A galvanometer of resistance 100 is converted to a voltmeter of range 10 ~V by connecting a resistance of 10 k . The resistance required to convert the same galvanometer to an ammeter of range 1 ~A is

Correct answer: D. 0.1 Here, V=10 V , aligned & R=10 k , & G=100 =0.1 k & V=I_g(G+R) ; I_g 1 mA aligned aligned & (I-I_g) S=I_g G & or S=(I_g G)/(I-I_g) aligned = 1 × 10^ -3 × 100 1-1 × 10^ -3 =0.1

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Current Electricity

AIIMS 2018 · 26 May · Physics · Current Electricity

A galvanometer of resistance 100 is converted to a voltmeter of...

A galvanometer of resistance

100 \Omega

is converted to a voltmeter of range

10 \mathrm{~V}

by connecting a resistance of

10 \mathrm{k} \Omega

. The resistance required to convert the same galvanometer to an ammeter of range

1 \mathrm{~A}

Options

A $0.4 \Omega$
B $0.3 \Omega$
C $1.2 \Omega$
D $0.1 \Omega$

Correct answer

D. $0.1 \Omega$

Step-by-step solution

Here,

V=10 \mathrm{ V} ,

\begin{aligned} & R=10 \mathrm{k} \Omega, \\ & G=100 \Omega=0.1 \mathrm{k} \Omega \\ & V=I_g(G+R) ; I_g \approx 1 \mathrm{ mA}\end{aligned}

\begin{aligned} & \left(I-I_g\right) S=I_g G \\ & \text{\text{ or } } \quad S=\frac{I_g G}{I-I_g}\end{aligned}

=\frac{1 \times 10^{-3} \times 100}{1-1 \times 10^{-3}}=0.1 \Omega

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