SATHEE: moving-charges-and-magnetism Question 45

moving-charges-and-magnetism Question 45

Question: Q. 7. The current flowing in the galvanometer $G$ when the key $K_{2}$ is kept open is $I$ . On closing the key $K_{2}$ , the current in the galvanometer becomes $I / n$ , where $n$ is integer.

Obtain an expression for resistance $R_{g}$ of the galvanometer in terms of $R, S$ and $n$ . To what form does this expression reduce when the value of $R$ is very large as compared to $S$ ?

U] [CBSE SQP 2014]

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Solution:

Ans. With key $K_{2}$ open, the current I in the galvanometer is given by

$I = \frac{E}{R + R_{G}}$

When $K_{2}$ is closed, the equivalent resistance, say $R^{'}$ , of the parallel combination of $S$ and $R_{G}$ is given by

$R^{'} = \frac{S R_{G}}{S + R_{G}}$

The total current, say $I^{'}$ drawn from the battery would now be

$I^{'} = \frac{E}{R + R^{'}}$

This current gets subdivided in the inverse ratio of $S$ and $R_{G}$ . Hence the current $I^{''}$ through $G$ , would now be given by

$\begin{matrix} I^{''} = \frac{S}{S + R_{G}} \cdot I^{'} = \frac{S}{(S + R_{G})} \cdot \frac{E}{(R + R^{'})} = \frac{S E}{(S + R_{G}) [R + \frac{S R_{G}}{S + R_{G}}]} \end{matrix}$

$\begin{aligned} = \frac{S E}{R S + R R_{G} + S R_{G}} But & I^{''} & = \frac{I}{n} = \frac{1}{n} (\frac{E}{R + R_{G}}) & ∴ & \frac{E}{n (R + R_{G})} & = \frac{S . E}{R S + R R_{G} + S R_{G}} 1 / 2 Or & n R S + n S R_{G} & = R S + R R_{G} + S R_{G} Or & (n - 1) R S & = R R_{G} - (n - 1) S R_{G} Or & (n - 1) R S & = R_{G} [R - (n - 1) S] & R_{G} & = \frac{(n - 1) R S}{R - (n - 1) S} \end{aligned}$

This is the required expression

When $R » S$ , we have

$R_{G} ≃ \frac{(n - 1) R S}{R} = (n - 1) S$

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Moving Charges And Magnetism

moving-charges-and-magnetism Question 45

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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