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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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