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(A) The circuit consists of two parallel branches connected across the $9\,V$ battery.Branch $1$ (left side) has a total resistance of $2\,\Omega + 4\

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$3$

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(A) The circuit consists of two parallel branches connected across the $9\,V$ battery.Branch $1$ (left side) has a total resistance of $2\,\Omega + 4\,\Omega = 6\,\Omega$. The current in this branch is $I_1 = \frac{9\,V}{6\,\Omega} = 1.5\,A$.The potential at point $A$ relative to the left junction (let's call it $C$) is $V_C - V_A = I_1 	imes 2\,\Omega = 1.5 	imes 2 = 3\,V$.Branch $2$ (right side) has a total resistance of $4\,\Omega + 2\,\Omega = 6\,\Omega$. The current in this branch is $I_2 = \frac{9\,V}{6\,\Omega} = 1.5\,A$.The potential at point $B$ relative to the left junction $C$ is $V_C - V_B = I_1 	imes 4\,\Omega = 1.5 	imes 4 = 6\,V$.Now,the potential difference between $A$ and $B$ is $|V_A - V_B| = |(V_C - 3) - (V_C - 6)| = |6 - 3| = 3\,V$.

$A$ network of four resistances is connected to a $9\,V$ battery,as shown in the figure. The magnitude of the voltage difference between the points $A$ and $B$ is .......... $V.$

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