$AB$ is a potentiometer wire of length $100\, cm$ and its resistance is $10 \,\Omega$. It is connected in series with a resistance $R = 40 \,\Omega$ and a battery of $e.m.f.$ $2 \,V$ and negligible internal resistance. If a source of unknown $e.m.f.$ $E$ is balanced by $40\, cm$ length of the potentiometer wire, the value of $E$ is ................. $V$
Medium
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(d) $E = \frac{e}{{(R + {R_h} + r)}}\,\frac{R}{L} \times l$$ = \frac{2}{{(10 + 40 + 0)}} \times \frac{{10}}{1} \times 0.4 = 0.16\,V$.
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