Physics - Current Electricity Question with Solution | TestHub

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 $r$. $\begin{array}{rl}& \Delta V=E-ir=iR\\ & \Rightarrow i=\frac{E}{(R+r)}\\ & \therefore \Delta V=iR=\frac{ER}{(R+r)}\end{array}$ $\because \Delta V\propto L$, i.e., the balance length is directly proportional to the potential drop. Therefore, $\frac{ER}{(R+r)}\propto L$ In case of internal resistance measurement by potentiometer, considering two balance lengths $\left(L_1,L_2\right)$ corresponding to shunt resistors $\left(R_1,R_2\right)$ respectively, $\Rightarrow \frac{V_1}{V_2}=\frac{L_1}{L_2}=\frac{\left\{E{R}_1/\left(R_1+r\right)\right\}}{\left\{E{R}_2/\left(R_2+r\right)\right\}}=\frac{R_1\left(R_2+r\right)}{R_2\left(R_1+r\right)}$. 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$