Question:

The current sensitivity of moving coil galvanometer is increased by 25%.This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be :

Updated On: Mar 23, 2025
  • +25%
  • -25%
  • -50%
  • Zero
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The Correct Option is A

Solution and Explanation

Given: \[ \tau = mB \] \[ K_{\theta} = \frac{IA \cdot NB}{I \cdot \theta} = \frac{A \cdot N \cdot B}{K} \quad \text{(1)} \] \[ \left( \frac{\theta}{\theta_1} \right) = \frac{I_1}{I_2} = \frac{A \cdot N \cdot B}{K} \quad \text{(2)} \] \[ \Rightarrow \lambda = a \cdot N \quad \text{(3)} \] \[ V_{\theta} = I \cdot R = \text{Current Sensitivity} \quad \text{(4)} \] \[ \text{Voltage sensitivity} \propto \text{Current sensitivity} \]

Correct Answer: (A)

The formulae involve relationships for current and voltage sensitivities. Equation (1) gives the relationship between the magnetic field \( B \) and the area of the coil \( A \). From Equation (2), we can deduce that voltage sensitivity is directly proportional to current sensitivity. This shows a clear relation where \( R \) remains constant.

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