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

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Aicharya · Accepted Answer

Consider the following diagram: The potential drop across the external shunt resistance $R$ is equal to the potential drop across the battery of emf $E$ and internal resistance $r$. $\begin{aligned} & \Delta V=E-i r=i R \ & \Rightarrow i=\frac{E}{(R+r)} \ & 	herefore \Delta V=i R=\frac{E R}{(R+r)}\end{aligned}$ $\because \Delta V \propto L$, i.e., the balance length is directly proportional to the potential drop. Therefore, $\frac{E R}{(R+r)} \propto L$ In case of internal resistance measurement by potentiometer, considering two balance lengths $\left(L_1, L_2ight)$ corresponding to shunt resistors $\left(R_1, R_2ight)$ respectively, $\Rightarrow \frac{V_1}{V_2}=\frac{L_1}{L_2}=\frac{\left\{E R_1 /\left(R_1+right)ight\}}{\left\{E R_2 /\left(R_2+right)ight\}}=\frac{R_1\left(R_2+right)}{R_2\left(R_1+right)}$. Here, $L_1=1 \mathrm{~m}, L_2=1.5 \mathrm{~m},$ $\Rightarrow \frac{1}{1.5}=\frac{3(6+r)}{6(3+r)}$ 6+2r =9+1.5r $\Rightarrow r=6 \Omega$

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1…

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