A cell,shunted by an 8 ; Omega resistance,is | vedclass

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(B) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The terminal potential difference $V$ across the cell when shunted by a resis

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$8$

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(B) Let $E$ be the $EMF$ of the cell and $r$ be its internal resistance. The terminal potential difference $V$ across the cell when shunted by a resistance $R$ is given by $V = E \left( \frac{R}{R+r} ight)$.Since the balancing length $l$ is directly proportional to the terminal potential difference $V$,we have $V \propto l$.For the first case: $V_1 \propto l_1 = 3 \; m$ with $R_1 = 8 \; \Omega$. Thus,$V_1 = k \cdot 3 = E \left( \frac{8}{8+r} ight)$.For the second case: $V_2 \propto l_2 = 2 \; m$ with $R_2 = 4 \; \Omega$. Thus,$V_2 = k \cdot 2 = E \left( \frac{4}{4+r} ight)$.Dividing the two equations: $\frac{3}{2} = \frac{E \left( \frac{8}{8+r} ight)}{E \left( \frac{4}{4+r} ight)} = \frac{8}{8+r} 	imes \frac{4+r}{4} = \frac{2(4+r)}{8+r}$.Solving for $r$: $3(8+r) = 4(4+r) \implies 24 + 3r = 16 + 4r \implies r = 8 \; \Omega$.

$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

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