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+//Problem 13.14: Use Norton’s theorem to determine the current flowing in the 3 ohm resistance of the network shown in Figure 13.36(a). The voltage source has negligible internal resistance.

+

+//initializing the variables:

+V = 24; // in volts

+R1 = 20; // in ohms

+R2 = 5; // in ohms

+R3 = 10; // in ohms

+R4 = 5/3; // in ohms

+R5 = 3; // in ohms

+

+//calculation:

+//The 3ohm resistance branch is short-circuited as shown in Figure 13.36(b).

+//Figure 13.36(c) is equivalent to Figure 13.36(b).

+Isc = V/R2

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

+r = R3*R2/(R3 + R2)

+//From the Norton equivalent network shown in Figure 13.36(f) the current in the 3ohm resistance is given by:

+I = (r/(r + R4 + R5))*Isc

+

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

+printf("\n the current in the 3ohm resistance is given by %.0f A",I)
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