In the given circuit,the terminal potential d | vedclass

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(B) The two resistors of $4 \,\Omega$ each are connected in parallel. Their equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{4}

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$2$

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(B) The two resistors of $4 \,\Omega$ each are connected in parallel. Their equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies R_{eq} = 2 \,\Omega$Now,the circuit consists of a cell of $EMF$ $E = 3 \,V$ and internal resistance $r = 1 \,\Omega$ connected in series with an external resistance $R_{eq} = 2 \,\Omega$.The total current $i$ in the circuit is:$i = \frac{E}{R_{eq} + r} = \frac{3}{2 + 1} = \frac{3}{3} = 1 \,A$The terminal potential difference $V$ of the cell is given by:$V = E - ir$$V = 3 - (1 	imes 1) = 3 - 1 = 2 \,V$

In the given circuit,the terminal potential difference of the cell is: (in $ V$)

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