A moving coil galvanometer has 50 turns and each turn has an area The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is . When a current flows through the galvanometer, a full scale

Question

A moving coil galvanometer has 50 turns and each turn has an area $2 \times 10^{- 4} m^{2} .$ The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is $10^{- 4}$ $N m r a d^{- 1}$ . 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 $50 Ω$ . 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 _______. [2018]

Smart Achievers · Accepted Answer

(5.56)

[ $Given: B = 0.02 T, C = 10^{- 4} {Nm rad}^{- 1}$

$θ = 0.2 rad$

$0 - 1.0 A$ and

$A = 2 \times 10^{- 4} m^{2}$ ]

We know, $C θ = N B A I_{g}$

$∴$ $I_{g} = \frac{C θ}{N B A} = \frac{10^{- 4} \times 0.2}{50 \times 2 \times 10^{- 4} \times 0.02} = 0.1 A$

To convert a galvanometer to an ammeter, a shunt is used in parallel to the galvanometer.

$I_{g} \times G = (I - I_{g}) S$

$∴$ $S = \frac{I_{g} G}{I - I_{g}} = \frac{0.1 \times 50}{1 - 0.1} = \frac{50}{9} = 5.56 Ω$

Solution

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Solution

Ans. (5.56) [Given: B=0.02 T, C=10-4 Nm rad-1 θ=0.2 rad N=50 and A=2×10-4 m2] We know, Cθ=NBAIg ∴ Ig=CθNBA=10-4×0.250×2×10-4×0.02=0.1 A To convert a galvanometer to an ammeter, a shunt is used in parallel to the galvanometer. Ig×G=(I-Ig)S ∴ S=IgGI-Ig=0.1×501-0.1=509=5.56 Ω

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