An unknown resistance $R_{1}$ is connected in series with a resistance of 10$\Omega$. This combination is connected to one gap of meter bridge while a resistance $R_{2}$is connected in the other gap. The balance point is at 50 cm, Now, when the 10$\Omega$ resistance is removed the balance point shifts 40cm. The value of $R_{1}$(in ohm) is

Question

Dr. Abubacker · Accepted Answer

(a) The balance condition of a meter bridge experiment
$\frac{R}{S} = \frac{l_{1}}{\left( 100 - l_{1} ight)}$
Here, $R = R_{1},\ S = R_{2}$
$	herefore\ \ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{\left( 100 - l_{1} ight)}$
Ist case
$\frac{R_{1} + 10}{R_{2}} = \frac{50}{50}\ \ \ \ \ \ \ \ldots\ldots\ldots(i)$
$\Longrightarrow R_{1} + 10 = R_{2}\ $
IInd case
$\frac{R_{1}}{R_{2}} = \frac{40}{60}$
$\Longrightarrow \ \ R_{2} = \frac{60}{40}R_{1}\ \ \ \ \ \ \ \ldots..(ii)$
So, Eqs. (i) and (ii) give
$R_{1} + 10 = \frac{60}{40}R_{1}$
$\Longrightarrow \frac{60}{40}R_{1} - R_{1} = 10\ $
$\Longrightarrow \frac{20}{40}R_{1} = 10\ $
$\Longrightarrow R_{1} = \frac{10 	imes 40}{20}$
$	herefore\ \ \ \ R_{1} = 20\Omega$

Explanation

Reviewed by Dr. Abubacker

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