A resistance of 4,Omega and a wire of length | vedclass

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(B) $1$. Calculate the current $I$ in the primary circuit:$I = \frac{V_{total}}{R_{total}} = \frac{10\,V}{4\,\Omega + 5\,\Omega + 1\,\Omega} = \frac{1

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$3$

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(B) $1$. Calculate the current $I$ in the primary circuit:$I = \frac{V_{total}}{R_{total}} = \frac{10\,V}{4\,\Omega + 5\,\Omega + 1\,\Omega} = \frac{10}{10} = 1\,A$.$2$. Calculate the potential drop across the potentiometer wire:The potential difference across the $5\,m$ wire is $V_{wire} = I 	imes R_{wire} = 1\,A 	imes 5\,\Omega = 5\,V$.$3$. Calculate the potential gradient $x$:$x = \frac{V_{wire}}{L} = \frac{5\,V}{5\,m} = 1\,V/m$.$4$. The balancing length is $l = 300\,cm = 3\,m$.The potential difference across the balancing length is $V_{bal} = x 	imes l = 1\,V/m 	imes 3\,m = 3\,V$.$5$. Since the two identical cells of $e.m.f.$ $E$ are connected in parallel,their equivalent $e.m.f.$ is $E_{eq} = E$.Therefore,$E = 3\,V$.

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