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+//Problem 30.02: Determine the current flowing in the 2 ohm resistor of the circuit shown in Figure 30.5 using Kirchhoff’s laws. Find also the power dissipated in the 3 ohm resistance.
+
+//initializing the variables:
+V = 8; // in volts
+R1 = 1; // in ohm
+R2 = 2; // in ohm
+R3 = 3; // in ohm
+R4 = 4; // in ohm
+R5 = 5; // in ohm
+R6 = 6; // in ohm
+
+//calculation:
+//Currents and their directions are assigned as shown in Figure 30.6.
+//Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary. loop ABCDE, and loop EDGF and loop DCHG
+//using kirchoff rule in 3 loops
+//three eqns obtained
+//R5*I1 + (R6 + R4)*I2 - R4*I3 = V
+//-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0
+// R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0
+//using determinants
+d1 = [V (R6 + R4) -1*R4; 0 (R6 + R1) R2; 0 (-1*(R3 + R4)) (R2 + R3 + R4)]
+D1 = det(d1)
+d2 = [R5 V -1*R4; -1*R1 0 R2; R3 0 (R2 + R3 + R4)]
+D2 = det(d2)
+d3 = [R5 (R6 + R4) V; -1*R1 (R6 + R1) 0; R3 (-1*(R3 + R4)) 0]
+D3 = det(d3)
+d = [R5 (R6 + R4) -1*R4; -1*R1 (R6 + R1) R2; R3 (-1*(R3 + R4)) (R2 + R3 + R4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D 
+//Current in the 2 ohm resistance
+I = I1 - I2 + I3
+//power dissipated in the 3 ohm resistance
+P3 = R3*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)current through 2 ohm resistor is %.2f A",I2)
+printf("\n (b)power dissipated in the 3 ohm resistor is %.2f W",P3)
+\ No newline at end of file