JEE Main 2023 · 31 January · evening · Single correct · +4 / −1
The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50 %. The percentage change in voltage sensitivity of the galvanometer will be :
- (a) 0 %
- (b) 75 %
- (c) 100 %
- (d) 50 %
Answer and solution
Answer (a) 0 %
Solution
Physics sir · 4 steps
More turns means more wire, so the resistance grows by exactly the same factor as the current sensitivity.
- Two sensitivities. Current sensitivity is S_I = θ/I = NAB/k. Voltage sensitivity is S_V = θ/V = θ/IR, which is the current sensitivity divided by the coil's resistance R.
S_V = S_I/R = NAB/kR
- Effect of more turns. A 50% rise in S_I needs N to become 1.5N. Each turn uses the same length of the same wire, so the wire length and therefore R also become 1.5 times larger.
- Ratio. S_V' = (1.5 S_I)/(1.5 R) = S_I/R = S_V. The two factors of 1.5 cancel.
- So. The voltage sensitivity does not change, so the percentage change is 0%, option (a).
The idea to keep. Voltage sensitivity is current sensitivity divided by coil resistance. Adding turns of the same wire raises both by the same factor, so voltage sensitivity stays the same.
Answer from the NTA final answer key. Solution by PrepWiser.