Question

The current sensitivity of a moving coil galvanometer is doubled by making the number of turns double. Then its voltage sensitivity will be:

• A. Double
• B. Half
• C. $\frac{1}{4}$times
• D. Remain unchanged

Answer 1

The Correct Option is D

We know the following relationships for a moving coil galvanometer:

1. Current sensitivity is given by: \[ S_I = \frac{\theta}{I} = \frac{NBA}{k} \] where: 
N = number of turns 
B = magnetic field 
A = area of coil 
k = restoring torque per unit twist

2. Voltage sensitivity is given by: \[ S_V = \frac{\theta}{V} = \frac{NBA}{kR} \] where R is the resistance of the coil.

When number of turns doubles: \[ N' = 2N \]

The current sensitivity becomes: \[ S_I' = \frac{(2N)BA}{k} = 2S_I \] (which matches the given condition)

However, the resistance also increases when turns double. Since resistance is proportional to length (and thus number of turns): \[ R' = 2R \]

Therefore, the new voltage sensitivity: \[ S_V' = \frac{(2N)BA}{k(2R)} = \frac{NBA}{kR} = S_V \] remains unchanged.

Final answer: The voltage sensitivity \(\boxed{\text{remains unchanged}}\).