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(D) $1$. Let the potential at node $H$ be $0 \ V$. Then the potential at $A$ is $E$.$2$. The current through branch $CF$ is $I_{CF} = 1 \ A$. The resi

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$12$

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(D) $1$. Let the potential at node $H$ be $0 \ V$. Then the potential at $A$ is $E$.$2$. The current through branch $CF$ is $I_{CF} = 1 \ A$. The resistance of branch $CF$ is $2 \ \Omega$. Thus,the potential difference $V_C - V_F = I_{CF} 	imes R_{CF} = 1 \ A 	imes 2 \ \Omega = 2 \ V$. Since $F$ is connected to $H$ (ground),$V_F = 0 \ V$,so $V_C = 2 \ V$.$3$. The branch $CD$ has resistance $1 \ \Omega$ and branch $DE$ has resistance $1 \ \Omega$. The total resistance of the path $CDE$ is $1 + 1 = 2 \ \Omega$. The current $I_{CDE} = V_C / 2 \ \Omega = 2 \ V / 2 \ \Omega = 1 \ A$.$4$. The total current leaving node $C$ towards $B$ is $I_{BC} = I_{CF} + I_{CDE} = 1 \ A + 1 \ A = 2 \ A$.$5$. The potential at $B$ is $V_B = V_C + I_{BC} 	imes R_{BC} = 2 \ V + (2 \ A 	imes 2 \ \Omega) = 2 \ V + 4 \ V = 6 \ V$.$6$. The current through branch $BG$ is $I_{BG} = V_B / R_{BG} = 6 \ V / 6 \ \Omega = 1 \ A$.$7$. The total current leaving the battery is $I_{total} = I_{BC} + I_{BG} = 2 \ A + 1 \ A = 3 \ A$.$8$. The potential at $A$ is $V_A = V_B + I_{total} 	imes R_{AB} = 6 \ V + (3 \ A 	imes 2 \ \Omega) = 6 \ V + 6 \ V = 12 \ V$.$9$. Since $V_H = 0 \ V$,the $emf$ $E = V_A - V_H = 12 \ V$.

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

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