initial commit / add all books

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diff --git a/608/CH32/EX32.01/32_01.sce b/608/CH32/EX32.01/32_01.sce
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+//Problem 32.01:A.c. sources of 100/_0° V and internal resistance 25 ohm and 50/_90° V and internal resistance 10 ohm, are connected in parallel across a 20 ohm load. Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source

+

+//initializing the variables:

+rv1 = 100; // in volts

+rv2 = 50; // in volts

+thetav1 = 0; // in degrees

+thetav2 = 90; // in degrees

+r1 = 25; // in ohm

+R = 20; // in ohm

+r2 = 10; // in ohm

+

+//calculation:

+//voltage

+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)

+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)

+//The circuit diagram is shown in Figure 32.7. Following the above procedure:

+//The network is redrawn with the 50/_90° V source removed as shown in Figure 32.8

+//Currents I1, I2 and I3 are labelled as shown in Figure 32.8.

+I1 = V1/(r1 + r2*R/(R + r2))

+I2 = (r2/(r2 + R))*I1

+I3 = (R/(r2 + R))*I1

+//The network is redrawn with the 100/_0° V source removed as shown in Figure 32.9

+//Currents I4, I5 and I6 are labelled as shown in Figure 32.9.

+I4 = V2/(r2 + r1*R/(r1 + R))

+I5 = (r1/(r1 + R))*I4

+I6 = (R/(r1 + R))*I4

+//Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown.

+//Current in the 20 ohm load,

+I20 = I2 + I5

+//Current in the 100/_0° V source

+IV1 = I1 - I6

+//Current in the 50/_90° V source

+IV2 = I4 - I3

+

+printf("\n\n Result \n\n")

+printf("\n (a)current in the 20 ohm load is %.3f + (%.3f)i A",real(I20), imag(I20))

+printf("\n (b)Current in the 100/_0° V source is %.3f + (%.3f)i A",real(IV1), imag(IV1))

+printf("\n (b)Current in the 50/_90° V source is %.3f + (%.3f)i A",real(IV2), imag(IV2))
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diff --git a/608/CH32/EX32.02/32_02.sce b/608/CH32/EX32.02/32_02.sce
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+//Problem 32.02:Use the superposition theorem to determine the current in the 4 ohm resistor of the network shown in Figure 32.11.

+

+//initializing the variables:

+V1 = 12; // in volts

+V2 = 20; // in volts

+R1 = 5; // in ohm

+R2 = 4; // in ohm

+R3 = 2.5; // in ohm

+R4 = 6; // in ohm

+R5 = 2; // in ohm

+

+//calculation:

+//Removing the 20 V source gives the network shown in Figure 32.12.

+//Currents I1 and I2 are shown labelled in Figure 32.12

+Re1 = (R4*R5/(R4 + R5)) + R3

+Re2 = Re1*R2/(Re1  + R2) + R1

+I1 = V1/Re2

+I2 = (R2/(Re1 + R2))*I1

+//Removing the 12 V source from the original network gives the network shown in Figure 32.14.

+//Currents I3, I4 and I5 are shown labelled in Figure 32.14.

+Re3 = (R1*R2/(R1 + R2)) + R3

+Re4 = Re3*R4/(Re3 + R4) + R5

+I3 = V2/Re4

+I4 = (R4/(Re3 + R4))*I3

+I5 = (R1/(R1 + R2))*I4

+//Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by

+Ir4 = I5 - I2

+

+printf("\n\n Result \n\n")

+printf("\ncurrent in the 4 ohm resistor of the network is %.3f A",Ir4)
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diff --git a/608/CH32/EX32.03/32_03.sce b/608/CH32/EX32.03/32_03.sce
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+//Problem 32.03: Use the superposition theorem to obtain the current flowing in the (4 + i3) ohm impedance of Figure 32.16.

+

+//initializing the variables:

+rv1 = 30; // in volts

+rv2 = 30; // in volts

+thetav1 = 45; // in degrees

+thetav2 = -45; // in degrees

+R1 = 4; // in ohm

+R2 = 4; // in ohm

+R3 = %i*3; // in ohm

+R4 = -1*%i*10; // in ohm

+

+//calculation:

+//voltage

+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)

+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)

+//The network is redrawn with V2 removed, as shown in Figure 32.17.

+//Current I1 and I2 are shown in Figure 32.17. From Figure 32.17,

+Re1 = R4*(R2 + R3)/(R4 + R3 + R2)

+Re2 = Re1 + R1

+//current

+I1 = V1/Re2

+I2 = (R4/(R2 + R3 + R4))*I1

+//The original network is redrawn with V1 removed, as shown in Figure 32.18

+//Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18,

+Re3 = R1*(R2 + R3)/(R1 + R3 + R2)

+Re4 = Re3 + R4

+I3 = V2/Re4

+I4 = (R1/(R2 + R3 + R1))*I3

+//If the network of Figure 32.18 is superimposed on the network of Figure 32.17, it can be seen that the current in the (4+i3) ohm impedance is given by

+Ir4i3 = I2 - I4

+

+printf("\n\n Result \n\n")

+printf("\ncurrent in the (4 + i3) ohm impedance of the network is %.3f + (%.3f)i A",real(Ir4i3), imag(Ir4i3))
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diff --git a/608/CH32/EX32.04/32_04.sce b/608/CH32/EX32.04/32_04.sce
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+//Problem 32.04: For the a.c. network shown in Figure 32.19 determine, using the superposition theorem, (a) the current in each branch, (b) the magnitude of the voltage across the(6 + i8) ohm impedance, and (c) the total active power delivered to the network.

+

+//initializing the variables:

+E1 = 5 + %i*0; // in volts

+E2 = 2 + %i*4; // in volts

+Z1 = 3 + %i*4; // in ohm

+Z2 = 2 - %i*5; // in ohm

+Z3 = 6 + %i*8; // in ohm

+

+//calculation:

+//The original network is redrawn with E2 removed, as shown in Figure 32.20.

+//Currents I1, I2 and I3 are labelled as shown in Figure 32.20.

+Ze1 = Z3*Z2/(Z3 + Z2)

+Ze2 = Ze1 + Z1

+//current

+I1 = E1/Ze2

+I2 = (Z2/(Z3 + Z2))*I1

+I3 = (Z3/(Z3 + Z2))*I1

+//The original network is redrawn with E1 removed, as shown in Figure 32.22

+//Currents I4, I5 and I6 are shown labelled in Figure 32.22 with I4 flowing away from the positive terminal of the E2 source.

+Ze3 = Z3*Z1/(Z3 + Z1)

+Ze4 = Ze3 + Z2

+I4 = E2/Ze4

+I5 = (Z1/(Z3 + Z1))*I4

+I6 = (Z3/(Z3 + Z1))*I4

+//If the network of Figure 32.18 is superimposed on the network of Figure 32.17, it can be seen that the current in the (4+i3) ohm impedance is given by

+i1 = I1 + I6

+i2 = I3 + I4

+i3 = I2 - I5

+//magnitude

+i1mag = (real(i1)^2 + imag(i1)^2)^0.5

+i2mag = (real(i2)^2 + imag(i2)^2)^0.5

+E1mag = (real(E1)^2 + imag(E1)^2)^0.5

+E2mag = (real(E2)^2 + imag(E2)^2)^0.5

+//phase

+phi1 = atan(imag(i1)/real(i1))

+phi2 = atan(imag(i2)/real(i2))

+//voltage across the(6 + i8) ohm impedance

+V6i8 = i3*Z3

+V6i8m = (real(V6i8)^2 + imag(V6i8)^2)^0.5

+//power

+P = (E1mag*i1mag*cos(phi1)) + (E2mag*i2mag*cos(phi2 - atan(imag(E2)/real(E2))))

+

+printf("\n\n Result \n\n")

+printf("\n(b)current in the (6 + i8) ohm resistor of the network is %.3f V",V6i8m)

+printf("\n(c)the total active power delivered to the network is %.3f W",P)
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diff --git a/608/CH32/EX32.05/32_05.sce b/608/CH32/EX32.05/32_05.sce
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+//Problem 32.05: Use the superposition theorem to determine, for the network shown in Figure 32.25, (a) the magnitude of the current flowing in the capacitor, (b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and (d) the total active power taken from the supply.

+

+//initializing the variables:

+rv1 = 50; // in volts

+rv2 = 30; // in volts

+thetav1 = 0; // in degrees

+thetav2 = 90; // in degrees

+R1 = 20; // in ohm

+R2 = 5; // in ohm

+R3 = -1*%i*3; // in ohm

+R4 = 8; // in ohm

+R5 = 8; // in ohm

+

+//calculation:

+//voltage

+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)

+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)

+//The network is redrawn with the V2 source removed, as shown in Figure 32.26.

+//Currents I1 to I5 are shown labelled in Figure 32.26. 

+//current

+Re1 = R4*R5/(R5 + R4) + R3

+Re2 = Re1*R2/(R2 + Re1)

+I1 = V1/(Re2 + R1)

+I2 = (Re1/(R2 + Re1))*I1

+I3 = (R2/(Re1 + R2))*I1

+I4 = (R4/(R4 + R5))*I3

+I5 = I3 - I4

+//The original network is redrawn with the V1 source removed, as shown in Figure 32.27.

+//Currents I6 to I10 are shown labelled in Figure 32.27

+Re3 = R1*R2/(R1 + R2)

+Re4 = Re3 + R3

+Re5 = Re4*R4/(Re4 + R4)

+Re6 = Re5  + R5

+I6 = V2/Re6

+I7 = (Re4/(Re4 + R4))*I6

+I8 = (R4/(Re4 + R4))*I6

+I9 = (R1/(R1 + R2))*I8

+I10 = (R2/(R1 + R2))*I8

+//current flowing in the capacitor is given by

+Ic = I3 - I8

+//magnitude of the current in the capacitor

+Icmag = (real(Ic)^2 + imag(Ic)^2)^0.5

+//

+i1 = I2 + I9

+i1mag = (real(i1)^2 + imag(i1)^2)^0.5

+//magnitude of the p.d. across the 5 ohm resistance is given by

+Vr5m = i1mag*R2

+//Active power dissipated in the 20 ohm resistance is given by

+i2 = I1 - I10

+i2mag = (real(i2)^2 + imag(i2)^2)^0.5

+phii2 = atan(imag(i2)/real(i2))

+Pr20 = R1*(i2mag)^2

+//Active power developed by the V1

+P1 = rv1*i2mag*cos(phii2)

+//Active power developed by V2 source

+i3 = I6 - I5

+i3mag = (real(i3)^2 + imag(i3)^2)^0.5

+phii3 = atan(imag(i3)/real(i3))

+if ((imag(i3)>0) & (real(i3)<0)) then

+    phii3 = phii3 + %pi

+end

+P2 = rv2*i3mag*cos(phii3 - (thetav2*%pi/180))

+//Total power developed

+P = P1 + P2

+

+printf("\n\n Result \n\n")

+printf("\n(a)the magnitude of the current flowing in the capacitor is %.2f A",Icmag)

+printf("\n(b) the p.d. across the 5 ohm resistance is %.3f V",Vr5m)

+printf("\n(c)the active power dissipated in the 20 ohm resistance is %.0f W",Pr20)

+printf("\n(d)the total active power taken from the supply is %.1f W",P)
\ No newline at end of file