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(C) In the loop $ABCDA$,the current $I_1 = \frac{4}{2+2} = 1\ A$ (clockwise).In the loop $EFGHE$,the current $I_2 = \frac{4}{3+3} = \frac{2}{3}\ A$ (c

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$8$

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(C) In the loop $ABCDA$,the current $I_1 = \frac{4}{2+2} = 1\ A$ (clockwise).In the loop $EFGHE$,the current $I_2 = \frac{4}{3+3} = \frac{2}{3}\ A$ (clockwise).To find $V_A - V_G$,we apply Kirchhoff's Voltage Law along the path $A-D-E-G$:$V_A - I_1(2\ \Omega) - 4\ V - I_2(3\ \Omega) = V_G$$V_A - (1\ A)(2\ \Omega) - 4\ V - (\frac{2}{3}\ A)(3\ \Omega) = V_G$$V_A - 2 - 4 - 2 = V_G$$V_A - 8 = V_G$$V_A - V_G = 8\ V$.

Three identical batteries each of $emf\ 4\ V$ and negligible internal resistance are connected as shown in the figure. The potential difference between points $A$ and $G$ $(V_A - V_G)$ is ................ $V$.

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