The current sensitivity of a moving coil galvanometer is doubled by making the number of turns double. Then its voltage sensitivity will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Remain unchanged
Current sensitivity of galvanometer is,
$\frac{\phi}{I}=\frac{N+B}{K}$ ...(1)
Shy, Voltage sensitivity is,
$\frac{\phi}{V}=\frac{NAB}{K}\frac{I}{V}=\frac{NAB}{K}\frac{1}{R}$
⇒ Voltage sensitivity = Current sensitivity × $\frac{1}{R}$
If the current sensitivity is double, resistance also becomes twice.
∴ Voltage sensitivity = $2×\frac{\phi}{I}×\frac{1}{2R}=\frac{\phi}{IR}$
∴ it remain unchanged.