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+//Problem 30.04: For the network shown in Figure 30.9, use Kirchhoff’s laws to determine the magnitude of the current in the (4 + i3)ohm impedance.
+
+//initializing the variables:
+rv1 = 10; // in volts
+rv2 = 12; // in volts
+rv3 = 15; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 0; // in degrees
+thetav3 = 0; // in degrees
+R1 = 4; // in ohm
+R2 = -1*5*%i; // in ohm
+R3 = 8; // in ohm
+R4 = 4; // in ohm
+R5 = %i*3; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+V3 = rv3*cos(thetav3*%pi/180) + %i*rv3*sin(thetav3*%pi/180)
+//Currents I1, I2 and I3 with their directions are shown in Figure 30.10.
+//Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF
+Z4 = R4 + R5
+//using kirchoff rule in 3 loops
+//three eqns obtained
+//R1*I1 + R2*I2 = V1 + V2
+//-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3
+// -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3
+//using determinants
+d1 = [(V1 + V2) R2 0; (V2 + V3) (R3 + R2) R3; V3 R3 (R3 + Z4)]
+D1 = det(d1)
+d2 = [R1 (V1 + V2) 0; -1*R3 (V2 + V3) R3; -1*R3 V3 (R3 + Z4)]
+D2 = det(d2)
+d3 = [R1 R2 (V1 + V2); -1*R3 (R3 + R2) (V2 + V3); -1*R3 R3 V3]
+D3 = det(d3)
+d = [R1 R2 0; -1*R3 (R3 + R2) R3; -1*R3 R3 (R3 + Z4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D 
+I3mag = (real(I3)^2 + imag(I3)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n magnitude of the current through (4 + i3)ohm impedance is %.2f A",I3mag)
+\ No newline at end of file