Question

In order to measure the internal resistance $r_{1}$ of a cell of emf $E$, a meter bridge of wire resistance $R_{0}=50\, \Omega$, a resistance $R _{0} / 2$, another cell of emf $E / 2$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $\ell=72\, cm$, then the value of $r _{1}=$ _____ $\Omega$

Answer 1

$R _{ AB }=50\, \Omega$
So, $R _{\Lambda P }=\frac{50}{100} \times 72=36 \,\Omega$
$I =\frac{\varepsilon}{ r _{1}+50+25}$ ...(i)
$-36 I -\frac{\varepsilon}{2}- Ir _{1}+\varepsilon=0$
Solving equation (i) and (ii)
$r _{1}=3 \,\Omega$