Physics · Current Electricity

JEE Advanced 2021 — Paper 2 — Question 7

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 $\ell=72 \mathrm{~cm}$ , then the value of $\mathrm{r}_{1}=$ $\qquad$ $\Omega$ .

Answer: 3

Numerical answer — enter this value.

$\mathrm{R}_{\mathrm{AB}}=50 \Omega$ So, $\mathrm{R}_{\mathrm{AP}}=\frac{50}{100} \times 72=36 \Omega$

\begin{gathered} \mathrm{I}=\frac{\varepsilon}{\mathrm{r}_{1}+50+25} \end{gathered}

$-36 \mathrm{I}-\frac{\varepsilon}{2}-\mathrm{Ir}_{1}+\varepsilon=0$

Solving equation (i) and (ii) $\mathrm{r}_{1}=3 \Omega$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.