Question

a moving coil galvanometer A has 200 turns and resistance 100 ohms another meter B has 100 turns and resistance 40 ohms all the other quantities are same in both the cases the current sensitivity of

Answer 1

Question is incomplete. I think correct question will be,

A moving coil galvanometer A has 200 turns and resistance 100 ohms another meter B has 100 turns and resistance 40 ohms all the other quantities are same in both the cases, then calculate the change in current sensitivity and voltage sensitivity.

Current sensitivity only depends upon the no. of turns and not on resistance. But Voltage sensitivity depends both on no. of turns and Resistance.

Change in Current intensity = $\frac{N - N' }{N}$

= (200 - 100)/200

= 1/2

= 0.5

In %, 0.5 × 100 = 50%.

Change in Voltage sensitivity = $\frac{\frac{N}{R} - \frac{N'}{R'}  }{\frac{N}{R} }$

= $\frac{\frac{200}{100} - \frac{100}{40}  }{\frac{200}{100} }$

= $\frac{2 - 2.5 }{2 }\\-0.25$

In %, -0.25 × 100 = -25%  (-ive means, that voltage sensitivity is decreasing.)

Hope it helps.

Answer 2

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