Physics · Current Electricity

JEE Advanced 2018 — Paper 2 — Question 9

A moving coil galvanometer has 50 turns and each turn has an area $2 \times 10^{-4} \mathrm{~m}^{2}$ . The magnetic field produced by the magnet inside the galvanometer is 0.02 T . The torsional constant of the suspension wire is $10^{-4} \mathrm{~N} \mathrm{~m} \mathrm{rad}^{-1}$ . When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad . The resistance of the coil of the galvanometer is $50 \Omega$ . This galvanometer is to be converted into an ammeter capable of measuring current in the range $0-1.0 \mathrm{~A}$ . For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is $\_\_\_\_$ .

Answer: 5.56

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{NiAB}=\mathrm{C} \theta$

$\mathrm{i}_{\mathrm{g}}=\frac{\mathrm{C} \theta_{\max }}{\mathrm{NAB}}=\frac{10^{-4} \times 0.2}{50 \times 2 \times 10^{-4} \times 0.02}=0.1 \mathrm{~A}$

$\therefore \mathrm{S}=\frac{\mathrm{i}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}}}{\left(\mathrm{I}-\mathrm{i}_{\mathrm{g}}\right)}=\frac{0.1 \times 50}{(1-0.1)}=\frac{50}{9}=5.555 \Omega$

$\therefore$ shunt resistance, $\mathrm{S}=5.56 \Omega$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments