A potentiometer wire of length L and a resist | vedclass

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(C) The current $I$ flowing through the potentiometer wire is given by the total e.m.f. divided by the total resistance of the primary circuit:$I = \f

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$\frac{E_0 r l}{(r + r_1) L}$

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(C) The current $I$ flowing through the potentiometer wire is given by the total e.m.f. divided by the total resistance of the primary circuit:$I = \frac{E_0}{r + r_1}$The potential difference $V$ across the entire length $L$ of the potentiometer wire is:$V = I r = \frac{E_0 r}{r + r_1}$The potential gradient $k$ along the wire is the potential difference per unit length:$k = \frac{V}{L} = \frac{E_0 r}{(r + r_1) L}$Since the unknown e.m.f. $E$ is balanced at a length $l$,the potential drop across length $l$ must equal $E$:$E = k l = \left( \frac{E_0 r}{(r + r_1) L} \right) l = \frac{E_0 r l}{(r + r_1) L}$

$A$ potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

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