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+//Problem 13.12: Use Norton’s theorem to determine the current flowing in the 10 ohm resistance for the circuit shown in Figure 13.34(a).

+

+//initializing the variables:

+V = 10; // in volts

+R1 = 2; // in ohms

+R2 = 8; // in ohms

+R3 = 5; // in ohms

+R4 = 10; // in ohms

+

+//calculation:

+//The 10ohm resistance branch is short-circuited as shown in Figure 13.34(b).

+//Figure 13.34(c) is equivalent to Figure 13.34(b). Hence

+Isc = V/R1

+//the resistance ‘looking-in’ at a break made between A and B is given by

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

+//From the Norton equivalent network shown in Figure 13.34(d) the current in the 10 ohm resistance is given by:

+I = (r/(r + R3 + R4))*Isc

+

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

+printf("\n the current in the 10 ohm resistance is given by %.3f A",I)
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