source = %.3f V \n',Ek);

+printf('1 mm correspponds to 1 micro Amp. Therefore, Resolution = %d micro V/mm \n',Rtotal);

+

+printf('For galvanometer A, 1 mm deflection for 300 mV. So, Sa = 1/300 mm/microV \n');

+Sa = 1/300; // Sensitivity of galvanometer A in mm/micro V

+printf('For galvanometer B, - \n');

+Rbtotal = Ri+Rmv;

+// A 5mm deflcetion is caused by 1200 micro V

+printf('Sb = 5/1200  mm/microV \n'); // mm/microV

+printf('Galvanometer B provides the amplified sensitivity i.e, Since, Sb>Sa \n');

diff --git a/3830/CH1/EX1.17/Ex1_17.sce b/3830/CH1/EX1.17/Ex1_17.sce
new file mode 100644
index 000000000..160ae65f0
--- /dev/null
+++ b/3830/CH1/EX1.17/Ex1_17.sce
@@ -0,0 +1,16 @@
+// Exa 1.17

+

+clc;

+clear;

+

+// Given

+

+// Refer Fig. 1.77

+Ifs = 10^-6; // Full scale deflection current in Amp

+Rm = 300; // Meter resistance in Ohms

+Erms = 10; // in Volts

+Idc = 1*10^-3; // in Amp

+// Solution

+S = 1/Ifs;

+Rs = 0.45* Erms/Idc - Rm;

+printf(' The value of multiplier resistance Rs = %.1f k Ohms \n',Rs/1000);

diff --git a/3830/CH1/EX1.18/Ex1_18.sce b/3830/CH1/EX1.18/Ex1_18.sce
new file mode 100644
index 000000000..6202c2377
--- /dev/null
+++ b/3830/CH1/EX1.18/Ex1_18.sce
@@ -0,0 +1,17 @@
+// Exa 1.18

+

+clc;

+clear;

+

+// Given

+

+Ifs = 10^-3; // Full scale deflection current in Amp

+Rm = 500; // Meter resistance in Ohms

+Range = 10; // Em = 10*Vrms 

+// Solution

+

+Sdc = 1/Ifs; // Dc Sensitivity in Ohms/Volt 

+Sac = 0.9*Sdc; // Ac Sensitivity in Ohms/Volt

+

+Rs = Sac * Range - Rm; 

+printf('The value of multiplier resistance Rs = %d Ohms \n',Rs);

diff --git a/3830/CH1/EX1.19/Ex1_19.sce b/3830/CH1/EX1.19/Ex1_19.sce
new file mode 100644
index 000000000..5cece7809
--- /dev/null
+++ b/3830/CH1/EX1.19/Ex1_19.sce
@@ -0,0 +1,32 @@
+// Exa 1.19

+

+clc;

+clear;

+

+// Given

+

+// Design of thermocouple voltmeter

+// Three ranges

+V1 = 5; // Volts

+V2 = 10; // Volts

+V3 = 25; // Volts

+Ifs = 50*10^-3; // Amp

+Rm = 200; // Ohms

+Imax = 5*10^-3; // Amps

+Rheater = 200; // Ohms

+

+// Solution

+

+printf(' To get a FSD for 5,10 and 25V, current through the heater must be limited to 5mA \n');

+

+printf('For a 5V range \n');

+Rs1 = V1/Imax - Rm;

+printf('Series resistance Rs = %d Ohms \n',Rs1);

+printf('For a 10V range \n');

+Rs2 = V2/Imax - Rm;

+printf('Series resistance Rs = %d Ohms \n',Rs2); 

+printf('For a 25V range \n');

+Rs3 = V3/Imax - Rm;

+printf('Series resistance Rs = %d Ohms \n',Rs3);  

+

+//The answer provided in the textbook for Rs3 is wrong

diff --git a/3830/CH1/EX1.2/Ex1_2.sce b/3830/CH1/EX1.2/Ex1_2.sce
new file mode 100644
index 000000000..0a9354c41
--- /dev/null
+++ b/3830/CH1/EX1.2/Ex1_2.sce
@@ -0,0 +1,18 @@
+// Exa 1.2

+

+clc;

+clear;

+

+// Given

+

+No_Div = 50; // No of divisions

+V = 100; // Max voltage measured (V)

+

+// Solution

+

+printf('Resolution is the samellest change in input that can be measured \n The meter can be read to 1/2 division \n');

+printf(' The resolution is 1/2 divisions and its value in volts is : ');

+// 100 Div = 100 V

+// 1 Div = 1 V

+// 0.5 Div = 0.5 V

+printf(' 0.5 V \n Thus, the resolution of instrument is 0.5 V \n'); 

diff --git a/3830/CH1/EX1.20/Ex1_20.sce b/3830/CH1/EX1.20/Ex1_20.sce
new file mode 100644
index 000000000..6d6cef8f4
--- /dev/null
+++ b/3830/CH1/EX1.20/Ex1_20.sce
@@ -0,0 +1,25 @@
+// Exa 1.20

+

+clc;

+clear;

+

+// Given

+

+// Refer Fig. 1.86

+Vdc = 12; // Volts

+Vac = 20; // Volts

+Vz = 12; // Volts

+Iz = 10*10^-3; // in Amps

+

+// Solution

+

+R =(Vac-Vdc)/Iz;

+printf('The value of resistance R = %d Ohms \n',R);

+P = Vdc*Iz;

+printf(' Power rating of the Zener = %d mW \n',P*1000);

+printf(' Factor 2 is the safety factor \n The power rating of the resistor taking a safety factor of 2 is P = ');

+psafety = 2*(Vdc - Vz)/R;

+printf(' %.2f W \n',psafety);

+printf(' 1/16 W resistor curve serves the purpose \n');

+

+//The answer provided in the textbook for power rating is wrong

diff --git a/3830/CH1/EX1.21/Ex1_21.sce b/3830/CH1/EX1.21/Ex1_21.sce
new file mode 100644
index 000000000..3db36aa53
--- /dev/null
+++ b/3830/CH1/EX1.21/Ex1_21.sce
@@ -0,0 +1,24 @@
+// Exa 1.21

+

+clc;

+clear;

+

+// Given

+

+// A Volt box design

+Vs = 100; // Input voltage (V)

+V2 = 5; // Output voltage (V)

+Rs = 10*10^6; // Desired sum of resistance(R1+R2) Ohms

+

+

+// Solution

+

+// By voltage divider formula, we get

+//  R2/(R1+R2) = V2/Vs ; 

+// i.e, By simplifying

+R2 = Rs*V2/Vs;

+

+R1 = Rs - R2;

+printf(' The desired values of R1 and R2 to satisfy Volt box requirements are %.1f M ohms and %.2f M ohms respectively \n ',R1/10^6,R2/10^6);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.22/Ex1_22.sce b/3830/CH1/EX1.22/Ex1_22.sce
new file mode 100644
index 000000000..bcb61442b
--- /dev/null
+++ b/3830/CH1/EX1.22/Ex1_22.sce
@@ -0,0 +1,18 @@
+// Exa 1.22

+

+clc;

+clear;

+

+// Given

+

+// A sine wave AC input 

+

+// Solution

+

+printf('For a square wave, the form factor is 1.0, that is, the average and rms value are same \n');

+printf(' For a sine wave, the form factor is 1.1, that is, the rms is 1.11 times the average \n');

+printf(' Since the meter is calibrated for a sine wave, for a 1.0 V rms value of square wave, it indicates 1.11 V \n');

+FFsq = 1.0;

+FFsi = 1.11;

+Perror = 100*(FFsi-FFsq)/FFsq;

+printf(' The percentage error in the meter indication = %d percent \n',Perror);

diff --git a/3830/CH1/EX1.23/Ex1_23.sce b/3830/CH1/EX1.23/Ex1_23.sce
new file mode 100644
index 000000000..13ee92511
--- /dev/null
+++ b/3830/CH1/EX1.23/Ex1_23.sce
@@ -0,0 +1,23 @@
+// Exa 1.23

+

+clc;

+clear;

+

+// Given

+

+// Dual slope integrating-type DVM

+C = 0.1*10^-6; // Capacitor in Farads

+R = 10*10^3; // Resistance in  Ohms

+Vr = 2; // Reference Voltage(Volts)

+Vmax = 10; // maximum output of circuit(Volts)

+

+// Solution

+

+Tc = C*R; // Integrator Time Constant

+Vo = Vr/Tc ; // Integrator output in Volt/sec

+

+Ti = Vmax/Vo;  //in sec

+

+printf(' The period of integration of dual slope integrating-type DVM = %d m sec \n',Ti*1000);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.24/Ex1_24.sce b/3830/CH1/EX1.24/Ex1_24.sce
new file mode 100644
index 000000000..56296cc57
--- /dev/null
+++ b/3830/CH1/EX1.24/Ex1_24.sce
@@ -0,0 +1,18 @@
+// Exa 1.24

+

+clc;

+clear;

+

+// Given

+

+Am = 20; // Capacitance in Farads

+dr= 5 ; // Percentage variation in capacitor value

+

+// Solution

+

+// A = Am (±) Am× dr/100 ; // A is guranteed value of capacitor

+

+A_upperlimit = Am*(1+ dr/100) ; 

+A_lowerlimit = Am*(1- dr/100) ; 

+

+printf(' The guranteed  limits of capacitance range from %.1f F to %.1f F \n',A_lowerlimit,A_upperlimit);

diff --git a/3830/CH1/EX1.25/Ex1_25.sce b/3830/CH1/EX1.25/Ex1_25.sce
new file mode 100644
index 000000000..e853a0ead
--- /dev/null
+++ b/3830/CH1/EX1.25/Ex1_25.sce
@@ -0,0 +1,17 @@
+// Exa 1.25

+

+clc;

+clear;

+

+// Given

+ 

+// A 0-250 range milliAmmeter

+Er = 2; // Percentage accuracy of Ammeter in terms of FSR

+I = 150; // Measurement of Ammeter in mA

+Ifsr = 250; // Full scale reading of milliAmmeter (mA)

+

+// Solution

+

+dV = Er/100 * Ifsr; // Error in FSR reading

+Lr = 100*dV/I;

+printf('The limiting error = %.2f percent \n',Lr);

diff --git a/3830/CH1/EX1.26/Ex1_26.sce b/3830/CH1/EX1.26/Ex1_26.sce
new file mode 100644
index 000000000..0377dfa2d
--- /dev/null
+++ b/3830/CH1/EX1.26/Ex1_26.sce
@@ -0,0 +1,23 @@
+// Exa 1.26

+

+clc;

+clear;

+

+// Given

+

+R = 50; // Resistance value (Ohms)

+dR = 0.2; // variation in Resistance value (Ohms)

+I = 4; // Current value measured (Amp)

+dI = 0.02; // variation in current measurements (Amp)

+

+// Solution

+

+Per_Limiting_Error_Resis = dR/R * 100;

+Per_Limiting_Error_Curr = dI/I * 100;

+

+P = I^2 * R;

+dP = Per_Limiting_Error_Curr*2 + Per_Limiting_Error_Resis;

+

+printf('The limiting error in resistance measurement = ± %.2f percent \n',Per_Limiting_Error_Resis);

+printf(' The limiting error in current measurement = ± %.2f percent \n',Per_Limiting_Error_Curr);

+printf(' The limiting error in power measurement =  %.2f percent \n',dP);

diff --git a/3830/CH1/EX1.27/Ex1_27.sce b/3830/CH1/EX1.27/Ex1_27.sce
new file mode 100644
index 000000000..3b819b980
--- /dev/null
+++ b/3830/CH1/EX1.27/Ex1_27.sce
@@ -0,0 +1,16 @@
+// Exa 1.27

+

+clc;

+clear;

+

+// Given

+

+// A moving coil Ammeter

+FSR = 10; // Full scale reading in Amp

+No_of_div = 100;

+

+// Solution

+

+one_scale_div = FSR/No_of_div ; // in Amp

+Resolution = 1/2 * one_scale_div ; // Since, the instrument can read upto half of the full-scale division(Amp)

+printf('Resolution = %d mA \n', Resolution*1000);

diff --git a/3830/CH1/EX1.28/Ex1_28.sce b/3830/CH1/EX1.28/Ex1_28.sce
new file mode 100644
index 000000000..f7d240103
--- /dev/null
+++ b/3830/CH1/EX1.28/Ex1_28.sce
@@ -0,0 +1,27 @@
+// Exa 1.28

+

+clc;

+clear;

+

+// Given

+

+// Limiting error for series and parallel combination of capacitors

+c1 = 99; // Capacitor value in Mf

+dc1 = 1; // Variation in capacitor value in Mf

+c2 = 49; // Capacitor value in Mf

+dc2 = 1; // Variation in capacitor value in Mf

+

+// Solution

+

+// C1 = c1(±) dc1;

+// C2 = c2(±) dc2;

+printf('For parallel combination, we have y = C1+C2 \n');

+dY_parallel = dc1 + dc2;

+printf(' Limiting error for parallel combination = (±) %d Mf \n',dY_parallel);    

+printf(' For series combination, we have 1/y = 1/C1+1/C2 \n');

+Yseries = c1*c2/(c1+c2);

+dYseries = (-dc1+c1)*(-dc2+c2)/(c1+c2-dc1-dc2);

+dY = Yseries - dYseries;

+printf(' Limiting error for series combination = (±) %.3f Mf \n',dY);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.29/Ex1_29.sce b/3830/CH1/EX1.29/Ex1_29.sce
new file mode 100644
index 000000000..9233e0b1d
--- /dev/null
+++ b/3830/CH1/EX1.29/Ex1_29.sce
@@ -0,0 +1,33 @@
+// Exa 1.29

+

+clc;

+clear;

+

+// Given

+

+// 3 resistances in series and parallel combination

+

+r1 = 200; // First resistance in (Ohms)

+dr1 = 5 ; // Percentage variation for first resistance

+r2 = 100; // Second resistance in (Ohms)

+dr2 = 5 ; // Percentage variation for second resistance

+r3 = 50; // Third resistance in (Ohms)

+dr3 = 5 ; // Percentage variation for third resistance

+

+// Solution

+

+printf('Lets say Rse be the series combination of resistances \n');// series

+Rse = r1+r2+r3;

+Relative_Error_series = r1/Rse * dr1 + r2/Rse * dr2 + r3/Rse * dr3; // in percentage

+Error_series_Ohms = Rse * Relative_Error_series/100;

+printf(' The Relative error for series combination(Rse) is %d percent which is equivalent to %.2f Ohms \n',Relative_Error_series,Error_series_Ohms);

+printf(' Lets say Rpa be the parallel combination of resistances \n');// parallel

+Rpa = r1*r2*r3/(r2*r3+r1*r2+r1*r3);  // lets say (x/y1+y2+y3)

+Error_x = dr1+dr2+dr3;

+Error_y1 = dr1+dr2;

+Error_y2 = dr2+dr3;

+Error_y3 = dr3+dr1;

+Error_Y = r1/Rse * Error_y1 + r2/Rse * Error_y2 + r3/Rse * Error_y3;

+Relative_Error_parallel = Error_x+ Error_Y;  // in percentage

+Error_parallel_Ohms = Rpa * Relative_Error_parallel/100 ;

+printf(' The Relative error for parallel combination(Rpa) is %d percent which is equivalent to %.4f Ohms \n',Relative_Error_parallel,Error_parallel_Ohms);

diff --git a/3830/CH1/EX1.3/Ex1_3.sce b/3830/CH1/EX1.3/Ex1_3.sce
new file mode 100644
index 000000000..a8993a58e
--- /dev/null
+++ b/3830/CH1/EX1.3/Ex1_3.sce
@@ -0,0 +1,19 @@
+// Exa 1.3

+

+clc;

+clear;

+

+// Given

+

+// A 3_1/2 digit DVM

+V = 19.99; // Max voltage in Volts

+

+// Solution

+

+printf('The maximum number of counts that can be made with 9 3_1/2 digit DVM is  1999 \n The samllest change in input that can be measured is 1 count \n');

+// 1 count in volts corresponds to resolution :-

+// 1999 counts = 19.99 V

+// 1 count = ?

+Resolution = 19.99/1999;

+

+printf(' Resolution = %d mV \n',round(Resolution*10^3));

diff --git a/3830/CH1/EX1.30/Ex1_30.sce b/3830/CH1/EX1.30/Ex1_30.sce
new file mode 100644
index 000000000..b16287b92
--- /dev/null
+++ b/3830/CH1/EX1.30/Ex1_30.sce
@@ -0,0 +1,44 @@
+// Exa 1.30

+

+clc;

+clear;

+

+// Given

+

+// Various Inductance Measurements

+L1 = 1.003; // First reading in mH

+L2 = 0.998; // second reading in mH

+L3 = 1.001; // third reading in mH

+L4 = 0.991; // fourth reading in mH

+L5 = 1.009; // Fifth reading in mH

+L6 = 0.996; // sixth reading in mH

+L7 = 1.005; // seventh reading in mH

+L8 = 0.997; // eight reading in mH

+L9 = 1.008; // nineth reading in mH

+L10 = 0.994; // tenth reading in mH

+n = 10; // total no of readings

+

+// Solution

+

+AM = (L1+L2+L3+L4+L5+L6+L7+L8+L9+L10)/n;

+printf('The arithmatic mean = %.4f mH \n',AM);

+

+// Deviation for each reading will be -

+d1 = L1 - AM; // deviation for 1st reading

+d2 = L2 - AM; // deviation for 2nd reading

+d3 = L3 - AM; // deviation for 3rd reading

+d4 = L4 - AM; // deviation for 4th reading

+d5 = L5 - AM; // deviation for 5th reading

+d6 = L6 - AM; // deviation for 6th reading

+d7 = L7 - AM; // deviation for 7th reading

+d8 = L8 - AM; // deviation for 8th reading

+d9 = L9 - AM; // deviation for 9th reading

+d10 = L10 - AM; // deviation for 10th reading

+

+Avg_deviation = (d1+d2+d3+d4+d5+d6+d7+d8+d9+d10)/n;

+printf(' The average deviation = %d mH \n',Avg_deviation);

+

+SD = sqrt((d1^2+d2^2+d3^2+d4^2+d5^2+d6^2+d7^2+d8^2+d9^2+d10^2)/(n-1));

+printf(' The standard deviation = %.3f mH \n',SD);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.31/Ex1_31.sce b/3830/CH1/EX1.31/Ex1_31.sce
new file mode 100644
index 000000000..d1af1d43c
--- /dev/null
+++ b/3830/CH1/EX1.31/Ex1_31.sce
@@ -0,0 +1,51 @@
+// Exa 1.31

+

+clc;

+clear;

+

+// Given

+

+// Various Current Measurements

+

+I1 = 41.7; // First reading in A

+I2 = 42; // second reading in A

+I3 = 41.8; // third reading in A

+I4 = 42; // fourth reading in A

+I5 = 42.1; // Fifth reading in A

+I6 = 41.9; // sixth reading in A

+I7 = 42; // seventh reading in A

+I8 = 41.9; // eight reading in A

+I9 = 42.5; // nineth reading in A

+I10 = 41.8; // tenth reading in A

+n=10; // Total no of observations

+I = [41.7;42;41.8;42;42.1;41.9;42;41.9;42.5;41.8];

+

+// Solution

+

+AM = (I1+I2+I3+I4+I5+I6+I7+I8+I9+I10)/n;

+printf('The arithmatic mean = %.4f A \n',AM);

+

+// Deviation for each reading will be -

+d1 = I1 - AM; // deviation for 1st reading

+d2 = I2 - AM; // deviation for 2nd reading

+d3 = I3 - AM; // deviation for 3rd reading

+d4 = I4 - AM; // deviation for 4th reading

+d5 = I5 - AM; // deviation for 5th reading

+d6 = I6 - AM; // deviation for 6th reading

+d7 = I7 - AM; // deviation for 7th reading

+d8 = I8 - AM; // deviation for 8th reading

+d9 = I9 - AM; // deviation for 9th reading

+d10 = I10 - AM; // deviation for 10th reading

+

+SD = sqrt((d1^2+d2^2+d3^2+d4^2+d5^2+d6^2+d7^2+d8^2+d9^2+d10^2)/(n-1));

+printf(' The standard deviation = %.3f A \n',SD);

+

+Y = 0.6745*SD;

+printf(' Probable error of one reading = %.3f A \n',Y);

+Vm = Y/sqrt(n-1);

+printf(' Probable error of mean = %.3f A \n',Vm);

+

+

+printf(' Range = %.1f A \n',max(I)-min(I));

+

+// The answers vary due to round off error

diff --git a/3830/CH1/EX1.32/Ex1_32.sce b/3830/CH1/EX1.32/Ex1_32.sce
new file mode 100644
index 000000000..294e5520b
--- /dev/null
+++ b/3830/CH1/EX1.32/Ex1_32.sce
@@ -0,0 +1,21 @@
+// Exa 1.32

+

+clc;

+clear;

+

+// Given

+

+E = 1.2 * 10^4 ; // Phosphor Young's Modulus (kg per mm^2)

+l = 400; // Length of strip (mm)

+w = 0.5; // Width of strip (mm)

+t = 0.08; // Thickness of strip (mm)

+Theta = 90; // In degrees

+

+

+// Solution

+

+T = (E*w*t^2)/(12*l);

+

+printf('By using the torque formula having E as youngs modulus, we get T = %.3f kg-mm \n',T);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.33/Ex1_33.sce b/3830/CH1/EX1.33/Ex1_33.sce
new file mode 100644
index 000000000..2ec650bee
--- /dev/null
+++ b/3830/CH1/EX1.33/Ex1_33.sce
@@ -0,0 +1,16 @@
+// Exa 1.33

+

+clc;

+clear;

+

+//Given

+

+W = 0.005; // Weight in Kg

+l = 2.4*10^-2; // Distance in m

+Td = 1.05*10^-4; // Deflection torque in kg-m

+

+// Solution

+

+Theta = asind(Td/(W*l));

+printf('Deflection torque is given by, Td = W*l*sin(theta)\n Therefore theta =  %.1f degrees \n',Theta);

+

diff --git a/3830/CH1/EX1.34/Ex1_34.sce b/3830/CH1/EX1.34/Ex1_34.sce
new file mode 100644
index 000000000..0c0e44eb3
--- /dev/null
+++ b/3830/CH1/EX1.34/Ex1_34.sce
@@ -0,0 +1,27 @@
+// Exa 1.34

+

+clc;

+clear;

+

+// Given

+

+I1 = 10; // Current which produces deflection of 90 degrees

+Theta1 = 90; // In degrees

+I2 = 5; // Current for which theta is to be calculated

+

+// Solution

+

+//The deflection which produces a current of 1A when instrument is spring controlled

+// Tc  ∝ theta

+// theta  ∝ I^2

+

+theta2 = (I2/I1)^2 * Theta1 ;

+printf('The deflection which produces a current of 1A when instrument is spring controlled  is equal to = %.1f degrees \n',theta2);

+//The deflection which produces a current of 1A when instrument is gravity controlled

+// Tc  ∝ sin(theta)

+// theta  ∝ I^2

+

+theta2_gravity = asind((I2/I1)^2 *sind(Theta1)) ;

+printf(' The deflection which produces a current of 1A when instrument is gravity controlled = %.2f degrees \n',theta2_gravity);

+

+// The value of I given as 1A in problem statement is incorrect to satisfy the problem answer(correct value is 5A)

diff --git a/3830/CH1/EX1.35/Ex1_35.sce b/3830/CH1/EX1.35/Ex1_35.sce
new file mode 100644
index 000000000..9929fc7ec
--- /dev/null
+++ b/3830/CH1/EX1.35/Ex1_35.sce
@@ -0,0 +1,21 @@
+// Exa 1.35

+

+clc;

+clear;

+

+// Given

+

+f = 450; // Resonating frequency in kHz

+C = 250; // Capacitor value at resonating frequency (pf)

+Q = 105; // Q-meter reading at resonance

+Rsh = 0.75; // Value of shunt resistance in ohms

+

+// Solution

+

+L = 1/((2*%pi*f*10^3)^2*C*10^-12);  // in H

+w=2*%pi*f*10^3;

+R = (w*L)/Q - Rsh;

+x= round(w*L/Q);

+printf(' The value of resistance,R = %.2f Ohms \n',double(w*L/Q)-Rsh);

+

+// The answer vary due to round off error

diff --git a/3830/CH1/EX1.36/Ex1_36.sce b/3830/CH1/EX1.36/Ex1_36.sce
new file mode 100644
index 000000000..ba88357b2
--- /dev/null
+++ b/3830/CH1/EX1.36/Ex1_36.sce
@@ -0,0 +1,21 @@
+// Exa 1.36

+

+clc;

+clear;

+

+// Given

+

+f1 = 1*10^6; // first resonant frequency in Hz

+C1 = 480*10^-12; // Capacitor value at f1 in Farad

+f2 = 2*10^6; // second resonant frequency in Hz

+C2 = 120*10^-12; // Capacitor value at f2 in Farad

+R = 10; // Resistance in Ohms

+

+// Solution

+

+Cd = (C1-4*C2)/3;  // Distributive capacitor

+Q = 1/((2*%pi*f1)*R*(C1+Cd));

+

+printf('The value of Cd and Q of the coil are %.1f pf and %.2f respectively',Cd*10^12,Q);

+

+//The answer provided in the textbook cannot be confirmed(The formulae for Cd mentions L2 variable whose value is not given in problem statement)

diff --git a/3830/CH1/EX1.4/Ex1_4.sce b/3830/CH1/EX1.4/Ex1_4.sce
new file mode 100644
index 000000000..c646177d7
--- /dev/null
+++ b/3830/CH1/EX1.4/Ex1_4.sce
@@ -0,0 +1,26 @@
+// Exa 1.4

+

+clc;

+clear;

+

+// Given

+

+S = 10*10^3; // Sensitivity of voltmeter in Ohms/Volt

+V = 75; // Reading in Volts

+Vmax = 100; // Max voltage in Volts

+I = 1.5*10^-3; // reading in Amp

+

+// Solution

+

+printf('Consider Fig.1.10, it shows Rm as meter of voltmeter drawing some current \n Thus, loading of the source happens i.e, loading effect \n');

+Rapparent = V/I;

+Rm = Vmax * S;

+// Rapparent = parallel combination of Rm and Rx

+// Therefore Rx can be given as

+Rx = (1/Rapparent - 1/Rm)^-1;

+printf(' True value of Rx = %.2f K Ohms \n',Rx);

+

+Error = 100* (Rx-Rapparent)/Rx ; // Error in percent

+printf(' The percentage error due to loading effect = %.1f percent \n',Error); 

+

+//The answers vary due to round off error

diff --git a/3830/CH1/EX1.5/Ex1_5.sce b/3830/CH1/EX1.5/Ex1_5.sce
new file mode 100644
index 000000000..f770ff743
--- /dev/null
+++ b/3830/CH1/EX1.5/Ex1_5.sce
@@ -0,0 +1,21 @@
+// Exa 1.5

+

+clc;

+clear;

+

+// Given

+

+Rm = 100; // Resistor value in Ohms

+I = 10; // Current in Amp

+Im = 1*10^-3; // Meter current in Amp

+

+// Solution

+

+Ish = I - Im;

+Vm = Im * Rm; // Vm = Vsh

+Vsh = Vm;

+Rsh = Vsh/Ish; 

+

+printf('The value of shunt resistance Rsh = %.2f Ohms \n',Rsh);

+

+//The answer provided in the textbook is wrong

diff --git a/3830/CH1/EX1.6/Ex1_6.sce b/3830/CH1/EX1.6/Ex1_6.sce
new file mode 100644
index 000000000..d18700c55
--- /dev/null
+++ b/3830/CH1/EX1.6/Ex1_6.sce
@@ -0,0 +1,27 @@
+// Exa 1.6

+

+clc;

+clear;

+

+// Given

+

+Imax = 100*10^-6; // Initial range of Ammeter in Amp

+Rm = 800; // Meter resistance in Ohms

+I1max = 0.1; // Range to be extended in Amp

+I2max = 10; // Range to be extended in Amp

+

+// Solution

+

+printf(' Referring Figs. 1.21 and 1.22 :- \n');

+

+n = I1max/Imax;

+Rsh = Rm/(n-1);

+printf('  Ra + Rb + Rc = Rsh; \n ');

+printf(' The value of Rsh by calculations = %.4f Ohms \n',Rsh);

+printf('  Referring calculations done in textbook,\n  we can get values of Ra,Rb and Rc as follows :- \n');

+Rc = Imax*(Rsh+Rm)/I2max;

+Rb = (Imax/I1max)*(Rsh+Rm) - Rc;

+Ra = Rsh-(Rb+Rc);

+printf('  Ra = %.3f Ohms, Rb = %.3f Ohms and Rc = %.3f Ohms \n',Ra,Rb,Rc);

+

+//The answer provided in the textbook is wrong for Rc and not at all given for Ra and Rb

diff --git a/3830/CH1/EX1.7/Ex1_7.sce b/3830/CH1/EX1.7/Ex1_7.sce
new file mode 100644
index 000000000..55aa7af4a
--- /dev/null
+++ b/3830/CH1/EX1.7/Ex1_7.sce
@@ -0,0 +1,26 @@
+// Formulae's from Example 1.6 are used here

+// Exa 1.7

+

+clc;

+clear;

+

+// Given

+

+// Referring circuit given in Fig. 1.23

+Rm = 1000; // Meter resistance in Ohms

+Im = 100*10^-6; // Meter current in Amp

+I1 = 1; // im Amp

+I2 = 0.1; // in Amp

+I3 = 10; // in Amp 

+

+// Solution

+

+n =  I3/I2;

+Rsh = Rm/(n-1);

+Rc = (Im/I2)*(Rsh+Rm);

+Rb = Rc - (Im/I1)*(Rsh+Rm);

+Ra = Rsh-(Rb+Rc);

+

+printf('The Values of Ra, Rb and Rc are %.2f Ohms, %.2f Ohms and %.2f Ohms respectively \n',Ra,Rb,Rc);

+

+//The answer provided in the textbook is wrong for Ra

diff --git a/3830/CH1/EX1.9/Ex1_9.sce b/3830/CH1/EX1.9/Ex1_9.sce
new file mode 100644
index 000000000..bb6c314c9
--- /dev/null
+++ b/3830/CH1/EX1.9/Ex1_9.sce
@@ -0,0 +1,30 @@
+// Exa 1.9

+

+clc;

+clear;

+

+// Given

+

+// Referring Fig.1.26

+Ifs = 50*10^-6; // Full scale deflection current in Amp

+Rm = 11; // Meter resistance in Ohms

+R1 = 3; // Range in Volts

+R2 = 10; // range in Volts

+R3 = 30; // Range in Volts

+

+// Solution

+

+S = 1/Ifs; // Sensitivity in Ohms/V

+

+printf('The values of multiplier resistances in the different ranges are :- \n');

+printf(' For 3-V range :');

+Rs1 = S*R1-Rm;

+printf('%d k Ohms \n',Rs1/1000);

+printf(' For 10-V range :');

+Rs2 = S*R2-Rm;

+printf('%d k Ohms \n',Rs2/1000);

+printf(' For 30-V range :');

+Rs3 = S*R3-Rm;

+printf('%d k Ohms \n',Rs3/1000);

+

+//The answer provided in the textbook is wrong for Rs1

diff --git a/3830/CH1/EX4.1/Ex4_1.sce b/3830/CH1/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..25e272882
--- /dev/null
+++ b/3830/CH1/EX4.1/Ex4_1.sce
@@ -0,0 +1,19 @@
+// Exa 4.1

+

+clc;

+clear;

+

+// Given

+

+// An oscilloscope

+

+R = 400; // Resistance(k Ohms)

+C = 0.025; // capacitance(micro Farad)

+T = 0.4; // Time period of saw-tooth output waveform(msec)

+

+// Solution

+

+printf('  The percentage of non linearity i.e deviation in output can be given as t/(4*R*C)\n  ');

+PD = (T*10^-3)/(4*R*10^3*C*10^-6) ;

+

+printf(' Therefore, by calculation, percent deviation = %d percent \n ',PD*100);

diff --git a/3830/CH2/EX2.1/Ex2_1.sce b/3830/CH2/EX2.1/Ex2_1.sce
new file mode 100644
index 000000000..6bd11c1fd
--- /dev/null
+++ b/3830/CH2/EX2.1/Ex2_1.sce
@@ -0,0 +1,18 @@
+// Exa 2.1

+

+clc;

+clear;

+

+// Given

+ 

+// A wien bridge oscillator under consideration

+R = 55*10^3; // Resistance in Ohms

+// R = R1 = R2  ... given

+C = 800*10^-12; // Capacitor in Farad

+// C = C2 = C1  .. given

+

+// Solution

+

+f = 1/(2*%pi*R*C) ;

+

+printf(' The frequency of oscillations = %.1f Hz \n',f);

diff --git a/3830/CH2/EX2.2/Ex2_2.sce b/3830/CH2/EX2.2/Ex2_2.sce
new file mode 100644
index 000000000..80d2014c6
--- /dev/null
+++ b/3830/CH2/EX2.2/Ex2_2.sce
@@ -0,0 +1,20 @@
+// Exa 2.2

+

+clc;

+clear;

+

+// Given

+

+// A wien bridge oscillator under consideration

+

+fo= 10^6 ; // frequency of oscillations in Hz

+

+// Solution

+

+printf(' Let R = 3 k Ohms \n');

+

+R = 3000; // Ohm's

+// since, fo = 1/(2*%pi*R*C); therefore,

+C = 1/(2*%pi*fo*R);

+

+printf('  Substituting that, the value of capacitor = %d pf \n',C*10^12);

diff --git a/3830/CH2/EX2.3/Ex2_3.sce b/3830/CH2/EX2.3/Ex2_3.sce
new file mode 100644
index 000000000..7c43f41da
--- /dev/null
+++ b/3830/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,19 @@
+// Exa 2.3

+

+clc;

+clear;

+

+// Given

+

+// A phase shift oscillator

+

+R = 800*10^3; // in Ohm's

+// R = R1 = R2 = R3  .. given

+C = 100*10^-12; // in Farad

+// C = C1 = C2 = C3  .. farad

+

+// Solution

+

+fo = 1/(2*%pi*R*C*sqrt(6));

+

+printf(' The frequency of oscillations = %d Hz \n',fo);

diff --git a/3830/CH2/EX2.4/Ex2_4.sce b/3830/CH2/EX2.4/Ex2_4.sce
new file mode 100644
index 000000000..419eb7297
--- /dev/null
+++ b/3830/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,21 @@
+// Exa 2.4

+

+clc;

+clear;

+

+// Given

+

+// Transistor Colpitts oscillator

+L = 100*10^-3; // Inductance(H)

+C1 = 0.005*10^-6; // Capacitor(F)

+C2 = 0.01*10^-6; // Capacitor(F)

+

+// Solution

+

+C = C1*C2/(C1+C2);

+printf(' By calculation, C = %.2f pf \n',C*10^12);

+fo = 1/(2*%pi*sqrt(L*C));

+

+printf('  The frequency of oscillator = %.1f kHz \n',fo*10^-3);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH2/EX2.5/Ex2_5.sce b/3830/CH2/EX2.5/Ex2_5.sce
new file mode 100644
index 000000000..f7c08efc7
--- /dev/null
+++ b/3830/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,23 @@
+// Exa 2.5

+

+clc;

+clear;

+

+// Given

+

+// A Hartley oscillator under consideration

+L1 = 100*10^-3; // Inductance(H)

+L2 = 1*10^-3; // Inductance(H)

+M = 50*10^-3; // Inductance(H)

+C = 100*10^-12; // Capacitor(F)

+

+// Solution

+

+L = L1+L2+2*M;

+printf(' By calculation, L = %d H \n',L*10^3);

+

+f = 1/(2*%pi*sqrt(L*C));

+

+printf('  The frequency of oscillation = %d kHz \n',f*10^-3);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH2/EX2.6/Ex2_6.sce b/3830/CH2/EX2.6/Ex2_6.sce
new file mode 100644
index 000000000..d06bcf648
--- /dev/null
+++ b/3830/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,20 @@
+// Exa 2.6

+

+clc;

+clear;

+

+// Given

+

+// An Amilifier under consideration

+Av = 40; // Voltage gain

+Vi = 0.1; // Input voltage without feedback(V)

+Vi_fb = 2.4; // Input voltage with feedback(V)

+

+// Solution

+

+A = Av*Vi_fb/Vi;

+

+// Av = A/(1-B*A) ; therefore,

+B = (1-A/Av)/A;

+

+printf(' The value of feedback ratio = %.6f \n ',B);

diff --git a/3830/CH3/EX3.2/Ex3_2.sce b/3830/CH3/EX3.2/Ex3_2.sce
new file mode 100644
index 000000000..a16d30ee1
--- /dev/null
+++ b/3830/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,18 @@
+// Exa 3.2

+

+clc;

+clear;

+

+// Given

+

+Noise = -90; // Minimum detectable signal (dbm)

+Ip = 300 ; // power level of third-order product(dbm)

+

+// Solution

+

+printf(' The expression for the dynamic range of the spectrum analyser = 2/3*(Ip-MDS) \n  So, by calculations :-\n');

+ 

+DR = 2/3*(Ip-Noise);

+printf('  Dynamic range %.1f dB \n',DR);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH3/EX3.3/Ex3_3.sce b/3830/CH3/EX3.3/Ex3_3.sce
new file mode 100644
index 000000000..fb4e806cf
--- /dev/null
+++ b/3830/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,19 @@
+// Exa 3.3

+

+clc;

+clear;

+

+// Given

+

+NF = 30; // Noise figure in dB

+BW = 1; // Bandwidth of 3 dB filter in kHz

+

+// Solution

+

+printf(' The noise level of the spectrum analyser is related to the noise figure and the IF bandwidth by the following equation - \n  MDS = -114 dbm + 10*log(BW/1MHz) + NF \n  so, by calculation :- ');

+

+MDS = -114 + 10*log10(BW*10^3/10^6)+NF;

+

+printf(' MDS = %d  dBm \n ' , MDS);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH4/EX4.1/Ex4_1.sce b/3830/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..25e272882
--- /dev/null
+++ b/3830/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,19 @@
+// Exa 4.1

+

+clc;

+clear;

+

+// Given

+

+// An oscilloscope

+

+R = 400; // Resistance(k Ohms)

+C = 0.025; // capacitance(micro Farad)

+T = 0.4; // Time period of saw-tooth output waveform(msec)

+

+// Solution

+

+printf('  The percentage of non linearity i.e deviation in output can be given as t/(4*R*C)\n  ');

+PD = (T*10^-3)/(4*R*10^3*C*10^-6) ;

+

+printf(' Therefore, by calculation, percent deviation = %d percent \n ',PD*100);

diff --git a/3830/CH4/EX4.2/Ex4_2.sce b/3830/CH4/EX4.2/Ex4_2.sce
new file mode 100644
index 000000000..69016bfc6
--- /dev/null
+++ b/3830/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,40 @@
+// Exa 4.2

+

+clc;

+clear;

+

+// Given

+

+f = 83.3 ; // frequency of  sinusoidal voltage in KHz

+

+// Solution

+// part a

+

+printf('Being sunchronised, the frequency of the saw-tooth wave will be a submultiple of the signal. \n');

+

+printf(' Frequency of saw-tooth curve = %.2f kHz \n',f/10);

+F = f/10;

+printf(' Period of the saw-tooth curve = %.1f microsec \n',(1/F)*10^3);

+

+// since, Sine wave y = A sin theta

+// but y/A = 0.5(since, end of trace was at position half the amplitide away from x-axis)

+theta = asind(1/2) ;

+printf(' The 10th wave is in short of a complete since wave by %d degrees \n',theta);

+printf(' Therefore, No of full waves of sine form seen on the screen are 9 11/12 waveforms \n');

+

+// Rise time +decay time = period of wave = 120 microsec

+T = 120 ; // period in microsec

+ Rise_by_decay = (119/12) / (10- 119/12);

+DecayTime = Rise_by_decay/T;

+printf(' Decay time = %.1f microsec \n',round(DecayTime));

+printf(' Rise time = %.1f microsec \n',T-DecayTime);

+

+// part b

+

+printf(' Since, increase time base frequency = 10/4 times the final value \n');

+

+L = (10/4)* theta ;

+printf(' Length of trace blanked in degrees due to flyback time = %d degrees \n ',L);

+T_new = T*4/10;

+printf('Period of new time base = %d microsec \n',T_new);

+printf(' Rise time as per new time base = %d microsec \n',T_new-1);

diff --git a/3830/CH4/EX4.3/Ex4_3.sce b/3830/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..6604a28fb
--- /dev/null
+++ b/3830/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,17 @@
+// Exa 4.3

+

+clc;

+clear;

+

+// Given 

+

+Va = 2500; // Applied voltage(Volts)

+e = 1.602*10^-19; // Charge of electron(C)

+m = 9.107*10^-31; // Mass of electron(Kg)

+

+// Solution

+

+// For Electron beam in the oscilloscope, its velocity is given as-

+V = sqrt(2*e*Va/m);

+

+printf(' The velocity of electron beam of an oscilloscope = %.3f * 10^6 m/sec \n',V/10^6);

diff --git a/3830/CH4/EX4.4/Ex4_4.sce b/3830/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..6ddf351ea
--- /dev/null
+++ b/3830/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,15 @@
+// Exa 4.4

+

+clc;

+clear;

+

+// Given

+

+Def_sensitivity = 0.05; // Deflection sensitivity in mm/V

+Spot_deflection = 5; // in mm

+

+// Solution

+

+AppliedVoltage = Spot_deflection/Def_sensitivity ;

+

+printf(' The applied voltage = %d V \n',AppliedVoltage);

diff --git a/3830/CH4/EX4.5/Ex4_5.sce b/3830/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..063cc090f
--- /dev/null
+++ b/3830/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,18 @@
+// Exa 4.5

+

+clc;

+clear;

+

+// Given

+

+// A CRT under consideration

+l = 20; // length of x-deflection plates in mm

+d = 5; // distance between x-deflection plates in mm

+s = 250; // distance between screen and center of plate in mm

+Va = 3000; // applied accelerating voltage in volts

+

+// Solution

+

+Def_sensitivity = l*s/(2*d*Va) ;

+printf(' The deflection sensitivity = %.5f mm/V \n',Def_sensitivity);

+printf('  The deflection factor = %.1f V/mm \n',1/Def_sensitivity);

diff --git a/3830/CH4/EX4.6/Ex4_6.sce b/3830/CH4/EX4.6/Ex4_6.sce
new file mode 100644
index 000000000..42401bbcf
--- /dev/null
+++ b/3830/CH4/EX4.6/Ex4_6.sce
@@ -0,0 +1,29 @@
+// Exa 4.6

+

+clc;

+clear;

+

+// Given

+

+l = 25; // length of x-deflection plates in mm

+d = 1; // distance between x-deflection plates in mm

+s = 200; // distance between screen and centre of plate in mm

+Va = 3000; // applied accelerating voltage in volts

+Lt = 100; // length of trace in mm

+

+// Solution

+

+// Deflection produced = y/Vd = s*l/(2*d*Va)

+

+y = 1/2 *(Lt);

+// Therefore, 

+Vd = 2*d*Va*y/(l*s) ;

+

+Vrms = Vd/sqrt(2) ;

+

+printf(' The Vrms of the applied sinusoidal voltage = %.1f V \n',Vd);

+

+Def_sensitivity = l*s/(2*d*Va) ;

+printf('  The deflection sensitivity = %.5f mm/V \n',Def_sensitivity);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH4/EX4.7/Ex4_7.sce b/3830/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..8359d25d7
--- /dev/null
+++ b/3830/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,20 @@
+// Exa 4.7

+

+clc;

+clear;

+

+// Given

+

+// Two sinusoidal voltage signals are applied to vertical and horizontal plates of CRO

+

+// Solution

+printf('Theta = asin(dvo/DV');

+// Referring fig(a) 

+Theta_a = asind(0) ; // dvo = 0

+printf(' Theta for trace shown in fig(a) = %d degrees\n',Theta_a);

+// Referring fig(b) 

+Theta_b = asind(3/6) ; // dvo = 3 and DV =6

+printf(' Theta for trace shown in fig(b) = %d degrees\n',Theta_b);

+// Referring fig(c) 

+Theta_c = asind(1/1) ; // dvo = DV = 1

+printf(' Theta for trace shown in fig(c) = %d degrees\n',Theta_c);

diff --git a/3830/CH4/EX4.8/Ex4_8.sce b/3830/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..df37267b9
--- /dev/null
+++ b/3830/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,15 @@
+// Exa 4.8

+

+clc;

+clear;

+

+// Given

+

+// Referring closed Lissajous pattern as shown in fig.

+wx = 2; // no of positive x-peak 

+wy = 3; // no of positive y-peak

+

+// Solution

+

+fy_fx = wy/wx ;

+printf(' Ratio of frequencies between vertical and horizontal signals = %.1f \n',fy_fx);

diff --git a/3830/CH4/EX4.9/Ex4_9.sce b/3830/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..068182b3e
--- /dev/null
+++ b/3830/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,19 @@
+// Exa 4.9

+

+clc;

+clear;

+

+// Given

+

+// Referring Lissajous pattern shown in figure

+wx = 1 ; // Sum of x-peak pattern

+wy = 2.5; // sum of y-peak pattern

+fx = 3; // frequency of horizontal signal

+

+X = wy/wx ; // X is ratio of fy/fx

+

+// Therefore, fy = 2.5*fx

+

+printf(' Frequency of vertical signal = %.1f kHz \n ',X*fx);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX1.8/Ex5_1.sce b/3830/CH5/EX1.8/Ex5_1.sce
new file mode 100644
index 000000000..ae7ba3075
--- /dev/null
+++ b/3830/CH5/EX1.8/Ex5_1.sce
@@ -0,0 +1,23 @@
+// Exa 5.1

+

+clc;

+clear;

+

+// Given

+

+E1 = 1/100; // exposure set for grid line impression(sec)

+E2 = 10; // second exposure duration(sec)

+R = 10^-4; // persistence of CRO screen(sec)

+I1 = 1; // Trace intensity for exposure 1(candle power)

+I2_normal = 4 ; // trace intensity for normal settings(candle power)

+

+// Solution

+

+printf(' The emission of light that would be received by photographic paper in both exposures must be the same \n  Also, the product of time and light is to be the same. \n');

+I_req = I1*E1/R;

+printf('  Hence, the image intensity required = %d \n' ,I_req );

+I_boost = I_req/I2_normal;

+printf('  Therefore, the intensity boost required = %d times \n' , I_boost);

+

+printf('  The light emitted is proportional to the kinetic energy of the electron while it strikes the screen, which is equal to sqrt(V) , where V is the velocity while striking \n');

+ printf('  Therefore, the accelerating voltage of the accelerating anode should br increased by %d times \n',sqrt(I_boost));

diff --git a/3830/CH5/EX5.10/Ex5_10.sce b/3830/CH5/EX5.10/Ex5_10.sce
new file mode 100644
index 000000000..fb6531326
--- /dev/null
+++ b/3830/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,19 @@
+// Exa 5.10

+

+clc;

+clear;

+ 

+// Given

+

+Va = 2000; // voltage applied to anodes(V)

+l = 50*10^-3; // length of horizontal plates(m)

+m = 9.1*10^-31; // mass of electron in kg

+e = 1.6*10^-19; // velocity of electron in m/s

+// Max transit time is T/4

+

+// Solution

+

+V = sqrt(2*Va*e/m);

+Fc = V/(4*l);

+printf(' The velocity of electron = %.3f * 10^8 m/s \n',V*10^-8);

+printf('  The cutoff frequency = %.3f MHz \n',Fc/10^6);

diff --git a/3830/CH5/EX5.2/Ex5_2.sce b/3830/CH5/EX5.2/Ex5_2.sce
new file mode 100644
index 000000000..423ef15ad
--- /dev/null
+++ b/3830/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,12 @@
+// Exa 5.2

+

+clc;

+clear;

+

+// Given 

+

+// Teo magnetic coils

+

+// Solution

+

+printf('When the bullet passes through the first coil, a pulse is generated \n This is applied to an AND gate. The other input to the AND gate is from a crystal oscillator\n When the bullet passes through the second coil, another pulse is generated, which is used to stop counting; this is the disable pulse \n Therefore, the number of counts accelerated during this interval is a measure of the time taken by the bullet to traversea distance d between the coils \n Therefore, velocity = d/t');

diff --git a/3830/CH5/EX5.3/Ex5_3.sce b/3830/CH5/EX5.3/Ex5_3.sce
new file mode 100644
index 000000000..57f809943
--- /dev/null
+++ b/3830/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,29 @@
+// Exa 5.3

+

+clc;

+clear;

+

+// Given

+

+fs = 10000; // frequency of modulated signal(Hz)

+fm = 200*10^3; // modulation frequency(Hz)

+Ri = 10; // Input resistance(ohms)

+e2_by_e1 = 1.3; // limit for lowest frequency(in %)

+

+// Solution

+

+F_lower = fm - fs ; 

+

+printf(' For a double-section filter, \n  e2/e1 =  1/sqrt(1+(w*Rf*Cf)^2) \n');

+// Therefore,

+

+function y=f(x) 

+    y =(1/(sqrt(2*%pi*F_lower*x)^2+1))-e2_by_e1/100;

+endfunction

+[x,v,info] = fsolve(0,f);

+printf('  The product of Rf*Cf = %.4f  sec \n ',x);

+printf(' Let Rf = 10^5 Ohms, so that attenuation is 10:1. Therefore, Cf = ');

+Cf = x*10^-7;

+printf(' %.3f pf \n ',Cf*10^12);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.4/Ex5_4.sce b/3830/CH5/EX5.4/Ex5_4.sce
new file mode 100644
index 000000000..24ab7b7b6
--- /dev/null
+++ b/3830/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,17 @@
+// Exa 5.4

+

+clc;

+clear;

+

+// Given

+

+// The Lissajous pattern

+Y2 = 2.5; // slope of the major axis(in div)

+Y1 = 1.2; // slope of the vertical axis(in div)

+

+// Solution

+

+printf(' The phase shift V2 and V1 can be given as sin(Theta) = Y1/Y2 \n  -where V1 and V2 are voltages applied to X and Y axis respectively \n ');

+

+Theta = asind(Y1/Y2) ;

+printf(' Since, the ellipse is lying in the I and the III quadrant, \n  The angle is theta or 360-theta , i.e, %.2f or %.2f \n',Theta,360-Theta);

diff --git a/3830/CH5/EX5.5/Ex5_5.sce b/3830/CH5/EX5.5/Ex5_5.sce
new file mode 100644
index 000000000..309ea2dfb
--- /dev/null
+++ b/3830/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,31 @@
+// Exa 5.5

+

+clc;

+clear;

+

+// Given

+

+S = 0.6*10; // sensitivity of oscillograph in V per cm

+A = 2; // Area of oscilloscope area in cm^2

+dx = 4; // x-axis deflection in cm

+dy = 3; // y-axis deflection in cm

+

+// solution

+

+printf(' Ref fig. 5.5(a and b) -If Ic leads Vc by 90 degree, C will be a lossless ideal capacitor, and it will have infinite resistance R. Therefore, Ic is leading Vc by <90 degree . Theta is loss of the capacitor \n ');

+printf(' Power factor = cos(theta) = 1  when theta = o degree) \n');

+

+pf = 1;

+

+Vcondenser = (1/sqrt(2)) * S * dx*200; // since one-two thousandth od C voltage is applied to the x-plates

+Icondenser = (1/sqrt(2)) * S * 1/100000 ; // since Y-plates are impressed with voltage 100000 times the magniture of  condenser I.

+

+Pcondenser = Vcondenser * Icondenser;

+

+printf(' If p.f =1, the ellipse could have a major axis of %d cm and a minimum axis of %d cm \n',2*dx,2*dy);

+

+printf(' Total area = %.2f cm^2 \n',%pi/4 * 2*dx*2*dy);

+

+printf(' power loss of the capacitor = %.4f W \n',Pcondenser*A/(12/%pi));

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.6/Ex5_6.sce b/3830/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..91ea8c739
--- /dev/null
+++ b/3830/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,19 @@
+// Exa 5.6

+

+clc;

+clear;

+

+// Given

+

+// A stationary Lissajous pattern

+Vy = 6 ; // max value on vertical axis

+Vx = 5; // max value on horizontal axis

+fx = 1500; // horizontal input frequency(Hz)

+

+// Solution

+

+// fy/fx = No of pts the target meets per bottom(x-axis)/No of pts the target meets per bottom(y-axis)

+

+fy = (Vy/Vx)*fx;

+

+printf('The frequency of vertical axis = %d Hz \n',fy);

diff --git a/3830/CH5/EX5.7/Ex5_7.sce b/3830/CH5/EX5.7/Ex5_7.sce
new file mode 100644
index 000000000..e96248365
--- /dev/null
+++ b/3830/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,17 @@
+// Exa 5.7

+

+clc;

+clear;

+

+// Given

+

+b2 =2.5 ; // Max no of divisions on y-axis

+b1 = 1.25; // point of intersection on y-axis(div)

+

+// Solution

+

+printf(' Let theta be the phase angle of V2 w.r.t V1 where V1 and V2 are the voltages applied to x and y axis respectively \n');

+// Sin theta = b1/b2;

+Theta = asind(b1/b2);

+

+printf('  Therefore, the phase angle of V2 w.r.t V1 = %d degrees \n. But another possible value is(360-theta) i.e. %d degrees \n',Theta,360-Theta);

diff --git a/3830/CH5/EX5.8/Ex5_8.sce b/3830/CH5/EX5.8/Ex5_8.sce
new file mode 100644
index 000000000..e591eb0a6
--- /dev/null
+++ b/3830/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,26 @@
+// Exa 5.8

+

+clc;

+clear;

+

+// Given

+

+Va = 2000; // Anode voltage(Volts)

+Vd = 100; // Deflecting plates volage(Volts)

+a =1.5*10^-2; // axial length in m

+Sd= 30*10^-2; // screen distance in m

+Ld = 5*10^-2; // deflecting plates length in m

+

+// Solution

+

+// Let,

+ x = 1.76*10^11 ; // e/m ratio in c/kg

+L = Sd + Ld/2 ;

+D = (Ld*L*Vd)/(2*a*Va) ; // Deflection produced(m)

+Vo = sqrt(2*x*Va); // velocity of electrons in m/kg

+

+

+printf(' The deflection produced on screen = %.3f cm \n',D*100);

+printf('  The velocity of the electrons when they enter the field of the deflecting plates = %.4f * 10^7 m/kg \n', Vo/10^7);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH5/EX5.9/Ex5_9.sce b/3830/CH5/EX5.9/Ex5_9.sce
new file mode 100644
index 000000000..e9fab8f1b
--- /dev/null
+++ b/3830/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,17 @@
+// Exa 5.9

+

+clc;

+clear;

+

+// Given

+

+// A saw-tooth waveform is applied to an average diode voltmeter(Refer Fig. 5.24)

+

+printf('For a saw-tooth waveform, the rms value = Vm/T  \n -where Vm js max voltage value and T being the time period \n');

+

+printf(' Average value Va, 0.433 Vrms \n ');

+

+printf('Similarly, Iav = 0.433*Vrms/R \n');

+

+printf(' Error = 100 * (0.433*Vrms - (0.433/R)*0.45 / (0.45*(Vrms/R))) = -3.8 percent \n');

+printf(' The meter reading is 3.8 percent less than the expected value \n'); 

diff --git a/3830/CH6/EX6.1/Ex6_1.sce b/3830/CH6/EX6.1/Ex6_1.sce
new file mode 100644
index 000000000..7a6e42123
--- /dev/null
+++ b/3830/CH6/EX6.1/Ex6_1.sce
@@ -0,0 +1,40 @@
+// Exa 6.1

+

+clc;

+clear;

+

+// Given data

+

+// Refering bridge shown in fig. 6.8 

+R1 = 1000; // Ohms

+R2 = 4000; // Ohms

+R3 = 100; // Ohms

+R4 = 400; // Ohms

+Rg = 100; // Ohms

+Si = 100; // Sensitivity in mm/microAmp

+V = 3; // Voltage applied

+R4_imbalance = 1; // resistance added in R4 to create imbalance

+

+// Solution

+

+printf('The bridge is originally in balance. Therefore, R1/R3 = R2/R4 \n');

+printf('Let there be imbalance in the bridge circuit because of increase in value of R4 value by 1 Ohm \n');

+printf('Therefore, R4 = 400+X Ohms \n');

+printf('Thevenins Resistance  Rth = (100*1000)/(100+1000) + (4000*(400+X))/(4400+X) \n'); // Rth = R1*R3/(R1+R3) + R2*R4/(R2+R4)

+printf('Neglecting X \n');

+// Therefore

+Rth = R1*R3/(R1+R3) + R2*R4/(R2+R4);

+printf('Rth becomes %d ohms \n',round(Rth));

+printf('Eth = [R3/(R1+R3) + R4/(R2+R4)]*E; \n');

+// Applying binomial expansion and neglecting X2 term, X is small 

+// Therefore

+X = R4_imbalance;

+

+Eth =  V*10*X/48400;

+printf('Applying binomial expansion, Eth = %.2f µV \n',round(Eth*10^6));

+Ig = Eth/(Rth+Rg); // Galvanometer current

+D = Ig*Si; // Deflection in mm

+printf('Galvanometer Current Ig = %.3f µA \n', Ig*10^6);

+printf('Galvanometer deflection D = %.2f mm \n',D*10^6);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH6/EX6.2/Ex6_2.sce b/3830/CH6/EX6.2/Ex6_2.sce
new file mode 100644
index 000000000..2bd384dcb
--- /dev/null
+++ b/3830/CH6/EX6.2/Ex6_2.sce
@@ -0,0 +1,27 @@
+// Exa 6.2

+

+clc;

+clear;

+

+// Given

+

+//Fig. 6.9 shows wheatstone bridge

+R1 = 1000; // Ohms

+R2 = 100; // Ohms

+R3 = 400; // Ohms

+Rx = 41; // Ohms(Unknown resistance)

+V = 1.5; // Voltage supplied

+Rg = 50; // Galvanometer resistance (ohms)

+Si = 2; // current sensitivity in mm/microAmp

+

+

+// Solution

+

+Rth = (R1*R3/(R1+R3)) + R2*Rx/(R2+Rx);

+Eth = V*(R3/(R1+R3) - Rx/(R2+Rx));

+Ig = Eth/(Rth+Rg);

+d = Ig*Si;

+printf('The thevenins equivalent resistance = %.1f Ohms \n',round(Rth));

+printf(' The thevenins equivalent voltage = %.1f mV \n',abs(Eth*10^3));

+printf(' The current through the galvanometer = %.2f micro Amp \n',abs(Ig*10^6));

+printf(' The deflection produced by the galvanometer caused by the imbalance in the circuit = %.2f mm \n',abs(d*10^6));

diff --git a/3830/CH6/EX6.3/Ex6_3.sce b/3830/CH6/EX6.3/Ex6_3.sce
new file mode 100644
index 000000000..c73deb1f3
--- /dev/null
+++ b/3830/CH6/EX6.3/Ex6_3.sce
@@ -0,0 +1,32 @@
+// Exa 6.3

+

+clc;

+clear;

+

+// Given

+

+//Fig 6.41 shows an AC bridge

+R1 = 800; // Ohms

+C1 = 0.4; // microFarad

+R2 = 500; // Ohms

+C2 = 1.0; // microFarad

+R3 = 1200; // Ohms

+

+

+// Solution

+

+// Z = R + j X;

+// Z1 = 800 + j/(w*C1)

+// Y2 = 1/R2 - j*(w*C2)

+//Z3 = 1200

+

+printf('At balance, Z1/Z4 = Z2/Z3 \n');

+

+printf(' Rearranging the equation, Z4 = Z1*Z3*Y2 \n') ;

+printf(' Equating the real and imaginary parts on both sides, \n');

+Z4 = R1*R3*1/R2;

+w = sqrt(C1*C2);

+printf(' The value of R in arm DA to produce a balance = %d ohms \n',Z4);

+printf(' The value of frequency at balance = %.4f Hz \n',w/(2*%pi));

+

+// The answers given in textbook for R and f are incorrect

diff --git a/3830/CH6/EX6.4/Ex6_4.sce b/3830/CH6/EX6.4/Ex6_4.sce
new file mode 100644
index 000000000..87530ec6a
--- /dev/null
+++ b/3830/CH6/EX6.4/Ex6_4.sce
@@ -0,0 +1,23 @@
+// Exa 6.4

+

+clc;

+clear;

+

+// Given

+// Referring Fig 6.42 to get expression for unknowns Rs and Ls

+

+

+// Solution

+

+printf('It is a bridged-T network. At balance,Z1+Z3+ Z1*Z3/Z2 = 0 \n ');

+printf('Z1 = 1/jwC \n ');

+printf('Z3 = 1/jwC \n ');

+printf('Z2 = R \n ');

+printf('Z4 = Rs+jwLs \n ');

+

+printf('substituting these values in the equation, equating real and imagnary parts, and simplifying,\n ');

+

+printf('1/jwC + 1/jwC - 1/(w^2*C^2*R)+Rs+jwLs = 0 \n ');

+printf('Therefore \n ');

+printf('Rs = 1/(w^2*C^2*R) \n ');

+printf('wLs = 2/wc \n ');

diff --git a/3830/CH6/EX6.5/Ex6_5.sce b/3830/CH6/EX6.5/Ex6_5.sce
new file mode 100644
index 000000000..592acea07
--- /dev/null
+++ b/3830/CH6/EX6.5/Ex6_5.sce
@@ -0,0 +1,26 @@
+// Exa 6.5

+

+clc;

+clear;

+

+// Given

+

+// Referring Fig. 6.43

+

+

+// Solution

+

+printf('This is also a bridged-T network. This circuit is used to compare different coils, Lp and Rp. Using the general equation for a bridged-T netwrok at balance,\n ');

+

+printf('Z1+Z3+ Z1*Z3/Z2+Z4= 0 \n ');

+printf('Z1 = 1/jwC \n ');

+printf('Z3 = 1/jwC \n ');

+printf('Z2 = Rp+ 1/jwLp \n ');

+printf('Z4 = R \n ');

+

+printf('substituting these values in the equation, equating real and imagnary parts, and simplifying \n ');

+

+printf('1/jwC + 1/jwC - 1/(w^2*C^2*R)+Rs+jwLs = 0 \n ');

+printf('Therefore \n ');

+printf('w*Lp = 1/(2*w*C) \n ');

+printf('Rp = 1/(R*(w*C)^2) \n ');

diff --git a/3830/CH6/EX6.6/Ex6_6.sce b/3830/CH6/EX6.6/Ex6_6.sce
new file mode 100644
index 000000000..a24b642c7
--- /dev/null
+++ b/3830/CH6/EX6.6/Ex6_6.sce
@@ -0,0 +1,15 @@
+// Exa 6.6

+

+clc;

+clear;

+

+// Given

+

+// Fig 6.44 shows R.L.C bridge

+

+// Solution

+

+printf('For a given RLC circuit the expressions for as follows :- \n ');

+printf('Resistance, Rx = (R2*R3)/R1 \n ');

+printf('Inductance Lx = R2*R3*C \n ');

+printf('Capacitance Cx = (C*R3)/R1 \n ')

diff --git a/3830/CH7/EX7.1/Ex7_1.sce b/3830/CH7/EX7.1/Ex7_1.sce
new file mode 100644
index 000000000..fada4fe29
--- /dev/null
+++ b/3830/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,30 @@
+// Exa 7.1

+

+clc;

+clear;

+

+// Given

+

+NonLinearity = 1 ; // in percentage

+P = 5; //Power rating in Watts

+StepSize = 50; // in Ohms

+Rmin = 10 ; // in Ohms

+Rmax = 10000 ; // in Ohms

+

+// Solution

+

+printf('Max Error in linearity - Non-linearity = 1 percent \n');

+printf(' Therefore, Rp/Rm should be less than 0.1 \n');

+// If Rp/Rm < 0.1

+// per_Error = 15 * (Rp/Rm)

+// Therefore

+Rp = (1/15)*Rmax;

+printf(' If Rp/Rm < 0.1 \n Therefore we can choose a potentiometer with a total resistance Rp = %.2f Ohms at the maximum. Any value of Rp less than %.2f Ohms  would be all right as far as the non-linearity is concerned  \n',Rp,Rp);

+

+printf(' However, lower the value of Rp lower will be the sensitivity. Therefore we choose 650 Ohms potentiometer from the family, which will have maximum sensitivity and at the same time have non-linearity less than 10 percent  \n');

+Rp_selected = 650; // Ohms

+

+Max_Ecx = sqrt(P*Rp_selected);

+s = Max_Ecx/360; //Sensitivity

+

+printf(' The senstivity of potentiometer = %.2f V/degree \n',s);

diff --git a/3830/CH7/EX7.10/Ex7_10.sce b/3830/CH7/EX7.10/Ex7_10.sce
new file mode 100644
index 000000000..b6a5869e2
--- /dev/null
+++ b/3830/CH7/EX7.10/Ex7_10.sce
@@ -0,0 +1,18 @@
+// Exa 7.10

+

+clc;

+clear;

+

+// Given

+

+// A copper resistance thermometer

+

+R1 = 15; // resistance in ohms at 20 °c

+T1 = 20; // temperature in °c

+T2 = 175; // max temperature in °c

+Alpha_T = 0.00425; // temperature coefficient of resistance at 25°c

+

+// Solution

+

+R2 = R1*(1+Alpha_T*(T2-T1)); // resistance at 175 °c

+printf(' The limiting value of resistance  = %.2f ohms \n',R2);

diff --git a/3830/CH7/EX7.11/Ex7_11.sce b/3830/CH7/EX7.11/Ex7_11.sce
new file mode 100644
index 000000000..2bed1894d
--- /dev/null
+++ b/3830/CH7/EX7.11/Ex7_11.sce
@@ -0,0 +1,17 @@
+// Exa 7.11

+

+clc;

+clear;

+

+// Given

+

+// A thermistor

+R1 = 120; // resistance in ohms at 25 °c

+T1 = 25; // temperature in °c

+T2 = 40; // temperature in °c

+Alpha_T = -0.05; // temperature coefficient of resistance over range 25-50°c

+

+// Solution

+

+R2 = R1*(1+Alpha_T*(T2-T1)); // resistance at 175 °c

+printf(' The resistance of thermistor at 40 °c = %d ohms \n',R2);

diff --git a/3830/CH7/EX7.12/Ex7_12.sce b/3830/CH7/EX7.12/Ex7_12.sce
new file mode 100644
index 000000000..fdeb3d0da
--- /dev/null
+++ b/3830/CH7/EX7.12/Ex7_12.sce
@@ -0,0 +1,20 @@
+// Exa 7.12

+

+clc;

+clear;

+

+// Given

+

+// A variable inductive transducer

+L1 = 2.5; // inductance in mH

+N1 = 50; // No of effective turns at L1

+N2 = 52; // No of effective turns at L2

+

+// Solution

+

+printf(' Since L directly proportional to N^2 \n');

+printf('  L1/N1^2 = L2/N2^2 \n ');

+printf(' Therefore, L2 i.e, \n ');

+

+L2 = L1* (N2/N1)^2;

+printf(' The inductance of coil when the effective turns of the coil are 52 = %.2f mH \n',L2);

diff --git a/3830/CH7/EX7.13/Ex7_13.sce b/3830/CH7/EX7.13/Ex7_13.sce
new file mode 100644
index 000000000..efeb4b5e8
--- /dev/null
+++ b/3830/CH7/EX7.13/Ex7_13.sce
@@ -0,0 +1,21 @@
+// Exa 7.13

+

+clc;

+clear;

+

+// Given

+

+// A variable reluctance-type inductive transducer

+L1 = 5; // Inductance of transducer in mH

+lg1 = 1.5; // Length of iron piece in mm

+d = 0.025; // Distance by which irno piece is moved towards electro magnet (mm)

+

+// Solution

+

+air_gap = lg1-d;

+printf(' Length of air gap = %.3f mm \n',air_gap);

+New_Inductance = L1 + lg1/air_gap;

+

+printf('  The coil inductance becomes = %.2f mH \n',New_Inductance);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH7/EX7.14/Ex7_14.sce b/3830/CH7/EX7.14/Ex7_14.sce
new file mode 100644
index 000000000..5c065b801
--- /dev/null
+++ b/3830/CH7/EX7.14/Ex7_14.sce
@@ -0,0 +1,18 @@
+// Exa 7.14

+

+clc;

+clear;

+

+// Given

+

+// An LVDT

+vo = 2.6; // Output voltage(volts) of LVDT

+d = 0.4; // displacement in mm

+

+// Solution

+

+printf(' The sensitivity s = RMS value of output voltage/Displacement \n');

+

+S = vo/d; // sensitivity

+

+printf('  Therefore, s = %.1f V/mm \n',S); 

diff --git a/3830/CH7/EX7.15/Ex7_15.sce b/3830/CH7/EX7.15/Ex7_15.sce
new file mode 100644
index 000000000..cc8ccd65d
--- /dev/null
+++ b/3830/CH7/EX7.15/Ex7_15.sce
@@ -0,0 +1,16 @@
+// Exa 7.15

+

+clc;

+clear;

+

+// Given

+

+// An LVDT

+Vo = 1.25; // Output voltage

+Dmax  = 0.0025;// max. deviation of linearity

+L = 0.75; // weight of load in kgf

+

+// Solution

+

+Linearity = (Dmax/Vo)*100;

+printf(' The linearity at a given load 0.65/kgf = %.1f percent \n',Linearity);

diff --git a/3830/CH7/EX7.16/Ex7_16.sce b/3830/CH7/EX7.16/Ex7_16.sce
new file mode 100644
index 000000000..96c28ba51
--- /dev/null
+++ b/3830/CH7/EX7.16/Ex7_16.sce
@@ -0,0 +1,21 @@
+// Exa 7.16

+

+clc;

+clear;

+

+// Given

+

+// An LVDT

+vo = 5; // secondary voltage(volts) of LVDT

+d = 12.5; // displacement in mm

+d0 = 8; // displacement from central position in mm

+

+// Solution

+

+printf(' The sensitivity s = RMS value of output voltage/Displacement \n');

+

+S = vo/d; // sensitivity

+

+printf('  Therefore, s = %.1f V/mm \n',S);

+

+printf('  Output voltage for a displacement of 8mm from its central position =  %.1f V \n',S*d0);

diff --git a/3830/CH7/EX7.17/Ex7_17.sce b/3830/CH7/EX7.17/Ex7_17.sce
new file mode 100644
index 000000000..92f6861cd
--- /dev/null
+++ b/3830/CH7/EX7.17/Ex7_17.sce
@@ -0,0 +1,24 @@
+// Exa 7.17

+

+clc;

+clear;

+

+// Given

+

+// An LVDT to measure deflection of bellows

+S1 = 40; // sensitivity in V/mm

+d = 0.125; // displacement in mm

+P1 = 0.8*10^6; // pressure in N/m^2

+Vo2 = 3.5 ; // Output of LVDT for pressure P2

+

+// Solution

+

+// output voltage for the pressure p1

+Vo1 = S1*d; // in volts

+

+L_senstivity = Vo1/P1;

+

+// For P2 calculations when V = 3.5

+P2 = Vo2/L_senstivity;

+

+printf('The sensitivity of LVDT and pressure when the output voltage of LVDT is 3.5 V \n are  %.2f * 10^-6 V/N/m^2 and %.1f * 10^5 N/m^2 respectively \n',L_senstivity*10^6,P2*10^-5);

diff --git a/3830/CH7/EX7.18/Ex7_18.sce b/3830/CH7/EX7.18/Ex7_18.sce
new file mode 100644
index 000000000..18f2e35d2
--- /dev/null
+++ b/3830/CH7/EX7.18/Ex7_18.sce
@@ -0,0 +1,20 @@
+// Exa 17.18

+

+clc;

+clear;

+

+// Given

+

+// Capacitive Transducer

+d = 0.05; // plate separation in mm

+C = 5*10^-12; // Capacitence in farad

+dell_C = 0.75*10^-12; // change in capacitence in farad

+

+// Solution

+

+// C = e*A/d;

+eA = C*d;

+

+//Now,

+dell_x = eA/dell_C;

+printf('The displacment that caused a change in capacitence is %.3f mm \n',dell_x);

diff --git a/3830/CH7/EX7.19/Ex7_19.sce b/3830/CH7/EX7.19/Ex7_19.sce
new file mode 100644
index 000000000..7b66078f8
--- /dev/null
+++ b/3830/CH7/EX7.19/Ex7_19.sce
@@ -0,0 +1,25 @@
+// Exa 17.19

+

+clc;

+clear;

+

+// Given

+

+// A Capacitive Transducer

+d = 2.5; // plate separation in mm

+A = 600; // Area (in mm^2)

+P = 8*10^5; // Pressure applied in N/m^2

+x = 0.5; // deflection produced in mm

+C = 400*10^-12; // Capacitence in farad

+

+// Solution

+

+// Since, C = e*A/d

+ e =C*d/A;

+

+printf('Since we have to find capacitence when no pressure is applied. At that time plate separation = %d mm \n', d-x);

+

+d1 = d-x; // plate separation(mm) after pressure applied 

+C1 = e*A/d1;

+

+printf(' The value of capacitence, C with d = 2mm  = %d micro farad \n', C1*10^12);

diff --git a/3830/CH7/EX7.2/Ex7_2.sce b/3830/CH7/EX7.2/Ex7_2.sce
new file mode 100644
index 000000000..c5652c2c9
--- /dev/null
+++ b/3830/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,22 @@
+// Exa 7.2

+

+clc;

+clear;

+

+// Given

+

+l = 50; // length of potentiometer in mm

+R = 5000; // Total resistance of potentiometer in Ohms

+Rt = 1850; // Resistance of potentiometer in Ohms

+

+// Solution

+

+R_length = R/l ; // Resistance per unit length

+R_normal = R/(l*10^-3*0.5);

+printf(' Resistance of normal position = %d Ohms \n',R_normal);

+R_change = R_normal - Rt;

+printf('  Change in resistance = %d Ohms \n',R_change);

+Displacement = R_change/R_length ;

+printf('  The linear displacement when the resistance of the potentiometer is 1850 ohms = %.2f mm \n',Displacement);

+

+// The answer provided in the textbook is wrong

diff --git a/3830/CH7/EX7.20/Ex7_20.sce b/3830/CH7/EX7.20/Ex7_20.sce
new file mode 100644
index 000000000..0f1e70d5b
--- /dev/null
+++ b/3830/CH7/EX7.20/Ex7_20.sce
@@ -0,0 +1,22 @@
+// Exa 7.20

+

+clc;

+clear;

+

+// Given

+

+// A Capacitence Transducer

+A = 5*10^-4; // Area in m^2

+C = 9.5*10^-12; // Capacitence in farad

+er = 81; //  Relative dielectric constant

+e0 = 8.854*10^-12; // Absolute dielectric constant in F/m

+

+// Solution

+

+// C = e0*er*A/d;

+// Therefore

+d = e0*er*A/C;

+printf('The plate separation d = %.2f mm \n',d*10^3);

+S = e0*er*A/d^2;

+

+printf(' Sensitivity s = %.3f * 10^-8 F/m \n',S*10^8);

diff --git a/3830/CH7/EX7.21/Ex7_21.sce b/3830/CH7/EX7.21/Ex7_21.sce
new file mode 100644
index 000000000..1b91649a4
--- /dev/null
+++ b/3830/CH7/EX7.21/Ex7_21.sce
@@ -0,0 +1,25 @@
+// Exa 7.21

+

+clc;

+clear;

+

+// Given

+

+// A 5-plate transducer

+n = 5; // no of plates

+l = 20*10^-3; // length of plate in m

+b = 20*10^-3; // breadth of plate in m

+d = 0.25*10^-3; // separation between plates in m

+

+// Solution

+

+A = l*b; // Area in mm^2

+er = 1; //  Relative dielectric constant

+e0 = 8.854*10^-12; // Absolute dielectric constant in F/m

+

+S = (n-1)*e0*er*A/d^2;

+

+printf('Sensitivity of the arrangement = %.3f * 10^-9 F/m \n',S*10^9);

+

+// The answer provided in the textbook is wrong

+

diff --git a/3830/CH7/EX7.3/Ex7_3.sce b/3830/CH7/EX7.3/Ex7_3.sce
new file mode 100644
index 000000000..437a8cc8e
--- /dev/null
+++ b/3830/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,15 @@
+// Exa 7.3

+

+clc;

+clear;

+

+// Given

+

+N = 50; // No of turns of potentiometer per mm

+Number_of_Resolution = 4; // No of resolutions of potentiometer

+

+// Solution

+

+Resolution = 1/N;

+printf(' Resolution of potentiometer = %.3f mm \n',Resolution);

+printf('  4 resolutions of potentiometer with one rotation = %.1f mm \n',10^3*Resolution/Number_of_Resolution);

diff --git a/3830/CH7/EX7.4/Ex7_4.sce b/3830/CH7/EX7.4/Ex7_4.sce
new file mode 100644
index 000000000..2822cd474
--- /dev/null
+++ b/3830/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,13 @@
+// Exa 7.4

+

+clc;

+clear;

+

+// Given

+

+G = 3.8; //  Gauge factor

+

+// Solution

+

+P = (G-1)/2;

+printf(' Poissons ratio of thin circular/wire of soft iron = %.1f \n',P);

diff --git a/3830/CH7/EX7.5/Ex7_5.sce b/3830/CH7/EX7.5/Ex7_5.sce
new file mode 100644
index 000000000..1efe69c70
--- /dev/null
+++ b/3830/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,18 @@
+// Exa 7.5

+

+clc;

+clear;

+

+// Given

+

+L = 0.1 ; // Initial length of wire in m

+R = 120; // Initial resistance of wire in ohms

+delta_L =  0.1*10^-3;// change in length of wire in m

+delta_R = 0.21; // change in resistance of wire in ohms

+

+// Solution

+

+e = delta_L/L;

+G = (delta_R/R)/e;

+

+printf(' The gauge factor of device = %.2f \n',G);

diff --git a/3830/CH7/EX7.6/Ex7_6.sce b/3830/CH7/EX7.6/Ex7_6.sce
new file mode 100644
index 000000000..6c1643d7a
--- /dev/null
+++ b/3830/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,19 @@
+// Exa 7.6

+

+clc;

+clear;

+

+// Given

+

+S = 1400; // Stress in Kgf/cm^2

+E = 2.1*10^6; // Youngs Modulus in Kgf/cm^2

+G = 2; // Gauge factor

+

+// Solution

+

+e = S/E;

+change_in_R = G*e;

+

+printf(' Percentage change in resistance of strain gauge = %.3f \n',change_in_R*100);

+

+// The answer provided in the textbook vary due to round off

diff --git a/3830/CH7/EX7.7/Ex7_7.sce b/3830/CH7/EX7.7/Ex7_7.sce
new file mode 100644
index 000000000..108673992
--- /dev/null
+++ b/3830/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,22 @@
+// Exa 7.7

+

+clc;

+clear;

+

+// Given

+

+Gf = 2 ; // Gauge factor of strain gauge

+S = 1000; // Stress in kg/cm^2

+E = 2*10^6; // Youngs Modulus in kg/cm^2

+

+// Solution

+

+e = S/E; // strain

+

+dR_R = e*Gf; // change in resistance

+ // Gf = 1+2u;

+// Therefore

+u = (Gf-1)/2; // poissons ratio

+

+printf('The percentage change in resistance of strain gauge = %.1f \n',dR_R*100);

+printf(' Poissons ratio  = %.2f \n',u);

diff --git a/3830/CH7/EX7.8/Ex7_8.sce b/3830/CH7/EX7.8/Ex7_8.sce
new file mode 100644
index 000000000..1bdb07e06
--- /dev/null
+++ b/3830/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,30 @@
+// Exa 7.8

+

+clc;

+clear;

+

+R = 200; // strain gauge resistance in Ohms

+G = 2.5; // Gauge factor

+RL = 400; // load resistance in Ohms

+V = 24; // input voltage in volts

+S = 140; // applied stress in mgf/m^2

+Y = 200; // Modulus of elasticity in GN/m^2

+

+// Solution

+

+V_normal = V*(R/(R+RL));

+

+printf('Voltage across strain gauge = %d V \n',V_normal);

+e = (S*10^-3)/Y;

+// Strain e =   dell_L/L

+//dell_R/R = G* dell_L/L;

+// so,

+dell_R = R*G*e;

+

+

+//strain gauge under strained condition

+V_strained = (R+dell_R) * V/(R+dell_R+RL);

+printf(' Voltage across strain gauge under strained condition = %.4f ohms \n',V_strained);

+

+dif = V_normal - V_strained;

+printf(' Change in output voltage = %.2f mV \n',abs(dif*10^3));

diff --git a/3830/CH7/EX7.9/Ex7_9.sce b/3830/CH7/EX7.9/Ex7_9.sce
new file mode 100644
index 000000000..0641c8d9f
--- /dev/null
+++ b/3830/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,26 @@
+// Exa 7.9

+

+clc;

+clear;

+

+// Given

+

+// A platinum resitance thermometer

+R1 = 120; // resistance in ohms at 25 °c

+T1 = 25; // temperature in °c

+T2 = 75; // temperature in °c

+Alpha_T = 0.00392; // temperature coefficient of resistance at 25°c

+R3 = 180; // resistance in ohms at unknown temp T3

+

+// Solution

+

+R2 = R1*(1+Alpha_T*(T2-T1)); // resistance at 75 °c

+printf(' The resistance at 75 °c = %.2f ohms \n',R2);

+

+// now, to get T3 corresponding to R3= 180 ohms

+

+// R3 = R2*(1+Alpha_T*(T3-T1));

+// Rearranging above equation to get T3 as 

+T3 = (R3/R1 -1)/Alpha_T + T1;

+

+printf(' The temperature corresponding to resistance 180 ohms = %.2f °c \n',T3);