JEE Advanced 2021 Paper 1 · Physics · Magnetic Effects of Current · medium

A moving-coil galvanometer of resistance 50 Ω gives full-scale deflection for a current of 2.0 mA. It is converted into a 0 to 1 A ammeter and…

One or more options correct

A moving-coil galvanometer of resistance

50\ \Omega

gives full-scale deflection for a current of

2.0\ \mathrm{mA}

. It is converted into a

0

1\ \mathrm{A}

ammeter and, separately, into a

0

10\ \mathrm{V}

voltmeter. Choose the correct statement(s).

A. The ammeter needs a shunt of about $0.10\ \Omega$ in parallel with the galvanometer.
B. The voltmeter needs a series resistance of $4950\ \Omega$ .
C. The voltmeter, connected across one of two $5.0\ \mathrm{k\Omega}$ resistors in series across an ideal $12\ \mathrm{V}$ source, reads $6.0\ \mathrm{V}$ .
D. The ammeter, inserted in series with a $10\ \Omega$ resistor across an ideal $10\ \mathrm{V}$ source, reads within $0.1\%$ of $1.000\ \mathrm{A}$ .

Show answer and solution

Answer: A, B

(A) The shunt is

S=\frac{I_gG}{I-I_g}=\frac{0.002\times50}{0.998}=0.1002\ \Omega

: correct. (B) The series resistance is

R=\frac{V}{I_g}-G=5000-50=4950\ \Omega

, so the voltmeter's total resistance is

5.0\ \mathrm{k\Omega}

: correct. (C) The voltmeter of

5\ \mathrm{k\Omega}

in parallel with one

5\ \mathrm{k\Omega}

resistor gives

2.5\ \mathrm{k\Omega}

, so the voltage divider gives

12\times\frac{2.5}{2.5+5}=4.0

V, not

6.0

V: incorrect. (D) The ammeter's resistance is

\frac{0.1002\times50}{50.1002}=0.0999\ \Omega

, so the current is

\frac{10}{10.0999}=0.990

A, about

1\%

below

1.000

A, not within

0.1\%

: incorrect. Hence (A) and (B).

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake