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(D) Let the potential at node $F$ be $0\,V$. Since $F, G, H$ are connected by wires,$V_F = V_G = V_H = 0\,V$.Given current through branch $CF$ is $1\,

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Both $(A)$ and $(B)$

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(D) Let the potential at node $F$ be $0\,V$. Since $F, G, H$ are connected by wires,$V_F = V_G = V_H = 0\,V$.Given current through branch $CF$ is $1\,A$,so $V_C - V_F = I_{CF} 	imes R_{CF} \implies V_C - 0 = 1\,A 	imes 2\,\Omega = 2\,V$.Now,consider the branch $CD-DE-EF$. The resistance of branch $DE$ is $1\,\Omega + 1\,\Omega = 2\,\Omega$ (in series).The current through branch $DE$ is $I_{DE} = \frac{V_C - V_E}{R_{DE}} = \frac{2\,V - 0\,V}{2\,\Omega} = 1\,A$.Now,applying Kirchhoff's Current Law at node $C$,the total current entering $C$ from $B$ is $I_{BC} = I_{CF} + I_{DE} = 1\,A + 1\,A = 2\,A$.Thus,the current through branch $DE$ is $1\,A$ and the current through branch $BC$ is $2\,A$.Therefore,both $(A)$ and $(B)$ are correct.

The figure shows a network of resistors and a battery. If $1\,A$ current flows through the branch $CF$,then find the current through the branches $DE$ and $BC$.

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