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+//Problem 13.11: A Wheatstone Bridge network is shown in Figure 13.32(a). Calculate the current flowing in the 32 ohm resistor, and its direction, using Th´evenin’s theorem. Assume the source of e.m.f. to have negligible resistance.

+

+//initializing the variables:

+E = 54; // in volts

+R1 = 2; // in ohms

+R2 = 14; // in ohms

+R3 = 3; // in ohms

+R4 = 11; // in ohms

+R5 = 32; // in ohms

+

+//calculation:

+//The 32ohm resistance branch is short-circuited as shown in Figure 13.32(b).

+//The p.d. between A and C,

+Vac = (R1/(R1 + R4))*E

+//The p.d. between B and C,

+Vbc = (R2/(R2 + R3))*E

+//Hence the p.d. between A and B

+Vab = Vbc - Vac

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

+r = R1*R4/(R1 + R4) + R2*R3/(R2 + R3)

+//The equivalent Th´evenin’s circuit is shown in Figure 13.32(f), the current in the 32 ohm resistance is given by:

+I = E/(r + R5)

+

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

+printf("\n the current in the 32 ohm resistance is given by %.0f A",I)
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