To determine the internal resistance of a cel | vedclass

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(C) The potential difference across the external shunt resistance $R$ is given by $V = \frac{E R}{R + r}$, where $E$ is the electromotive force (emf)

Vedclass · Accepted Answer

$6 \, \Omega$

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(C) The potential difference across the external shunt resistance $R$ is given by $V = \frac{E R}{R + r}$, where $E$ is the electromotive force (emf) and $r$ is the internal resistance of the cell.Since the potentiometer measures the potential difference, the balance length $L$ is directly proportional to the potential difference $V$, i.e., $L \propto V$.Therefore, $\frac{L_1}{L_2} = \frac{V_1}{V_2} = \frac{R_1(R_2 + r)}{R_2(R_1 + r)}$.Given: $L_1 = 1 \, m$, $R_1 = 3 \, \Omega$, $L_2 = 1.5 \, m$, $R_2 = 6 \, \Omega$.Substituting the values:$\frac{1}{1.5} = \frac{3(6 + r)}{6(3 + r)}$$\frac{2}{3} = \frac{6 + r}{2(3 + r)}$$2(2)(3 + r) = 3(6 + r)$$4(3 + r) = 18 + 3r$$12 + 4r = 18 + 3r$$r = 18 - 12 = 6 \, \Omega$.Thus, the internal resistance of the cell is $6 \, \Omega$.

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \, m$ when the cell is shunted by $3 \, \Omega$ resistance and at a length $1.5 \, m$ when the cell is shunted by $6 \, \Omega$ resistance. The internal resistance of the cell is:

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