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(B) The circuit consists of two parallel branches connected across a $12 \,V$ battery.Branch $1$ (upper): Contains $5 \,\Omega$ and $20 \,\Omega$ resi

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$6$

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(B) The circuit consists of two parallel branches connected across a $12 \,V$ battery.Branch $1$ (upper): Contains $5 \,\Omega$ and $20 \,\Omega$ resistors in parallel,which are in series with a $4 \,\Omega$ resistor.Equivalent resistance of $5 \,\Omega$ and $20 \,\Omega$ in parallel is $R_p = \frac{5 	imes 20}{5 + 20} = \frac{100}{25} = 4 \,\Omega$.Total resistance of the upper branch is $R_{upper} = 4 \,\Omega + 4 \,\Omega = 8 \,\Omega$.Current in the upper branch is $I_{upper} = \frac{V}{R_{upper}} = \frac{12 \,V}{8 \,\Omega} = 1.5 \,A$.The potential difference across the $4 \,\Omega$ resistor is $V_4 = I_{upper} 	imes 4 \,\Omega = 1.5 \,A 	imes 4 \,\Omega = 6 \,V$.

$A$ circuit containing five resistors is connected to a battery with a $12 \,V$ emf as shown in the figure. The potential difference across the $4 \,\Omega$ resistor is ........... $V$.

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