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(B) Let the potential at point $H$ be $0\,V$. Then the potential at $A$ is $E$. Branch $CF$ has a resistance of $2\,\Omega$ and a current of $1\,A$ fl

Vedclass · Accepted Answer

$12$

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(B) Let the potential at point $H$ be $0\,V$. Then the potential at $A$ is $E$. Branch $CF$ has a resistance of $2\,\Omega$ and a current of $1\,A$ flows through it. Thus,the potential difference across $CF$ is $V_C - V_F = 1\,A 	imes 2\,\Omega = 2\,V$. Since $F$ is connected to $H$ (ground),$V_F = 0\,V$,so $V_C = 2\,V$. Now,consider the branch $CD-DE$. The resistance is $1\,\Omega + 1\,\Omega = 2\,\Omega$. The current through this branch is $I_{CD} = (V_C - V_E) / 2\,\Omega = (2\,V - 0\,V) / 2\,\Omega = 1\,A$. The total current flowing through $BC$ is $I_{BC} = I_{CF} + I_{CD} = 1\,A + 1\,A = 2\,A$. The potential at $B$ is $V_B = V_C + I_{BC} 	imes 2\,\Omega = 2\,V + (2\,A 	imes 2\,\Omega) = 6\,V$. The current through branch $BG$ is $I_{BG} = V_B / 6\,\Omega = 6\,V / 6\,\Omega = 1\,A$. The total current flowing from the battery is $I = I_{BC} + I_{BG} = 2\,A + 1\,A = 3\,A$. The potential at $A$ is $V_A = V_B + I 	imes 2\,\Omega = 6\,V + (3\,A 	imes 2\,\Omega) = 12\,V$. Since $V_A = E$,the $emf$ of the battery is $12\,V$.

The figure shows a network of resistors and a battery. If $1\,A$ current flows through the branch $CF$,then find the $emf$ $E$ of the battery. (in $,V$)

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