Key formulas for a moving coil galvanometer (MCG)

For a galvanometer, $au = N B I A$ and at equilibrium, $N B I A = C\theta$ where:

• $N$ = number of turns
• $B$ = magnetic field
• $A$ = area of coil
• $I$ = current
• $C$ = torsional constant
• $\theta$ = deflection

Hence, $\frac{\theta}{I} = \frac{N B A}{C}$ This is the current sensitivity: $S_i = \frac{\theta}{I} = \frac{N B A}{C}$

If galvanometer resistance is $G$, then with applied voltage $V$, $I = \frac{V}{G}$ so voltage sensitivity is $S_v = \frac{\theta}{V} = \frac{\theta}{I}\cdot \frac{I}{V} = \frac{N B A}{C} \cdot \frac{1}{G} = \frac{N B A}{C G}$

Check statement A

Torsional constant $C$ is defined by $\tau = C\theta$ Since $\theta$ is dimensionless, $[C] = [\tau] = [\text{torque}] = [\text{force} \times \text{distance}] = [M L^2 T^{-2}]$

So A is correct.

Check statement B

Current sensitivity: $S_i = \frac{N B A}{C}$ Voltage sensitivity: $S_v = \frac{N B A}{C G} = \frac{S_i}{G}$

Thus increasing current sensitivity does not necessarily increase voltage sensitivity, because if resistance $G$ also changes suitably, $S_v$ may remain unchanged or even decrease.

So B is correct.

Check statement C

Voltage sensitivity is $S_v = \frac{N B A}{C G}$

If $N$ is doubled, resistance of the coil $G$ also generally doubles (same wire, more turns means more wire length, hence more resistance).

Therefore, $S_v \propto \frac{N}{G}$ and if both $N$ and $G$ double, $S_v$ remains unchanged, not doubled.

Hence C is false.

Check statement D

To convert a galvanometer into an ammeter, a small shunt resistance is connected in parallel, not a large one.

So D is false.

Check statement E

Current sensitivity is $S_i = \frac{N B A}{C}$ so it is directly proportional to $N$, not inversely proportional.

Hence E is false.

Final evaluation of options

Correct statements are: $A, B$

Therefore the correct option is B: A, B Only.