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Question

Physics Question on Moving charges and magnetism

Current sensitivity of a moving coil galvanometer is 5div/mA5 \,div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20div/V20 \,div/V. The resistance of the galvanometer is

A

500Ω500 \, \Omega

B

40Ω40 \, \Omega

C

250Ω250 \, \Omega

D

25Ω25 \, \Omega

Answer

250Ω250 \, \Omega

Explanation

Solution

Current sensitivity
IS=NBACI_S = \frac{NBA}{C}
Voltage sensitivity
VS=NBACRGV_S = \frac{NBA}{CR_G}
So, resistance of galvanometer
RG=ISVS=5×120×103=500020=250ΩR_G = \frac{I_S}{V_S} = \frac{5 \times 1}{20 \times 10^{-3}} = \frac{5000}{20} = 250 \, \Omega