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(C) The circuit consists of two parallel branches connected across a $2\,V$ battery. Each branch contains three resistors of $5\,\Omega$ each,connecte

Vedclass · Accepted Answer

$\frac{4}{3}\,V$

Vedclass · Answer

(C) The circuit consists of two parallel branches connected across a $2\,V$ battery. Each branch contains three resistors of $5\,\Omega$ each,connected in series. The total resistance of each branch is $R_{eq} = 5\,\Omega + 5\,\Omega + 5\,\Omega = 15\,\Omega$. The potential difference across each resistor in a series circuit is proportional to its resistance. Since all resistors are identical $(5\,\Omega)$,the total potential difference of $2\,V$ is divided equally among the three resistors in each branch. Thus,the potential difference across each resistor is $V_{resistor} = \frac{2\,V}{3} = \frac{2}{3}\,V$. The path between points $A$ and $B$ consists of two resistors of $5\,\Omega$ each. Therefore,the potential difference between $A$ and $B$ is $V_{AB} = V_{resistor1} + V_{resistor2} = \frac{2}{3}\,V + \frac{2}{3}\,V = \frac{4}{3}\,V$.

The potential difference between points $A$ and $B$ of the adjoining figure is

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