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(C) The circuit consists of two parallel branches connected across a $10\,V$ battery.Branch $1$ (top) has resistances $1\,\Omega$ and $3\,\Omega$ in s

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$5$

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(C) The circuit consists of two parallel branches connected across a $10\,V$ battery.Branch $1$ (top) has resistances $1\,\Omega$ and $3\,\Omega$ in series. Total resistance $R_1 = 1 + 3 = 4\,\Omega$.Current in branch $1$,$I_1 = \frac{10\,V}{4\,\Omega} = 2.5\,A$.The potential at $A$ relative to $C$ is $V_C - V_A = I_1 	imes 1\,\Omega = 2.5\,V$,so $V_A = V_C - 2.5$.Branch $2$ (bottom) has resistances $3\,\Omega$ and $1\,\Omega$ in series. Total resistance $R_2 = 3 + 1 = 4\,\Omega$.Current in branch $2$,$I_2 = \frac{10\,V}{4\,\Omega} = 2.5\,A$.The potential at $B$ relative to $C$ is $V_C - V_B = I_2 	imes 3\,\Omega = 7.5\,V$,so $V_B = V_C - 7.5$.Now,calculating the potential difference $V_A - V_B$:$V_A - V_B = (V_C - 2.5) - (V_C - 7.5) = -2.5 + 7.5 = 5\,V$.

$A$ battery of $emf$ $10\,V$ is connected to resistances as shown in the figure. The potential difference between $A$ and $B$,$(V_A - V_B)$ is ................ $V$.

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