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In the circuit shown in Fig. 1.21, calculate the value of the unknown resistance R and the current flowing through it when the current in branch OC is zero.

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If current through R-ohm resistor is I amp, AO branch carries the same current, since, current through the branch CO is zero. This also means that the nodes C and O are at the equal potential. Then, equating voltage-drops, we have VAO = VAC

This means branch AC carries a current of 4I.

This is current of 4 I also flows through the branch CB. Equating the voltage-drops in branches OB and CB, 

1.5 × 4 I =R I, giving R = 6 Ω 

At node A, applying KCL, a current of 5 I flows through the branch BA from B to A. Applying KVL around the loop BAOB, I = 0.5 Amp.

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