Question

In order to measure the internal resistance r1 of a cell of emf E, a meter bridge of wire resistance R0 = 50 Ω, a resistance $\frac{R_0}{2}$, another cell of emf $\frac{E}{2}$ (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r 1 = ____

Answer 1

Current will flow in the main circuit
$\begin{array}{l}I = \frac{E}{r_{1}+\frac{3R_{0}}{2}}\end{array}$
$\begin{array}{l}+E- IR_{0}\times 0.72-Ir_{1}-\frac{E}{2}=0\end{array}$
$\begin{array}{l}\frac{E}{2}=\frac{2E}{2r_{1}+3R_{0}}\times [0.72R_{0}+r_{1}]\end{array}$
$\begin{array}{l}2r_{1}+3R_{0}= 4[0.72R_{0}+r_{1}]\end{array}$
0.12R0 = 2r1
r1 = 3Ω