In order to measure internal resistance r₁ of a cell of emf E , a meter bridge of wire resistance

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3.Current Electricity

normal

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 $l=72 cm$, then the value of $r_1=$ . . . . $\Omega$

A

$2$

B

$3$

C

$4$

D

$5$

(IIT-2021)

Solution

$i\left(\frac{R_0}{2}+0.28 R_0\right)=\frac{E_0}{2}$

$i \times 0.78 R_0=\frac{E_0}{2}$

$i=\frac{E_0}{2 \times 0.78 R_0}=\frac{E_0}{ r _1+\frac{3}{2} R _0}$

$r _1+1.5 R _0=1.56 R _0$

$r _1=0.06 R _0$

$=0.06 \times 50=3 \Omega$

Standard 12

Physics

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