JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Current Electricity

A moving coil galvanometer has 50 turns and each turn has an area

2 \times 1 0^{- 4} m^{2}

. The magnetic field produced by the magnet inside the galvanometer is

0.02 T

. The torsional constant of the suspension wire is

1 0^{- 4} N m r a d^{- 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

50Ω

. This galvanometer is to be converted into an ammeter capable of measuring current in the range

0 - 1.0 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

Solution

At full scale deflection,

N i A B = C θ

, so

i_{g} = \frac{C θ}{N A B} = \frac{1 0 ^{- 4} \times 0.2}{50 \times 2 \times 1 0 ^{- 4} \times 0.02} = 0.1

A. For an ammeter with range 0-1.0 A, the shunt resistance is

S = \frac{i _{g} R _{g}}{I - i _{g}} = \frac{0.1 \times 50}{1.0 - 0.1} = \frac{5}{0.9} = 5.56

Ω.

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Current Electricity

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