Question

In the moving coil galvanometer, if number of turns in the coil is doubled, then the current sensitivity ______ and the voltage sensitivity ______.

• A. remains unchanged, will be doubled
• B. will be halved, will be doubled
• C. will be doubled, remains unchanged
• D. will be halved, remains unchanged

Hint:

Increasing turns is a great way to detect smaller \textbf{currents}, but it won't help you detect smaller \textbf{voltages} because the extra wire adds extra resistance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Current Sensitivity ($I_s$) is the deflection per unit current ($NBA/k$). Voltage Sensitivity ($V_s$) is the deflection per unit voltage ($NBA/kR$).
Step 2: Formula Application:
When the number of turns $N$ is doubled ($N' = 2N$): $I_s \propto N \implies I_s' = 2 I_s$ (Doubled).
Step 3: Explanation:
When $N$ doubles, the length of the wire doubles, which also doubles the resistance ($R' = 2R$). $V_s = \frac{NBA}{kR}$. Since both $N$ and $R$ double, they cancel each other out. $V_s' = \frac{(2N)BA}{k(2R)} = V_s$ (Unchanged).
Step 4: Final Answer:
Current sensitivity doubles; voltage sensitivity remains unchanged.