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

To understand why the voltage sensitivity of a moving coil galvanometer remains unchanged when the number of turns is doubled, let's analyze the situation: The current sensitivity (SI) of a galvanometer is defined as the deflection per unit current, given by: 

SI = $\frac{\theta}{I}$,

where $\theta$ is the deflection and $I$ is the current. The voltage sensitivity (SV) is defined as the deflection per unit voltage, formulated as:

SV = $\frac{\theta}{V}$.

Since $V = IR$, we rewrite the voltage sensitivity as:

SV = $\frac{\theta}{IR}$ = $\frac{\theta/I}{R}$ = $\frac{SI}{R}$.

When the number of turns (N) in the coil is doubled, assuming other factors like area (A), magnetic field (B), and resistance (R) remain constant, the current sensitivity doubles because it is directly proportional to the number of turns:

$SI' = 2 \cdot SI$.

However, the voltage sensitivity is given by:

SV' = $\frac{SI'}{R}$ = $\frac{2 \cdot SI}{R}$ = $2 \cdot \frac{SI}{R}$.

But doubling the number of turns also increases the coil's resistance approximately by the same factor (assuming uniform wire thickness and material):

$R' = 2 \cdot R$.

So the voltage sensitivity becomes:

SV'' = $\frac{2 \cdot SI}{2 \cdot R}$ = $\frac{SI}{R}$ = $SV$.

This shows that the voltage sensitivity remains unchanged. Hence, the correct answer is that the voltage sensitivity will "Remain unchanged".

Answer 2

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 \({\text{remains unchanged}}\).