diff options
Diffstat (limited to '3554')
73 files changed, 1674 insertions, 0 deletions
diff --git a/3554/CH1/EX1.1/Ex1_1.sce b/3554/CH1/EX1.1/Ex1_1.sce new file mode 100644 index 000000000..4f94dba7e --- /dev/null +++ b/3554/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,15 @@ +//Exa 1.1a
+
+clc;
+clear all;
+
+Yn=80;//voltage across a resistor(Volts)
+Xn=79;//Measured voltage (Volts)
+
+//solution
+e=Yn-Xn; //absolute error
+Pe=(Yn-Xn)/Yn *100;//% error
+A=1-abs((Yn-Xn)/Yn); //relative accuracy
+a=100*A;
+printf('Absolute Error = %d V \n Percentage Error = %.2f percent\n Relative accuracy = %.4f \n Percentage of accuracy = %0.2f percent \n',e,Pe,A,a);
+disp("");
diff --git a/3554/CH1/EX1.2/Ex1_2.sce b/3554/CH1/EX1.2/Ex1_2.sce new file mode 100644 index 000000000..777c954b6 --- /dev/null +++ b/3554/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,13 @@ +//Exa 1.2
+
+clc;
+clear all;
+
+//Refering to table 1.1- Set of 10 measurements that were recorded in the laboratory.
+
+X={98;101;102;97;101;100;103;98;106;99}; //From table 1.1
+
+//solution
+X_n= mean(X); //Average value
+Prec=1-abs((X(6)-X_n)/X_n);//precision of 6th reading
+printf('The precision of 6th measurement = %0.3f \n',Prec);
diff --git a/3554/CH1/EX1.3/Ex1_3.sce b/3554/CH1/EX1.3/Ex1_3.sce new file mode 100644 index 000000000..97002fa04 --- /dev/null +++ b/3554/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,39 @@ +// Exa 1.3a
+
+clc;
+clear all;
+
+Sv=1000; //voltmeter sensitivity(ohm/V)
+Vt=80; //Voltage across unknown resistance (V)
+It=10; //Current through unknown resistance (mA)
+Scale=150; //Volts
+
+//solution
+
+//Neglecting milliammeter resistance
+Rt=Vt/It; //Total circuit resistance(K ohm)
+Rv=Sv*Scale/1000; //Voltmeter resistance(K ohm/V)
+Rx=Rt*Rv/(Rv-Rt); //actual value of unknown resistance(K ohm)
+err=(Rx-Rt)/Rx *100;
+printf('Apparent value of resistance = %d K ohm \n Actual value of resistance = %.2f K ohm \n Percentage error = %.1f percent \n',Rt,Rx,err);
+disp("");
+
+// Exa 1.3b
+
+Sv=1000; //voltmeter sensitivity(ohm/V)
+Vt=30; //Voltage across unknown resistance (V)
+It=600; //Current through unknown resistance (mA)
+Scale=150; //Volts
+
+//solution
+
+//Neglecting milliammeter resistance
+Rt=Vt/(It*10^-3); //Total circuit resistance(ohm)
+Rv=Sv*Scale; //Voltmeter resistance(ohm/V)
+Rx=Rt*Rv/(Rv-Rt); //actual value of unknown resistance(ohm)
+err=(Rx-Rt)/Rx *100;
+printf('Apparent value of resistance = %d ohm \n Actual value of resistance = %.3f ohm \n Percentage error = %.3f \n',Rt,Rx,err);
+disp("In Example1.3a, a well calibrated voltmeter may give a misleading resistance when connected across two points in a high resistance circuit.")
+disp("The same voltmeter, when connected in a low resistance circuit(Examole 1.3b) may give a more dependable reading. This shows that voltmeters have a loading effect in the circuit during measurement.");
+// In the 1.3b example, the answer mentioned in the textbook for Rx and percent error is isncorrect.
+
diff --git a/3554/CH1/EX1.4/Ex1_4.sce b/3554/CH1/EX1.4/Ex1_4.sce new file mode 100644 index 000000000..c22cf88c6 --- /dev/null +++ b/3554/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,31 @@ +// Exa 1.4
+
+clc;
+clear all;
+
+// Given data
+x1= 49.7;
+x2= 50.1;
+x3= 50.2;
+x4= 49.6;
+x5= 49.7;
+
+// solution
+
+X_mean= (x1+x2+x3+x4+x5)/5; // Arithmatic mean
+
+d1= x1-X_mean;
+d2= x2-X_mean; // deviation from each value
+d3= x3-X_mean;
+d4=x4-X_mean;
+d5=x5-X_mean;
+
+d_total= d1+d2+d3+d4+d5; //Algebraic sum of deviations
+
+printf('The arithmatic mean is %.2f \n \n',X_mean);
+printf(' Deviation from x1 is %.2f \n ',d1);
+printf('Deviation from x2 is %.2f \n ',d2);
+printf('Deviation from x3 is %.2f \n ',d3);
+printf('Deviation from x4 is %.2f \n ',d4);
+printf('Deviation from x5 is %.2f \n \n',d5);
+printf(' The algebraic sum of deviation is %d \n',d_total);
diff --git a/3554/CH1/EX1.5/Ex1_5.sce b/3554/CH1/EX1.5/Ex1_5.sce new file mode 100644 index 000000000..7c323e56a --- /dev/null +++ b/3554/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,26 @@ +// Exa 1.5
+//Data taken from Exa 1.4 as stated
+
+clc;
+clear all;
+
+// Given data
+
+x1= 49.7;
+x2= 50.1;
+x3= 50.2;
+x4= 49.6;
+x5= 49.7;
+n= 5; // number of x values
+
+// solution
+
+X_mean= (x1+x2+x3+x4+x5)/5; // Arithmatic Mean
+d1= x1-X_mean;
+d2= x2-X_mean; // deviation from each value
+d3= x3-X_mean;
+d4=x4-X_mean;
+d5=x5-X_mean;
+
+D_av= (abs(d1)+abs(d2)+abs(d3)+abs(d4)+abs(d5))/n; //Average deviation
+printf('The average deviation = %.3f \n',D_av);
diff --git a/3554/CH1/EX1.6/Ex1_6.sce b/3554/CH1/EX1.6/Ex1_6.sce new file mode 100644 index 000000000..1b18337d9 --- /dev/null +++ b/3554/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,27 @@ +// Exa 1.6
+//Data taken from Eg 1.4 as stated
+
+clc;
+clear all;
+
+// Given data
+
+x1= 49.7;
+x2= 50.1;
+x3= 50.2;
+x4= 49.6;
+x5= 49.7;
+n= 5; // number of x values
+
+// solution
+
+X_mean= (x1+x2+x3+x4+x5)/5; // Arithmatic Mean
+d1= x1-X_mean;
+d2= x2-X_mean; // deviation from each value
+d3= x3-X_mean;
+d4=x4-X_mean;
+d5=x5-X_mean;
+
+Std_dev= sqrt((d1^2+d2^2+d3^2+d4^2+d5^2)/(n-1)); //Standard deviation
+printf('The standard deviation = %.2f \n',Std_dev);
+
diff --git a/3554/CH1/EX1.7/Ex1_7.sce b/3554/CH1/EX1.7/Ex1_7.sce new file mode 100644 index 000000000..e466deaba --- /dev/null +++ b/3554/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,19 @@ +// Exa 1.7
+
+clc;
+clear all;
+
+// Given data
+
+Range= 600; //volgmeter range(volts)
+Accu= 0.02; //Accuracy
+X= 250; //voltage to be measured(volts)
+
+// Solution
+
+Mag= Accu * Range; //magnitude of limiting error
+X_mag = Mag/X * 100; // limiting error at 250V inpercentag
+
+printf('Limiting error when instrument is used to measure at 250V = %.1f percentage \n',X_mag);
+
+
diff --git a/3554/CH1/EX1.8/Ex1_8.sce b/3554/CH1/EX1.8/Ex1_8.sce new file mode 100644 index 000000000..9662b2138 --- /dev/null +++ b/3554/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,23 @@ +// Exa 1.8
+
+clc;
+clear all;
+
+// Given data
+
+X= 100; // Range of voltmeter(V)
+x= 70; // Measured value on voltmeter(V)
+Y= 150; // Range of milliammeter
+y= 80; // Measurex d value on milliammeter
+Accu= 0.015; // Accuracy of instruments
+
+// Solution
+
+X_mag= Accu*X; //Magnitude of limiting error for voltmeter
+Y_mag= Accu*Y; // Magnitude of limiting error for milliammeter
+x_mag= X_mag/x; // limiting error at 70V
+y_mag= Y_mag/y; // limiting error at 80mA
+
+disp("Limiting error for the power calculation is the sum of the individual limiting errors involved");
+printf(' Therefore, limiting error = %.3f percentage \n',(x_mag+y_mag)*100);
+// The answer vary due to round off error
diff --git a/3554/CH10/EX10.1/Ex10_1.sce b/3554/CH10/EX10.1/Ex10_1.sce new file mode 100644 index 000000000..1f381d119 --- /dev/null +++ b/3554/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,24 @@ +// Exa 10.1
+
+clc;
+clear all;
+
+// Given data
+
+// 1st measurement
+f1=1; // in MHZ
+C1=500; // in pf
+// 2nd measurement
+f2=2; //in MHz
+C2=110; // in pf
+
+// Solution
+// Using equation 10.2(page no. 278) to calculate distributed Capacitance
+
+Cs=(C1-4*C2)/3; // Distributed capacitance in pf
+printf('The value of distributed capacitance = %d pf \n',Cs);
+// using equation of resonant frequency given as f1=1/(2*%pi*sqrt(L*(C1+Cs));
+// Therefore
+L=1/(4*(%pi)^2*f1^2*(C1+Cs)); // Inductor value
+
+printf(' The value of L(inductor) is =%.3f micro H \n',L*10^6);
diff --git a/3554/CH10/EX10.2/Ex10_2.sce b/3554/CH10/EX10.2/Ex10_2.sce new file mode 100644 index 000000000..227a1bcd5 --- /dev/null +++ b/3554/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,24 @@ +// Exa 10.2
+
+clc;
+clear all;
+
+// Given data
+
+f1=2; // in MHz
+f2=6; // in MHz
+C1=500; // in pf
+C2=50; // in pf
+
+// Solution
+
+disp("Given that f2=3*f1");
+disp("Therefore by using equation 10.1");
+disp(" 1/(2*%pi*sqrt(L*(C2+Cs)) = 3/(2*%pi*sqrt(L*(C1+Cs))");
+disp("Therefore");
+disp("C1+Cs=9(C2+Cs)");
+//Therefore Cs is given as
+Cs=(C1-9*C2)/8;// Self capacitance in pf
+printf(' \nThe value of the self capacitance is = %.2f pf \n', Cs);
+
+
diff --git a/3554/CH11/EX11.1/Ex11_1.sce b/3554/CH11/EX11.1/Ex11_1.sce new file mode 100644 index 000000000..973ab4783 --- /dev/null +++ b/3554/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,17 @@ +// Exa 11.1
+
+clc;
+clear all;
+
+// Given data
+
+// Wheatstone's bridge circuit
+R1=10; // k Ohms
+R2=15; // k Ohms
+R3=40; // k Ohms
+
+// Solution
+// From the equation (11.4) of balanced bridge we have
+
+Rx=R2*R3/R1; // Unknown resistance Rx
+printf(' The unknown resistance Rx is = %d k Ohms \n',Rx);
diff --git a/3554/CH11/EX11.10/Ex11_10.sce b/3554/CH11/EX11.10/Ex11_10.sce new file mode 100644 index 000000000..458d8f06a --- /dev/null +++ b/3554/CH11/EX11.10/Ex11_10.sce @@ -0,0 +1,23 @@ +// Exa 11.10
+
+clc;
+clear all;
+
+// Given data
+
+// Wien's bridge
+R1=3.1; // k Ohms
+C1=5.2; // micro farads
+R2=25; // k Ohms
+f=2.5; // kHz
+R4=100;// k Ohms
+
+// Solution
+
+w=2*%pi*f; // Angular frequency
+// Substituting the value of C3 from Eq. 11.22(page no. 330) in Eq.11.21(pagr no. 330) to get value of R3 as follows
+R3=R4/R2 *(R1+1/(w^2*R1*C1^2));
+// Also we can get C3 from Eq. 11.22(page no. 330)
+C3=1/(w^2*C1*R1*R3);
+printf(' The parallel resistance of %.1f K ohms and capacitance of %.1f pf\n causes a Wien bridge to null with values of given component values. \n',R3,C3*10^6);
+
diff --git a/3554/CH11/EX11.2/Ex11_2.sce b/3554/CH11/EX11.2/Ex11_2.sce new file mode 100644 index 000000000..aa19bc4ee --- /dev/null +++ b/3554/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,24 @@ +// Exa 11.2
+
+clc;
+clear all;
+
+// Given data
+//Refering fig. 11.5 - Unbalanced Wheatstone bridge
+
+R1=1; // in k Ohms
+R2= 2.5; // in k Ohms
+R3=3.5; // in k Ohms
+R4=10; // in k Ohms
+V= 6; // Applied Voltage(V)
+Rg=0.3; // Galvanometer resistance in k Ohms
+
+// Solution
+
+// Eth=Ea-Eb ; \\ Thevenin's equivalent voltage
+Eth=V*(R4/(R2+R4) - R3/(R1+R3));
+Rth=(R1*R3/(R1+R3)) + (R2*R4/(R2+R4)) ;
+// Refering the equivalent circuit connected along with the galvanometer as shown in fig. 11.6
+Ig=Eth/(Rth+Rg) ; // Current through galvanometer
+printf(' The current through galvanometer is = %.2f micro Amp \n',round(Ig*10^3));
+//The answer vary due to round off error
diff --git a/3554/CH11/EX11.3/Ex11_3.sce b/3554/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..14520fe33 --- /dev/null +++ b/3554/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,19 @@ +// Exa 11.3
+
+clc;
+clear all;
+
+// Given data
+
+// Refering Fig. 11.9(page no.311) - slightly unbalanced Wheatstone bridge
+R= 700; // in Ohms
+Dell_R= 35; // in Ohms
+E=10; // Supplied voltage(V)
+Rg=125;//Internal resistance of galvanometer(Ohms)
+
+// Solution
+
+Eth= E*Dell_R/(4*R) ; // Thevenin's equivalent voltage(V)
+Rth=R; // Thevenin's equivalent resistance(Ohms)
+Ig= Eth/(Rth+Rg); // Current through galvanometer(Amp)
+printf(' The current through galvanometer by the approximation method is %.1f micro Amp \n',Ig*10^6);
diff --git a/3554/CH11/EX11.4/Ex11_4.sce b/3554/CH11/EX11.4/Ex11_4.sce new file mode 100644 index 000000000..240568637 --- /dev/null +++ b/3554/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,19 @@ +// Exa 11.4
+
+clc;
+clear all;
+
+// Given data
+// Refering Fig. 11.12(page no.315)- Kelvin's bridge
+
+Ra_b=1000;// The ratio of Ra to Rb
+R1= 5; // in Ohms
+
+// Solution
+
+// We have R1=0.5*R2
+R2=R1/0.5;
+
+//From the eqation for Kelvin'd bridge- Rx*Ra=Rb*R2
+Rx=R2*(1/1000); // Unknown resistance
+printf(' The value of Rx = %.2f Ohm \n ',Rx);
diff --git a/3554/CH11/EX11.5/Ex11_5.sce b/3554/CH11/EX11.5/Ex11_5.sce new file mode 100644 index 000000000..8d39131b8 --- /dev/null +++ b/3554/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,24 @@ +// Exa 11.5
+
+clc;
+clear all;
+
+// Given data
+// Refering circuit in Fig. 11.15(a) and graph in 11.15(b) on page no.317
+
+R1=5; // k Ohms
+R2=5; //k Ohms
+R3= 5; // k Ohms
+E=6; // Applied voltage(V)
+
+// Solution
+
+// The value of Rv when bridge is balanced is calculated as
+Rv=R2*R3/R1;
+printf(' The value of Rv = %d K Ohms \n' , Rv);
+disp(" From the graph, the temperature at which bridge is balanced is = 80 degree celsius");
+disp(" From the graph, the resistance Rv for balancing bridge at 60 degree celcius comes out to be 4.5 k Ohms ");
+// Therefore
+Rv1=4.5; // Resistance Rv at 60 degree celcius(K ohms)
+es=E*(R3/(R1+R3) - Rv1/(R2+Rv1) ); // Error signal
+printf(' The amplitude of error signal at 60 degree celsius is = %.3f V \n',es);
diff --git a/3554/CH11/EX11.6/Ex11_6.sce b/3554/CH11/EX11.6/Ex11_6.sce new file mode 100644 index 000000000..fb39d7b71 --- /dev/null +++ b/3554/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,20 @@ +// Exa 11.6
+
+clc;
+clear all;
+
+// Given data
+f=2; // kHz
+C3=100; // micro farads
+R1=10; // k Ohms
+R2=50; // k Ohms
+R3=100; // k Ohms
+
+// Solution
+
+// Using equations 11.12(a) and 11.12(b) (page no. 321)to find values of Rx and Cx
+
+Rx=R2*R3/R1;
+Cx=R1/R2 *C3;
+
+printf(' The equivalent circuit consist of resistance Rx of %d K ohms \n in series with a capacitor Cx of %d micro farads \n',Rx,Cx);
diff --git a/3554/CH11/EX11.7/Ex11_7.sce b/3554/CH11/EX11.7/Ex11_7.sce new file mode 100644 index 000000000..f9c3a66ba --- /dev/null +++ b/3554/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,19 @@ +// Exa 11.7
+
+clc;
+clear all;
+
+// Given data
+
+C1=0.01; // micro farads
+R1=470; // k Ohms
+R2=5.1; // k Ohms
+R3=100; // k Ohms
+
+// Solution
+// Using equation 11.15 given on page no. 324 to find Rx and Lx
+
+Rx=R2*R3/R1;
+Lx=R2*R3*C1;
+
+printf(' The series equivalent of the unknown impedence consist of series combination\n of Rx = %.2f k Ohms and Lx= %.1f H \n' , Rx, Lx);
diff --git a/3554/CH11/EX11.8/Ex11_8.sce b/3554/CH11/EX11.8/Ex11_8.sce new file mode 100644 index 000000000..b9511ba11 --- /dev/null +++ b/3554/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,21 @@ +// Exa 11.8
+
+ clc;
+clear all;
+
+// Given data
+
+w=3000; // Angular frequency in rad/s
+R2=10*10^3; // Ohms
+R1= 2*10^3; // Ohms
+C1=1*10^-6; // farads
+R3=1*10^3; // Ohms
+
+// Solution
+
+// Using equations 11.19 and 11.18 (page no.326)to find values of Rx and Lx
+
+Rx=w^2*R1*R2*R3*C1^2/(1+w^2*R1^2*C1^2);
+Lx=R2*R3*C1/(1+w^2*R1^2*C1^2);
+
+printf(' The series equivalent inductance and resistance of the network consist of\n Rx of %.2f k Ohms and Lx of %d mH \n',Rx/1000,Lx*10^3);
diff --git a/3554/CH11/EX11.9/Ex11_9.sce b/3554/CH11/EX11.9/Ex11_9.sce new file mode 100644 index 000000000..1d9f63b6a --- /dev/null +++ b/3554/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,26 @@ +// Exa 11.9
+
+clc;
+clear all;
+
+// Given data
+
+// Refering Fig. 11.26(page no.328) - an AC bridge(SCHERING'S BRIDGE)
+
+R1= 1; // k Ohms
+C1=0.5; // micro farads
+R2=2; // k Ohms
+C3=0.5; // micro farads
+f= 1000; // Hz
+
+// Solution
+// Using Equations 11.20(a) and 11.20(b) given on page no. 328 we get value Rx and Cx
+
+Rx=C1/C3*R2;// in k Ohms
+Cx=R1/R2 * C3; // in micro farads
+
+D=2*%pi*f*Cx*10^-6*Rx*10^3; // Dissipation factor
+
+printf(' The unknown capacitance Cx is equal to %.2f micro farads\n ',Cx);
+printf(' The dissipation factor = %.4f \n ',D);
+
diff --git a/3554/CH12/EX12.1/Ex12_1.sce b/3554/CH12/EX12.1/Ex12_1.sce new file mode 100644 index 000000000..820e88065 --- /dev/null +++ b/3554/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,15 @@ +// Exa 12.1
+
+clc;
+clear all;
+
+// Given data
+
+Chart_speed=40; // in mm/sec
+Time_base=5; // in mm
+
+// Solution
+
+Period= Time_base/Chart_speed;
+frequ=1/Period;
+printf(' The frequency of the signal = %d cycles/sec \n',frequ);
diff --git a/3554/CH12/EX12.2/Ex12_2.sce b/3554/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..42d6ef687 --- /dev/null +++ b/3554/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,17 @@ +// Exa 12.2
+
+clc;
+clear all;
+
+// Given data
+
+fre=20; // in Hz
+Time_base=5*10^-3; // in
+
+// Solution
+
+Period=1/fre; // in sec
+// Since period= timebase/ chart speed;
+Chart_speed=Time_base/Period; // in mm/sec
+
+printf(' The chart speed used to record one complete cycle on 5mm of recording paper =%.1f mm/sec \n',Chart_speed*10^3);
diff --git a/3554/CH13/EX13.1/Ex13_1.sce b/3554/CH13/EX13.1/Ex13_1.sce new file mode 100644 index 000000000..eeb9f7f5c --- /dev/null +++ b/3554/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,20 @@ +// Exa 13.1
+
+clc;
+clear all;
+
+// Given data
+// Refer circuit given in Fig no.13.2(b) given on page no.381
+
+Shaft=3; // Shaft stroke in inches
+Wiper=0.9;// in inches
+R=5; // Total resistance(R1+R2) in K ohms
+Vt=5; // Applied voltage in volts
+
+// Solution
+
+R2=Wiper/Shaft * R ;// in k Ohms
+// Since Vo/Vt=R2/(R1+R2);
+// Therefore
+Vo=R2/(R) *Vt; // Output Voltage(R1+R2)
+printf(' The output voltage = %.1f V \n',Vo);
diff --git a/3554/CH13/EX13.2/Ex13_2.sce b/3554/CH13/EX13.2/Ex13_2.sce new file mode 100644 index 000000000..64174d9e4 --- /dev/null +++ b/3554/CH13/EX13.2/Ex13_2.sce @@ -0,0 +1,22 @@ +// Exa 13.2 + +clc; +clear; + +// Given data +Ra=5; // (R1+R2) in k Ohms +Rb=5;// (R3+R4) in k Ohms +Vt=5; // Applied voltage (V) +Shaft=5; // Shaft distance in inches + +// Solution + +disp(" As given, wiper moves 0.5 inch towards A from the centre, it will have moved 3 inches from B"); +Wiper=3; // Wiper movement from B in inches +Wiper1=2.5;//Wiper movement from A in inches +R2=Wiper/Shaft * R; // in k Ohms +R4=Wiper1/Shaft * R; // in k Ohms +//Ve=VR2-VR4 +Vc=(R2/(Ra)) *Vt - (R4/(Rb)) * Vt; + +printf(' The new value of Vc= %.1f V \n',Vc);
\ No newline at end of file diff --git a/3554/CH13/EX13.3/Ex13_3.sce b/3554/CH13/EX13.3/Ex13_3.sce new file mode 100644 index 000000000..e61b81efd --- /dev/null +++ b/3554/CH13/EX13.3/Ex13_3.sce @@ -0,0 +1,16 @@ +// Exa 13.3
+
+clc;
+clear all;
+
+// Given data
+
+K=2; // Gauge factor
+strain=1*10^-6; // Ratio of change in length to original length
+R=130; // Resistance in Ohms
+
+// Solution
+
+// As K = ratio of dell R/R to dell L/L
+Dell_R =K*R*strain ; // Change in resistance
+printf(' The change in resistance = %d micro Ohms \n',Dell_R*10^6);
diff --git a/3554/CH13/EX13.4/Ex13_4.sce b/3554/CH13/EX13.4/Ex13_4.sce new file mode 100644 index 000000000..6c1a6e03b --- /dev/null +++ b/3554/CH13/EX13.4/Ex13_4.sce @@ -0,0 +1,22 @@ +// Exa 13.4
+
+clc;
+clear all;
+
+// Given data
+
+R= 4;// Resistance of thermistor in k Ohms
+R1=0.003;// Meter resistance in k Ohms
+Rc=0.017; // in k Ohms
+Vt=15;// in Volts
+
+// Solution
+
+// From fig. 13.2(b)- graph of Temp vs Resistance we find that,thermistor resistance at 25 degree Celsius is 4 K ohms and at 65.5556 degree Celsius it is 950 ohms.
+R_25= 4;// in k Ohms
+R_65=0.95; // in k Ohms
+
+I1=Vt/(R_25+R1+Rc); // current at 25 degree Celsius(A)
+I2=Vt/(R_65+R1+Rc); // current at 65.556 degree Celsius(A)
+
+printf(' The current meter reading at 25 degree Celsius = %.2f mA and at 150 degree fahrenheit = %.1f mA \n', I1,I2);
diff --git a/3554/CH13/EX13.5/Ex13_5.sce b/3554/CH13/EX13.5/Ex13_5.sce new file mode 100644 index 000000000..9f2b4bfa7 --- /dev/null +++ b/3554/CH13/EX13.5/Ex13_5.sce @@ -0,0 +1,25 @@ +// Exa 13.5
+
+clc;
+clear all;
+
+// Given data
+
+Input=6.3; // V
+Output=5.2; // V
+Range= 9.5; // inches
+
+// Solution
+
+// 0.5 inches core displacement produces 5.2 V
+// Therefore, a 0.45 inch movement produces voltage as
+V1=0.45*5.2/0.5;
+// Similarly - 0.30 inches core movement produces voltage as
+V2=-0.30*-5.2/(-0.5); // V
+// Also -0.25 inch core movement produces voltage as
+V3=-0.25*-5.2/(-0.5); // V
+
+printf('The core movement of 0.45 inch produces voltage of %.2f V and\n movement of -0.30 inch produces voltage of %.2f V \n' ,V1,V2);
+printf(' The core movement of -0.25 inch from the centre produces voltage of %.1f V \n',V3);
+
+
diff --git a/3554/CH13/EX13.6/Ex13_6.sce b/3554/CH13/EX13.6/Ex13_6.sce new file mode 100644 index 000000000..03a3a4fd8 --- /dev/null +++ b/3554/CH13/EX13.6/Ex13_6.sce @@ -0,0 +1,18 @@ +// Exa 13.6
+
+clc;
+clear all;
+
+// Given data
+
+K=0.32; // Coupling co efficient
+Op=1;// Output in oz.in.
+
+// Solution
+
+// 1 oz.in.= 1 oz.in. * (1 ft/12 in.) * (1 lb/16 oz) * (1.3561/1 ft lb) = 7.06*10^-3 J ;
+
+Elec_mech= 7.06*10^-3; // Electrical energy converted to mechanical energy(J)
+Ee=Elec_mech/K; // Applied Electrical energy
+printf(' The electrical energy of %.2f mJ must be applied \n',Ee*10^3);
+// The answer mentioned in the book is incorrect
diff --git a/3554/CH13/EX13.7/Ex13_7.sce b/3554/CH13/EX13.7/Ex13_7.sce new file mode 100644 index 000000000..fadd3c985 --- /dev/null +++ b/3554/CH13/EX13.7/Ex13_7.sce @@ -0,0 +1,22 @@ +// Exa 13.7
+
+clc;
+clear all;
+
+// Given data
+// Refering circuit in fig. 13.7(a) and graph in fig.13.7(b) on page no.422
+
+I=10; // mA
+V=30;// Volts
+Illumination=400;// in l/m^2
+
+// Solution
+
+disp(" From graph(13.7(b) , cell resistance at 400 l/m^2 is 1 K ohms");
+Rcell=1; // K ohms
+
+R1=V/I - Rcell; //Required series resistance
+
+Rdark=100;//Cells dark resistance in K ohms
+Idark=V/(R1+Rdark); // Dark current
+printf(' The required series resistance and dark current level are %d K ohms amd %.1f mA respectively \n',R1,Idark);
diff --git a/3554/CH14/EX14.1/Ex14_1.sce b/3554/CH14/EX14.1/Ex14_1.sce new file mode 100644 index 000000000..1e65631f7 --- /dev/null +++ b/3554/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,45 @@ +// Exa 14.1
+
+ clc;
+clear all;
+
+
+// Given data
+
+fa=800; // The highest frequency(Hz)
+Vp=2; //Volts
+
+
+// Solution
+disp("Let C1=0.1 micro farads")
+C1=0.1; // micro farads
+// Then Rf is given as
+Rf=1/(2*%pi*C1*10^-6*fa);// Ohms
+printf(' Calculated value of Rf = %.3f k Ohms. selecting nearest higher value of 2.2 k Ohms \n ',Rf/1000);
+
+fb=20*fa;
+R1=1/(2*%pi*C1*10^-6*fb);// Ohms
+printf('The calculated value of R1 = %.1f Ohms. Let R1=100 Ohms \n',R1);
+
+// Since R1*C1=Rf*Cf
+Cf=R1*C1*10^-6/2200;//Rf is taken as 2.2 k Ohms as stated above
+printf(' The value of Cf = %.5f micro farads. Let Cf=0.0047 micro farads \n',Cf*10^6);
+
+Rom=(1/100+1/2200)^-1;
+printf(' Rom = %.1f Ohms \n',Rom);
+
+t=0:0.1*10^-3:1.5*2.50*10^-3;
+
+Vin=Vp*sin(2*%pi*fa*t);//Input Voltage equation
+xlabel("Time(sec)");
+ylabel("Voltage(V)");
+title("Input Voltage");
+plot2d(t,Vin);
+figure(1);
+
+Vo=-2200*0.1*10^-6*Vp*2*%pi*fa*cos(2*%pi*800*t);//Output voltage equation
+xlabel("Time(sec)");
+ylabel("Voltage(V)");
+title("Output Voltage");
+plot2d(t,Vo);
+// The answers vary due to round off error
diff --git a/3554/CH14/EX14.2/Ex14_2.sce b/3554/CH14/EX14.2/Ex14_2.sce new file mode 100644 index 000000000..8a74078ab --- /dev/null +++ b/3554/CH14/EX14.2/Ex14_2.sce @@ -0,0 +1,23 @@ +// Exa 14.2 + +clc; +clear; + +// Given data + +Va=2;// Volts +Vb=1;// Volts +Vc=3; // Volts +Ra=3;// k Ohms +Rb=3;// k Ohms +Rc=3;// k Ohms +Rf=1;// k Ohms +Rom=270;// Ohms +Supply=15;// Volts + +// Solution + +disp(" Assuming that the opamp is initially nulled"); +// Using equation 14.8 to determine the output voltage +Vo=-(Rf/Ra *Va+Rf/Rb *Vb+Rf/Rc *Vc); +printf(' The output voltage = %d Volts \n',Vo);
\ No newline at end of file diff --git a/3554/CH14/EX14.3/Ex14_3.sce b/3554/CH14/EX14.3/Ex14_3.sce new file mode 100644 index 000000000..4e2a7a681 --- /dev/null +++ b/3554/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,29 @@ +// Exa 14.3
+
+clc;
+clear all;
+
+// Given data
+
+R1=2.2;// k Ohms
+Rf=10;// k Ohms
+R=120;// (Ra=Rb=Rc) k Ohms
+E=5; // Volts
+Vcc=15; // Volts
+Rt=120; // k Ohms at reference temperature of 25 degree celsius
+Tco=- 1; // Temperature co-efficient in K/degree celsius
+
+// Given data
+
+disp(" At 25 degree celsius, Ra=Rb=Rc=120 K. Therefore, the bridge is balanced and Va=Vb.Therefore, Vo=0.");
+Delta_R=Tco*(0-25);
+// For 0 degree celsius
+printf(' At 0 degree celsius the change delta_R in the resistance of the thermistor is %d k Ohms \n ',Delta_R);
+
+Vo=-(Delta_R)*E*Rf/(2*(2*R+Delta_R)*R1);
+printf(' The output voltage at 0 degree celsius = %.2f Volts \n ',Vo);
+// For 100 degree celsius
+Delta_R1=Tco*(100-25);
+Vo1=-(Delta_R1)*E*Rf/(2*(2*R+Delta_R1)*R1);
+
+printf(' The output voltage at 100 degree celsius = %.2f Volts \n ',Vo1);
diff --git a/3554/CH14/EX14.4/Ex14_4.sce b/3554/CH14/EX14.4/Ex14_4.sce new file mode 100644 index 000000000..f9c5fa2a4 --- /dev/null +++ b/3554/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,21 @@ +// Exa 14.4
+// Refer circuit 14.25 given on page no. 484
+
+clc;
+clear all;
+
+// Given data
+
+E=10;// Volts
+R=50;// Unstrained gauge resistance(Ohms)
+Gain=100;// Amplifier gain
+Vo=1.5;// Output Voltage
+
+// Solution
+
+// Using the formula: Vo=E*(Delta_R/R)*gain
+
+Delta_R=Vo*R/(E*Gain);// Change in resistance
+
+printf('The change in resistance =%.2f Ohms\n This means that Rt1 and Rt3 decrease by 0.07 ohms \n and Rt2 and Rt4 increase by 0.07 ohms when a certain weight is placed on the scale platform\n',Delta_R);
+// The answer mentioned in the textbook is incorrect as R=50 Ohms and not 100 Ohms.
diff --git a/3554/CH15/EX15.1/Ex15_1.sce b/3554/CH15/EX15.1/Ex15_1.sce new file mode 100644 index 000000000..dcdb5d451 --- /dev/null +++ b/3554/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,19 @@ +//Exa 15.1
+
+clc;
+clear all;
+
+
+// Given data
+Fh=2;// kHz
+Af=2;// Pass band gain
+
+// Solution
+
+disp(" Let C1= 0.01 micro farads ");
+C=0.01;//micro farads
+R=1/(2*%pi*Fh*C);// k Ohms
+printf(' The calculated value of R is %.3f K ohms. Nearest practical value for R1 is 8.2 k Ohms\n',R);
+//Af=1+Rf/R1;
+// As Af=2. So, Rf=R1
+disp(" In this case , Rf=R1= 10 k Ohms is selected ");
diff --git a/3554/CH15/EX15.10/Ex15_10.sce b/3554/CH15/EX15.10/Ex15_10.sce new file mode 100644 index 000000000..e395a330e --- /dev/null +++ b/3554/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,35 @@ +// Exa 15.10
+
+clc;
+clear all;
+
+// Given data
+// Second order inverting Butterworth low pass filter
+// Refering Table 15.1 and 15.3 in page no 517 and 538 respectively
+
+Af=6;// DC gain
+Fc=1.5;// KHz
+Q=10;
+
+// Solution
+
+disp(" According to Table 15.1, the inverting configurations would normally be used to give an inverting low pass output. However, to obtain a gain of 6, an inverting uncommitted opamp has to br used, hence the non-inverting filter configuration must be used.");
+
+// From table 15.4 given on page no 538
+R2=316/Q;
+R3=100/(3.16*Q-1);
+// R1 treated as open circuit
+printf(' \n The R1 is open while R2 and R3 are %.1f ,k Ohms %.1f k Ohms respectively \n',R2,R3);
+// From equations 15.54 given on page no 538 we get R4 and R5
+R4=(5.03)*10^7/(Fc*10^3);//Ohms
+R5=R4;
+printf(' \n The calculated value of R4=R5=%.2f k Ohms \n',R4/1000);
+disp(" use R4=R5=33 k Ohms");
+
+disp(" Let R6=1.8 K ohms");
+R6=1.8; // K ohms
+R7=R6*Af;
+R8=(1/R6 + 1/R7)^-1;
+
+printf(' The values of R6 and R7 are %.1f k Ohms, %.3f K ohms respectively \n',R6,R7);
+printf(' The value of R8 = %.3f k Ohms \n ',R8);
diff --git a/3554/CH15/EX15.11/Ex15_11.sce b/3554/CH15/EX15.11/Ex15_11.sce new file mode 100644 index 000000000..e8e877d74 --- /dev/null +++ b/3554/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,30 @@ +// Exa 15.11
+
+clc;
+clear all;
+
+// Given data
+
+Fc=4;// kHz
+Q=8;
+
+// Solution
+
+disp(" The FLT-U2 can be used as a notch filter by summing the inverted output of the bandpass filter designed with the input signal by means of the uncommitted opamp.");
+
+// From table 15.3 given on page no 538
+R2=100;// k Ohms
+R3=100/((3.40*Q)-1);
+// R1 treated as open circuit
+printf(' The R1 is open while R2 and R3 are %.1f , %.2f K ohms respectively \n',R2,R3);
+
+// From equations 15.54 given on page no 538 we get R4 and R5
+R4=(5.03)*10^7/(Fc*10^3);
+R5=R4;
+printf(' The calculated value of R4=R5=%.2f k Ohms(12 k Ohms) \n',R4/1000);
+disp(" Let R6=R7=R8=10 K ohms ");
+R=10000;//R=R6=R7=R8=10 k Ohms
+R9=(1/R+1/R+1/R)^-1;
+printf(' The value of R9 =%.2f K ohms \n',R9/1000);
+disp(" The complete circuit diagram is shown in fig. 15.26 on page no. 541.");
+// The value of R3 vary due to round off error.
diff --git a/3554/CH15/EX15.2/Ex15_2.sce b/3554/CH15/EX15.2/Ex15_2.sce new file mode 100644 index 000000000..3dfba4a17 --- /dev/null +++ b/3554/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,16 @@ +// Exa 15.2
+
+clc;
+clear all;
+
+// Given data
+
+Wc=20*10^3; // Angular cutoff frequency in rad/s
+C=0.01*10^-6; //in farads
+
+// Solution
+
+// As Wc=1/(R*C);
+R=1/(Wc*C);
+
+printf(' The value of resistance required = %d k Ohms \n',R/1000);
diff --git a/3554/CH15/EX15.3/Ex15_3.sce b/3554/CH15/EX15.3/Ex15_3.sce new file mode 100644 index 000000000..6f43e0acf --- /dev/null +++ b/3554/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,23 @@ +// Exa 15.3
+
+clc;
+clear all;
+
+// Given data
+
+Fh= 2*10^3;// Cutoff frequency in Hz
+
+// Solution
+
+disp(" Let C2=C3=0.0033 micro farads ");
+
+// Fh=1/(2*%pi*R*C); where R=R2=R3 and C2=C3=C;
+C=0.0033*10^-6; // farads
+// Therefore
+R=1/(2*%pi*Fh*C);
+printf(' Calculated value of R = %.1f K ohms. Let, R=R2=R3=22 k Ohms is selected\n',R/1000);
+// Since Rf/R1=0.586, therefore Rf=0.586*R1;
+// Let fix value of R1 as
+R1=10*10^3; // Ohms
+Rf=0.586*R1;
+printf(' The remaining components after calculation comes out to be as Rf= %.2f k Ohms and R1= %d k Ohms \n',Rf/1000,R1/1000);
diff --git a/3554/CH15/EX15.4/Ex15_4.sce b/3554/CH15/EX15.4/Ex15_4.sce new file mode 100644 index 000000000..211767976 --- /dev/null +++ b/3554/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,16 @@ +// Exa 15.4
+
+clc;
+clear all;
+
+// Given data
+// Second order filter
+
+R=47*10^3; // Ohms(R2=R3=R)
+C=0.0022*10^-6; // farads(C2=C3=C)
+
+// Solution
+
+Fl=1/(2*%pi*R*C); //low cutoff frequency(Hz)
+printf(' The low cutoff frequency for a high pass filter =%.2f kHz\n',Fl/1000);
+
diff --git a/3554/CH15/EX15.5/Ex15_5.sce b/3554/CH15/EX15.5/Ex15_5.sce new file mode 100644 index 000000000..5e95c3259 --- /dev/null +++ b/3554/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,36 @@ +// Exa 15.5
+
+clc;
+clear all;
+
+// Given data
+
+Fl=100;// lower cutoff frequency in Hz
+Fh=1000;// higher cutoff frequency in Hz
+Af=4;// pass band gain
+
+// Solution
+
+// Wide bandpass filter design
+// 1. For a low pass filter Fh=1 KHz =1/(2*%pi*R*C);
+
+disp(" For a low pass filter section");
+disp(" Let C1=0.01 micro farads ");
+C1=0.01;// micro farads
+R1=1/(2*%pi*Fh*C1*10^-6);
+printf('The value of resistor = %.1f K ohms \n',R1/1000);
+
+// 2. For a high pass filter Fl=100 Hz=1/(2*%pi*R*C);
+disp(" For a high pass filter section");
+disp(" Let C2=0.01 micro farads ");
+C2=0.01;// micro farads
+R2=1/(2*%pi*Fl*C2*10^-6);
+printf(' The value of resistor = %d K ohms \n',R2/1000);
+
+disp(" Since the pass band gain is 4, the gain of the high pass and low pass filter sections are set each equal to 2. Therefore, R1=Rf=10 K ohms for both sections.");
+
+// Q for filter
+Fc=sqrt(Fl*Fh);
+
+Q=Fc/(Fh-Fl);
+printf(' The value of Q =%.2f which is less than 10, as expected for a wide band pass filter\n',Q);
diff --git a/3554/CH15/EX15.6/Ex15_6.sce b/3554/CH15/EX15.6/Ex15_6.sce new file mode 100644 index 000000000..92ce4e67b --- /dev/null +++ b/3554/CH15/EX15.6/Ex15_6.sce @@ -0,0 +1,26 @@ +// Exa 15.6
+
+clc;
+clear all;
+
+//Given data
+// Refering fig. 15.17- Narrow band pass filter
+
+Fc=1; // kHz
+Q=5; //Quality factor
+Avo=8; //Voltage gain
+Fc1=1.5;//New centre frequency(kHz)
+
+// Solution
+
+disp(" Let C1=C2=C3=C(say)=0.01 micro farads");
+C=0.01;//micro farads
+// But
+R1=Q/(2*%pi*Fc*10^3*C*10^-6*Avo); // From eqn. 15.45 on page no.530
+R2=Q/(2*%pi*Fc*10^3*C*10^-6*(2*Q^2-Avo));// From eqn. 15.47 on page no. 530
+R3=Q/(%pi*Fc*10^3*C*10^-6); // From eqn. 15.46 on agr no. 530
+printf(' The Values of R1, R2 and R3 are %.3f k Ohms(approx 10 k Ohms), %.3f k Ohms(approx 2 k Ohms and %.3f k Ohms(aprox 159 k Ohms) respectively\n',R1/1000,R2/1000,R3/1000);
+// To change Fc to Fc1 we simply have to change R2 to R21 given as
+R21=2000*(Fc/Fc1)^2;// since R2=2 k Ohms(approx)
+printf(' The calculated value of new R2 to change Fc from 1 kHz to 1.5 kHz keeping Avo(Voltage gain) and BW constant is = %.2f ohms (approx 820 Ohms) \n',R21);
+
diff --git a/3554/CH15/EX15.7/Ex15_7.sce b/3554/CH15/EX15.7/Ex15_7.sce new file mode 100644 index 000000000..ef83d5ce0 --- /dev/null +++ b/3554/CH15/EX15.7/Ex15_7.sce @@ -0,0 +1,32 @@ +// Exa 15.7
+
+clc;
+clear all;
+
+// Given data
+// Refering fig. 15.20- Wide band reject filter
+
+Fl=100;// Hz
+Fh=1000;// Hz
+
+// Solution
+// 1. For a high pass filter Fh=1 KHz =1/(2*%pi*R*C);
+disp(" For a high pass filter section");
+disp(" Let C1=0.01 micro farads ");
+C1=0.01;// micro farads
+R1=1/(2*%pi*Fh*C1*10^-6);
+printf(' The value of resistor = %.1f k Ohms \n',R1/1000);
+
+// 2. For a low pass filter Fl=100 Hz=1/(2*%pi*R*C);
+disp(" For a low pass filter section");
+disp(" Let C2=0.01 micro farads ");
+C2=0.01;// micro farads
+R2=1/(2*%pi*Fl*C2*10^-6);
+printf(' The value of resistor = %.1f k Ohms \n',R2/1000);
+
+disp(" Since the pass band gain is 4, the gain of the high pass and low pass filter sections are set each equal to 2. Therefore, R1=Rf=10 k Ohmss for both section");
+disp(" Further, the gain of the summing amplifier is set to 1, therefore R2=R3=R4=10 k Ohms"); // K ohms
+R=10000;//Ohms(R=R2=R3=R4)
+Rom=(1/R+1/R+1/R)^-1;
+printf(' The value of Rom = %.1f k Ohms\n',Rom/1000);
+// There is a printing mistake as c=0.1 micro fard is printed instead of 0.01 micro farad.
diff --git a/3554/CH15/EX15.8/Ex15_8.sce b/3554/CH15/EX15.8/Ex15_8.sce new file mode 100644 index 000000000..b7e84a363 --- /dev/null +++ b/3554/CH15/EX15.8/Ex15_8.sce @@ -0,0 +1,19 @@ +// Exa 15.8
+
+clc;
+clear all;
+
+// Given data
+// Active notch filter
+
+Fn=50; //Notch out frequency(Hz)
+
+// Solution
+
+disp(" Let C=0.047 micro farads");
+C=0.047; // micro farads
+R=1/(2*%pi*Fn*C*10^-6);
+
+printf(' The value of R is calculated as %d k Ohms \n',round(R/1000));
+disp("For R/2, two 68 k Ohms resistors connected in parallel are used and for the 2C components, two 0.047 micro farad capacitors connected in parallel are used.");
+
diff --git a/3554/CH15/EX15.9/Ex15_9.sce b/3554/CH15/EX15.9/Ex15_9.sce new file mode 100644 index 000000000..fadd5f844 --- /dev/null +++ b/3554/CH15/EX15.9/Ex15_9.sce @@ -0,0 +1,18 @@ +// Exa 15.9
+
+clc;
+clear all;
+
+// Given data
+// Refering Fig. 15.2(a)- All pass filter
+f=2.5;// Input frequency in kHz
+
+// Solution
+
+disp(" Let C=0.01 micro farads and R= 15 k Ohms");
+C=0.01;// micro farads
+R=15;// k Ohms
+Phi=2*atan(2*%pi*f*C*R); // phase angle in radians
+
+printf(' This means that the output voltage Vo has the same frequency and amplitude as the input voltage but lags it by - %d degrees\n',Phi*180/%pi);
+
diff --git a/3554/CH16/EX16.1/Ex16_1.sce b/3554/CH16/EX16.1/Ex16_1.sce new file mode 100644 index 000000000..7632424b4 --- /dev/null +++ b/3554/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,16 @@ +// Exa 16.1
+
+clc;
+clear all;
+
+// Given data
+
+fd=75; // Frequency deviation in KHz
+fm=5;// Frequency of modulating signal in kHz
+
+// Solution
+
+// From equation 16.5 (page no. 590) we calculate Mi as
+Mi=fd/fm; // Modulation index
+
+printf(' The modulation index =%d \n',Mi);
diff --git a/3554/CH17/EX17.1/Ex17_1.sce b/3554/CH17/EX17.1/Ex17_1.sce new file mode 100644 index 000000000..554f1d687 --- /dev/null +++ b/3554/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,31 @@ +// Exa 17.1
+
+clc;
+clear all;
+
+// Given data
+// A 5 bit resistive divider
+
+n=5;// since 5 bit resistive divider
+Ip1=[1 1 0 1 1];// Digital input 1(1st element of array is MSB)
+Ip2=[1 0 1 1 0];//Digital input 2(1st element of array is MSB)
+V1=10;// Voltage corresponding to binary 1
+V0=0;// Voltage corresponding to binary 0
+
+// Solution
+
+LSB_weight=1/(2^n - 1);
+printf('The LSB weight = %.4f \n ',LSB_weight);
+LSB2_weight=2^(2-1)/(2^n-1);
+printf('The 2nd LSB weight = %.4f \n ',LSB2_weight');
+LSB3_weight=2^(3-1)/(2^n-1);
+printf('The 3rd LSB weight = %.4f \n ',LSB3_weight');
+LSB_op=V1*LSB_weight;// Change in output voltage due to change in LSB
+printf('The change in output voltage due to change in LSB = %.4f V \n ',LSB_op);
+LSB2_op=V1*LSB2_weight;
+printf('THe 2nd LSB causes a change in output voltage of %.4f V \n ',LSB2_op);
+LSB3_op=V1*LSB3_weight;
+printf('THe 3rd LSB causes a change in output voltage of %.4f V \n ',LSB3_op);
+Va=(V1*2^4+V1*2^3+V0*2^2+V1*2^1+V1*2^0)/(2^n-1);// output voltage for digital Ip1
+Vb=(V1*2^4+V0*2^3+V1*2^2+V1*2^1+V0*2^0)/(2^n-1);
+printf('The output voltage for a digital input 1 and 2 are %.2f V and %.3f V respectively \n ',Va,Vb);
diff --git a/3554/CH17/EX17.2/Ex17_2.sce b/3554/CH17/EX17.2/Ex17_2.sce new file mode 100644 index 000000000..2141dcc59 --- /dev/null +++ b/3554/CH17/EX17.2/Ex17_2.sce @@ -0,0 +1,20 @@ +
+// Exa17.2
+
+clc;
+clear all;
+
+// Given data
+
+n=5;// 5 bit ladder
+V=10;// For binary 1
+
+// Solution
+// refering table 17.4(page no. 615)-Various Output voltage for corresponding MSB
+
+disp("The output voltage for each bit is as follows:");
+disp("");
+for i=1:n
+MSB(i)=V/2^i; //voltage corresponding to MSB i
+printf(' %d MSB Va = V/2^%d = %.4f V \n ',i,i, MSB(i));
+end
diff --git a/3554/CH17/EX17.3/Ex17_3.sce b/3554/CH17/EX17.3/Ex17_3.sce new file mode 100644 index 000000000..00703a724 --- /dev/null +++ b/3554/CH17/EX17.3/Ex17_3.sce @@ -0,0 +1,18 @@ +// Exa 17.3
+
+clc;
+clear all;
+
+// Given data
+
+Vin=5;// Input voltage(Volts)
+Rin=2.5;// k Ohms
+Rf=1;//k Ohms
+
+// Solution
+
+Iin=Vin/Rin;//Input current(mA)
+If=Iin;
+Vout=-If*Rf;
+
+printf('The output voltage = %d Volts \n',Vout);
diff --git a/3554/CH17/EX17.4/Ex17_4.sce b/3554/CH17/EX17.4/Ex17_4.sce new file mode 100644 index 000000000..db85b5236 --- /dev/null +++ b/3554/CH17/EX17.4/Ex17_4.sce @@ -0,0 +1,22 @@ +// Exa 17.4
+
+clc;
+clear all;
+
+// Given data
+Vref=5;//Reference voltage(V)
+R=5;// k Ohms
+
+// Solution
+
+disp("From fig. 17.18(c) , for a 4-bit D/A converter I=Vref/R* (D3+D2*2^-1+D1*2^-2+D0*^-3)");
+//16-input combinations are as follows
+Ip={[0 0 0 0];[0 0 0 1];[0 0 1 0];[0 0 1 1];[0 1 0 0];[0 1 0 1];[0 1 1 0];[0 1 1 1];[1 0 0 0];[1 0 0 1];
+[1 0 1 0];[1 0 1 1];[1 1 0 0];[1 1 0 1];[1 1 1 0];[1 1 1 1]};//[D3 D2 D1 D0 bits]
+
+disp(" Input Bits Output Current(mA) percent Fraction of maximum ");
+for i=1:16
+Iout(i)=Vref/R * (Ip(i,1)+Ip(i,2)*2^-1+Ip(i,3)*2^-2+Ip(i,4)*2^-3);
+
+printf(' %d %d %d %d %.3f %.3f \n',Ip(i,1),Ip(i,2),Ip(i,3),Ip(i,4),Iout(i),(Iout(i)/1.875)*100);//1.875(mA) is the highest output current
+end
diff --git a/3554/CH2/EX2.1/Ex2_1.sce b/3554/CH2/EX2.1/Ex2_1.sce new file mode 100644 index 000000000..68775e314 --- /dev/null +++ b/3554/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,21 @@ +// Exa 2.1
+
+clc;
+clear all;
+
+// Given data
+
+N= 100; // Number of turns
+W=20; // Width of coil(mm)
+D= 30; // Depth of coil(mm)
+B= 0.1; // Flux density (wb/m^2)
+I= 10; // Current in coil(mA)
+K= 2*10^-6; // Spring constant(Nm/degree)
+
+// Solution
+A= W*10^-3*D*10^-3; // Area of coil(m^2)
+Td= B*N*A*I*10^-3; // Deflecting torque(Nm)
+disp("As deflecting torque = restoring torque(K*Theta)");
+Theta= Td/K;
+printf(' The defecting torque = %.1f * 10^-6 Nm \n ', Td*10^6);
+printf('Therefore, the deflection = %d degrees \n ' , Theta);
diff --git a/3554/CH20/EX20.1/Ex20_1.sce b/3554/CH20/EX20.1/Ex20_1.sce new file mode 100644 index 000000000..16c5732f1 --- /dev/null +++ b/3554/CH20/EX20.1/Ex20_1.sce @@ -0,0 +1,15 @@ +// Exa 20.1
+
+clc;
+clear all;
+
+// Given data
+
+V1=20;// superimposed small AF voltage(V)
+V2=30;//Bridge balance voltage(V)
+R1=100;// Bridge arm resistor(ohms)
+
+// Solution
+
+RF_pwr=(V2^2-V1^2)/(4*R1);
+printf('RF test power = %.2f W \n',RF_pwr);
diff --git a/3554/CH20/EX20.2/Ex20_2.sce b/3554/CH20/EX20.2/Ex20_2.sce new file mode 100644 index 000000000..2ee1741cd --- /dev/null +++ b/3554/CH20/EX20.2/Ex20_2.sce @@ -0,0 +1,18 @@ +// Exa 20.2
+
+clc;
+clear all;
+
+// Given data
+
+M=200;// mass in grams
+Heat =1;//Sp. Heat of water(cal/gm degree)
+T1=30;//Initial temperature (degree Celsius)
+T2=40;//Final temperature (degree Celsius)
+
+// Solution
+
+Pwr_rad=4.18*M*Heat*(T2-T1); // in Watt
+printf(' The power radiated = %.2f kW \n',Pwr_rad/1000);
+
+// The answer in the textbook is mentioned as 8.3 kW but by calculation it comes out to be 8.36 kW.
diff --git a/3554/CH20/EX20.3/Ex20_3.sce b/3554/CH20/EX20.3/Ex20_3.sce new file mode 100644 index 000000000..cd3a49072 --- /dev/null +++ b/3554/CH20/EX20.3/Ex20_3.sce @@ -0,0 +1,13 @@ +// Exa 20.3
+
+clc;
+clear all;
+
+// Given data
+
+Vmax=8;//Maximum value of voltage
+Vmin=2;//minimum value of voltage
+
+//Solution
+SWR=(Vmax+Vmin)/(Vmax-Vmin);//Standing wave ratio
+printf('Standing Wave Ratio = %.2f \n ',SWR);
diff --git a/3554/CH21/EX21.1/Ex21_1.sce b/3554/CH21/EX21.1/Ex21_1.sce new file mode 100644 index 000000000..b2a2d589f --- /dev/null +++ b/3554/CH21/EX21.1/Ex21_1.sce @@ -0,0 +1,16 @@ +// Exa 21.1
+
+clc;
+clear all;
+
+// Given data
+Emax=20; //Max value of variable(mA)
+Emin=4;//Min value of variable(mA)
+Em=13;//Measured value of variable
+Eref=10;//Set(ref) point of variable(mA)
+
+// Solution
+//Ep=(Em-Eref)/(Emax-Emin)*100; // Percentage error from page no.(703)
+//Therefore
+Ep=(Em-Eref)/(Emax-Emin)*100;
+printf('The value of Ep = %.2f percent. Since Ep is positive, the measurement is above the set point \n ',Ep);
diff --git a/3554/CH3/EX3.1/Ex3_1.sce b/3554/CH3/EX3.1/Ex3_1.sce new file mode 100644 index 000000000..2a081ccf3 --- /dev/null +++ b/3554/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,16 @@ +// Exa 3.1
+
+clc;
+clear all;
+
+// Given data
+
+Rm= 100; // Internal resistance in Ohm's
+Im= 1; // Full scale deflecfion current in milliAmpere
+I= 100; // Total current in milli Ampere
+
+// Solution
+
+Rsh= (Im*Rm)/(I-Im); // Shunt resistance
+printf('The value of shunt resistance = %.2f Ohm \n', Rsh);
+
diff --git a/3554/CH3/EX3.2/Ex3_2.sce b/3554/CH3/EX3.2/Ex3_2.sce new file mode 100644 index 000000000..aef9fea92 --- /dev/null +++ b/3554/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,38 @@ +// Exa 3.2
+
+clc;
+clear all;
+
+// Given data
+
+//Refer fig. 3.4
+Rm=100;//Ohms
+Im=50; ///micro Amp
+
+// Solution
+
+//For 0-1mA range
+ // Ish*Rsh=Im*Rm;
+ disp("The four linear equations are as follows:");
+ printf(' R1+R2+R3+R4 = %.2f \n ',50*100/950); //say-equaion (3.1)
+
+ //For 0-10 mA range
+ printf('R1+R2+R3-(50/9950)*R4= % .3f \n ',100*50/9950); //say-equation(3.2)
+
+ //For 0-50 mA range
+ printf('R1+R2-(50/49950)*R3-(50/49950)*R4 = %.3f \n ',100*50/49950); //say-equation(3.3)
+
+ //For 0-100mA range
+ printf('R1-(50/99950)*R2-(50/99950)*R3-(50/99950)*R4 = %.3f \n ',50*100/99950);//say-equation(3.4)
+
+ //converting it into matrix form
+ A=[1 1 1 1;1 1 1 -(50/9950);1 1 -(50/49950) -(50/49950);1 (-50/99950) (-50/99950) (-50/99950)];
+
+ B=[-50*100/950 ; -100*50/9950 ; -100*50/49950 ; -50*100/99950];
+
+ [R,y]=linsolve(A,B);A*R+B;//linear equaion solving function
+
+ disp("The value of R1,R2 R4 and R4 are given as follows-");
+ printf(' R1 = %.5f Ohms \n R2= %.5f Ohms \n R3= %.5f Ohms \n R4= %.5f Ohms \n ',R(1),R(2),R(3),R(4));
+
+ // The value of R3 vary due to round off errors
diff --git a/3554/CH4/EX4.1/Ex4_1.sce b/3554/CH4/EX4.1/Ex4_1.sce new file mode 100644 index 000000000..07723237a --- /dev/null +++ b/3554/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,13 @@ +// Exa 4.1
+
+clc;
+clear all;
+
+// Given data
+
+Iful = 200; // Fullscale deflection current in micro Amperes
+Sen= 1/(Iful*10^-3) ; // Sensitivity of Voltmeter(K Ohms/V)
+
+// Solution
+
+printf(' The sensitivity of the voltmeter = %d k Ohms/V \n',Sen);
diff --git a/3554/CH4/EX4.10/Ex4_10.sce b/3554/CH4/EX4.10/Ex4_10.sce new file mode 100644 index 000000000..fc5ca250b --- /dev/null +++ b/3554/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,17 @@ +// Exa 4.10
+
+clc;
+clear all;
+
+// Given data
+
+Vin=10; // Input RMS voltage(V)
+Ifsd=1; // Full scale deflection current(mA)
+Rm=250;// Internal resistance of voltmeter(ohms)
+
+// Solution
+
+Sdc=1/(Ifsd*10^-3); // DC sensitivity(K ohm/V)
+Sac=0.9*Sdc; //AC sensitivity(k Ohm/V)
+Rs=Sac*Vin-Rm;// Multiplier resistor(Ohm)
+printf(' The value of multiplier resistor = %.2f k Ohms \n',Rs/1000);
diff --git a/3554/CH4/EX4.11/Ex4_11.sce b/3554/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..8e122562d --- /dev/null +++ b/3554/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,22 @@ +// Exa 4.11
+
+clc;
+clear all;
+
+// Given data
+Rm=100; // Meter resistance(Ohms)
+Ifsd=1; // Full scale deflection current(mA)
+Rh=2000; // Half of full scale deflection resistance(Ohms)
+V=3; // Internal battery voltage(V)
+
+// Solution
+// Using equations 4.1 and 4.2 given on page no. 104
+
+R1=Rh-Ifsd*10^-3*Rh/V ;// Current limiting resistance(Ohms)
+R2= Ifsd*10^-3*Rm*Rh/(V-Ifsd*10^-3*Rh); // Zero adjust resistance(Ohms)
+ V1= V-0.05*V; // Voltage after 5 percent drop in battery voltage
+R3=Ifsd*10^-3*Rh*Rm/(V1-Ifsd*10^-3*Rh);// Maximum value of R2 to compensate drop in battery
+
+printf(' The values of R1 and R2 are %.1f Ohms and %d Ohms respectively \n ',R1,R2);
+printf('The maximum value of R2 to compensate for a 5 percentage drop \n in battery voltage is =%.2f Ohms \n',R3);
+
diff --git a/3554/CH4/EX4.2/Ex4_2.sce b/3554/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..e59f683e3 --- /dev/null +++ b/3554/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,16 @@ +// Exa 4.2
+
+clc;
+clear all;
+
+// Given data
+
+Iful= 50; // Fullscale deflection current in micro Amperes
+Rm= 500; // Internal resistance in Ohms
+V= 10; // Full range voltage of instrument(Volts)
+
+// Solution
+
+Rs= V/(Iful *10^-6)-Rm; // Multiplier resistance
+
+printf('The value of multiplier resistance = %.1f k Ohms\n',Rs/1000);
diff --git a/3554/CH4/EX4.3/Ex4_3.sce b/3554/CH4/EX4.3/Ex4_3.sce new file mode 100644 index 000000000..3a33e3ea9 --- /dev/null +++ b/3554/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,37 @@ +// Exa 4.3
+
+clc;
+clear all;
+
+// Given data
+// Refer Fig. 4.3 on page no. 77
+
+Rm=50; // Internal resistance of Voltmeter(ohms)
+Ifsd=2; // full sclae deflection current(mA)
+
+//Solution
+
+// For 10V range(V4 position of switch)
+V1=10;//Volts
+Rt1=V1/(Ifsd*10^-3); //total resistance in k Ohms
+R4=Rt1-Rm;
+printf('The value of R4 = %d Ohms \n',R4);
+// For a 50V range(V3 position of switch)
+V2=50;//Volts
+Rt2=V2/(Ifsd*10^-3);
+R3=Rt2-(R4+Rm);
+printf(' The value of R3 = %d k Ohms \n',R3/1000);
+
+// For 100V range(V2 position of switch)
+V3=100;//Volts
+Rt3=V3/(Ifsd*10^-3); //total resistance in k Ohms
+R2=Rt3-(R3+R4+Rm);
+printf(' The value of R2 = %d k Ohms \n',R2/1000);
+// For a 250V range(V3 position of switch)
+V4=250;//Volts
+Rt4=V4/(Ifsd*10^-3);
+R1=Rt4-(R2+R3+R4+Rm);
+printf(' The value of R1 = %d k Ohms \n',R1/1000);
+
+
+
diff --git a/3554/CH4/EX4.4/Ex4_4.sce b/3554/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..4e3df5cda --- /dev/null +++ b/3554/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,18 @@ +// Exa 4.4
+
+clc;
+clear all;
+
+// Given data
+
+Iful= 200; // Full scale deflection current in micro Amperes
+Rm= 100;// Internal resistance of the movement in Ohms
+Range= 50; // Voltage range
+
+// Solution
+
+S= 1/(Iful * 10^-6); // Sensitivity of voltmeter is ohms/volt
+// Rs=S*Range-Rm ;
+Rs=S*Range-Rm; // Multiplier resistance
+
+printf(' The value of multiplier resistance = %.1f K Ohms \n', Rs/1000);
diff --git a/3554/CH4/EX4.5/Ex4_5.sce b/3554/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..65f78b882 --- /dev/null +++ b/3554/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,28 @@ +// Exa 4.5
+
+clc;
+clear all;
+// Refer circuit diagram in Fig. 4.5 page no.79
+
+// Given data
+
+Ifsd=50; // Full scale deflection current (micro Amp)
+Rm= 1000; // Internal resistance in Ohms
+V1= 5; // Range of voltmeter 1 (V)
+V2=10; //Range of voltmeter 2 (V)
+V3=50;// Range of voltmeter 3 (V)
+
+// Solution
+
+S= 1/(Ifsd*10^-6); // Sensitivity of voltmeter in Ohms/V
+ // The value of multiplier resistance for different ranges
+
+// For 5V range
+Rs1= S*V1-Rm;
+
+// For 10V range
+Rs2= S*V2-Rm;
+// For 50V range
+Rs3=S*V3-Rm;
+
+printf(' The value of multiplier resistance for 0-5V , 0-10V and 0-50V range are \n %d k Ohms, %d k Ohms, %d k Ohms respectively \n ',Rs1/1000,Rs2/1000,Rs3/1000);
diff --git a/3554/CH4/EX4.6/Ex4_6.sce b/3554/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..f5784d267 --- /dev/null +++ b/3554/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,36 @@ +// Exa 4.6
+
+clc;
+clear all;
+// Referring Fig. 4.6- Example on loading effect from page no.81
+
+// Given data
+R1=10000; // Ohms
+R2=10000;// Ohms
+V=100; // Applied Voltage
+
+// Solution
+
+VR2= R2/(R1+R2)* V;// True Voltage across R2 resistance
+printf('True voltage across R2 = %d V \n ',VR2);
+
+// Case-1 : Using a voltmeter 1 having sensitivity of 1000 Ohms/V
+
+S1=1000; // Sensitivity in Ohms/volt
+R21=S1*VR2; //R2 resistance on its 50 V range(Ohms)
+Req1=R21*R2/(R21+R2);// Equivalent resistance across R2(ohms)
+printf('Connecting the meter 1 across R2 causes an equivalent parallel resistance given by %.2f k Ohms \n ',Req1/1000);
+V21=Req1/(Req1+R2) * V;
+printf('Now the voltage across the total combination is given by %.2f V \n ',V21);
+
+// Case-2 : Using a voltmeter having sensitivity of 20,000 Ohms/V
+
+S22=20000; // Sensitivity in Ohms/volt
+R22=S22*VR2;// R2 resistance on its 50V range(Ohms)
+Req2=R22*R2/(R22+R2);// Equivalent resistance across R2(ohms)
+printf('Connecting the meter 2 across R2 causes an equivalent parallel resistance given by %.2f k Ohms \n ',Req2/1000);
+V22=Req2/(Req2+R2) * V;
+printf('Now the voltage across the total combination is given by %.2f V \n ',V22);
+
+disp(" This example shows that a high sensitivity voltmeter(i.e voltmeter 2 in this case) should be used to get accurate readings");
+// The answers vary due to riund off error.
diff --git a/3554/CH4/EX4.7/Ex4_7.sce b/3554/CH4/EX4.7/Ex4_7.sce new file mode 100644 index 000000000..2ca177149 --- /dev/null +++ b/3554/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,40 @@ +// Exa 4.7
+
+clc;
+clear all;
+// Referring circuit given in fig. 4.7 on page no.81
+
+S1=1000; // Sensitivity of meter 1 (Ohms/volt)
+S2=20000;// Sensitivity of meter 2(Ohms/volt)
+Rm1=200;// Meter resistance(Ohms)
+Rm2=1500;// Meter resistance(Ohms)
+V1=10; // Range of voltmeter 1(Volts)
+V2=10;
+Ra=25000; // in Ohms
+Rb=5000;// in Ohms
+V=30; // Applied Voltage(V)
+
+//Solution
+
+VRb= Rb/(Ra+Rb) * V; // Voltage across Rb
+printf('The voltage across the resistance Rb, without either meter connected = %d V\n ',VRb);
+
+// For meter 1
+Rt1=S1* V1; // Total resistance of meter1
+
+Req1= Rb*Rt1/(Rb+Rt1); // Total resistance across Rb
+VRb1= Req1/(Req1+Ra) * V; // Voltage reading across Rb with meter1
+printf('The voltage across Rb when meter 1 is used is = %.2f V \n',VRb1);
+Err1=(VRb-VRb1)/VRb *100; // Voltmeter 1 error
+printf(' Voltmeter 1 error in percentage = %.1f \n ',Err1);
+
+// For meter 2
+
+Rt2=S2* V2; // Total resistance of meter 2
+
+Req2= Rb*Rt2/(Rb+Rt2); // Total resistance across Rb
+VRb2= Req2/(Req2+Ra) * V; // Voltage reading across Rb with meter2
+printf('The voltage across Rb when meter 2 is used is = %.1f V \n',VRb2);
+
+Err2=(VRb-VRb2)/VRb *100; // Voltmeter 2 error
+printf(' Voltmeter 2 error in percentage = %d \n ',Err2);
diff --git a/3554/CH4/EX4.8/Ex4_8.sce b/3554/CH4/EX4.8/Ex4_8.sce new file mode 100644 index 000000000..8fdcb708a --- /dev/null +++ b/3554/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,51 @@ +// Exa 4.8
+
+clc;
+clear all;
+
+// Given data
+
+Ra= 45; // in k Ohms
+Rb=5; // in k Ohms
+V=50; // Supplied Voltage(V)
+S=20; // sensitivity in k Ohms/V
+
+// Solution
+
+VRb=Rb/(Ra+Rb) * V;
+printf('The voltage drop across Rb without the voltmeter connected is = %d V\n',VRb);
+
+// On the 5V range
+Range1 = 5; // Volts
+
+Rm1=S*Range1;// k Ohms
+Req1=Rm1*Rb/(Rm1+Rb); // k Ohms
+VRb1=Req1/(Req1+Ra) *V; // Voltage across Rb on 5V range
+printf(' The voltmeter reading on 5V range is = %.3f V\n',VRb1);
+Err1=(VRb-VRb1)/VRb * 100;
+printf(' Percentage error on 5V range in percentage = %.2f \n',Err1);
+
+// On 10V range
+
+Range2 = 10; // Volts
+
+Rm2=S*Range2;// k Ohms
+Req2=Rm2*Rb/(Rm2+Rb); // k Ohms
+VRb2=Req2/(Req2+Ra) *V; // Voltage across Rb on 10V range
+printf(' The voltmeter reading on 10V range is = %.3f V\n',VRb2);
+Err2=(VRb-VRb2)/VRb * 100;
+printf(' Percentage error on 10V range in percentage = %.3f \n',Err2);
+
+// On 30V range
+
+Range3 = 30; // Volts
+
+Rm3=S*Range3;// k Ohms
+Req3=Rm3*Rb/(Rm3+Rb); // k Ohms
+VRb3=Req3/(Req3+Ra) *V; // Voltage across Rb on 30V range
+printf(' The voltmeter reading on 30V range is = %.3f V \n',VRb3);
+Err3=(VRb-VRb3)/VRb * 100;
+printf(' Percentage error on 30V range in percentage = %.1f \n',round(Err3));
+
+disp(" In this example, the 30V range introduces the least error due to loading. However, the voltage being measured causes only a 10% full scale deflection, whereas on the 10V range the applied voltage causes approximately a one third of the fullscale deflection with less than 3% error.");
+//The answers vary due to round off error
diff --git a/3554/CH4/EX4.9/Ex4_9.sce b/3554/CH4/EX4.9/Ex4_9.sce new file mode 100644 index 000000000..27b204e21 --- /dev/null +++ b/3554/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,18 @@ +// Exa 4.9
+
+clc;
+clear all;
+
+// Given data
+// As per values given in Fig.4.19(page no.94)
+
+ein=10; // Input RMS voltage(V)
+Ifsd=1; // Full scale deflection current(mA)
+Rm=200;// Internal resistance of voltmeter(Ohms)
+
+// Solution
+
+Range=0.45*ein; // Range of Voltmeter
+Sdc=1/(Ifsd*10^-3); // DC Sensitivity of meter movement(k Ohm/V)
+Rs=Sdc* Range-Rm;// Multiplier resistance(Ohm)
+printf(' The value of the multiplier resistor = %.1f k Ohms\n',Rs/1000);
diff --git a/3554/CH5/EX5.1/Ex5_1.sce b/3554/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..560e11c56 --- /dev/null +++ b/3554/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,18 @@ +// Exa 5.1
+
+clc;
+clear all;
+
+// Given data
+
+R=100; // in k Ohms
+C=1; //in micro farads
+ei=1;// Applied voltage to integrator(V)
+t1=1; // time in Sec
+
+// Solution
+
+// Using equation 5.1 from page no. 118
+eo=ei*t1/(R*10^3*C*10^-6); // Output voltage from integrator
+printf(' The voltage at output of integrator after 1 sec is = %d Volts \n',eo);
+
diff --git a/3554/CH5/EX5.2/Ex5_2.sce b/3554/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..c80c2476d --- /dev/null +++ b/3554/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,20 @@ +// Exa 5.2
+
+clc;
+clear all;
+
+// Given data
+// With reference to data given in Exa 5.1
+
+ei=1; // Applied input voltage to integrator(V)
+t1=1; // sec
+
+// Given data
+
+er=5;// Reference voltage applied at time t1 to integrator(V)
+
+// Solution
+// Using equation 5.3 from page no. 118
+
+t2=ei/er * t1;// Time interval t2(sec)
+printf(' The time interval of t2 is = %.1f sec \n',t2);
diff --git a/3554/CH5/EX5.3/Ex5_3.sce b/3554/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..3262a4955 --- /dev/null +++ b/3554/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,18 @@ +// Exa 5.3
+
+clc;
+clear all;
+
+// Given data
+// 3 1/2 digit display
+
+V1=1; // Volts
+V2=10;//Volts
+
+// Solution
+disp("Number of full digits is 3.");
+n=3;//Full digits
+Reso=1/10^n;
+printf(' Resolution = %.3f . Hence, meter cannot distinguish two values if their difference is less than %.3f \n ',Reso,Reso);
+printf('For full scale reading of 1V, the resolution is %.3f V \n ',V1*Reso);
+printf('For full scale reading of 10V, the resolution is %.2f V \n ',V2*Reso);
diff --git a/3554/CH5/EX5.4/Ex5_4.sce b/3554/CH5/EX5.4/Ex5_4.sce new file mode 100644 index 000000000..2b5cfb0f8 --- /dev/null +++ b/3554/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,22 @@ +// Exa 5.4
+
+clc;
+clear all;
+
+// Given data
+
+n=4; // Number of full digits
+V1=12.98;//Reading 1 to be measured(V)
+V2=0.6973;//Reading 2 to be measured(V)
+
+// Solution
+
+Reso=1/10^n; //Resolution
+printf(' Resolution is %.4f \n ',Reso);
+disp("There are 5 digit places in 4 and 1/2 digits display, therefore 12.98 would be displayed as 12.980");
+disp("");
+printf(' Resolution on 1V range = %.4f. Any reading upto the 4th decimal can be displayed \n ',1*Reso);
+disp("Therefore. 12.98 would be displayed as 12.980 and 0.6973 will be displayed as 0.6973");
+disp("");
+printf(' Resolution on 10V range = %.3f. Therefore. 12.98 would be dislayed as 12.98 \n ',10*Reso);
+disp("Therefore on a 10V range,the reading of 0.6973 would be displayed as o.697 instead of 0.6973");
diff --git a/3554/CH7/EX7.1/Ex7_1.sce b/3554/CH7/EX7.1/Ex7_1.sce new file mode 100644 index 000000000..5e3c80d31 --- /dev/null +++ b/3554/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,19 @@ +// Exa 7.1
+
+clc;
+clear all;
+
+// Given data
+// Referring to waveform shown in fig 7.50 on page 211
+
+V_attn= 0.5; // Vertical attenuator(V/div)
+div=3; // No of vertical divisions
+
+// Solution
+
+// Using equation : Vp-p=(volts/div) * (no. Of div/1);
+
+Vp_p=V_attn * div/1 ;
+
+printf(' The peak to peak amplitude of the signal = %.1f Volts \n',Vp_p);
+
diff --git a/3554/CH7/EX7.2/Ex7_2.sce b/3554/CH7/EX7.2/Ex7_2.sce new file mode 100644 index 000000000..964e9bda6 --- /dev/null +++ b/3554/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,19 @@ +// Exa 7.2
+
+clc;
+clear all;
+
+//Given data
+// Refering waveform shown in fig 7.50 on page no. 211
+
+div=4; // No of horizontal divisions for One cycle
+// Given data
+time_div= 2; // Time per div control in micro sec/div
+
+// Solution
+
+// The period of signal is given as T=(time/div) *(No of div/ cycle);
+T=time_div *10^-6 * div/1 ; // Time period is calculated over 1 cycle
+F= 1/T; // Frequency is inverse of time period
+
+printf(' The frequency of signal = %d kHz \n',F/1000);
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