Question

A moving coil voltmeter has an internal resistance of 50 \(\Omega\). The scale of the meter is divided into 100 equal divisions. When a potential of 1 V is applied to terminals of the voltmeter, a deflection of 100 divisions is obtained. However, it is desired that when a potential of 500 V is applied to the terminals, a deflection of 100 divisions should be obtained. The value of resistance that needs to be connected in series to achieve this is ________ \(\Omega\).

Hint:

When designing a voltmeter with a given deflection scale, use Ohm’s law to determine the total resistance required for the desired voltage and deflection.

Answer 1

Correct Answer: 24950

The current corresponding to 1 V deflection (at 100 divisions) is: \[ I_1 = \frac{1}{50} = 0.02\ \text{A} \] For the desired deflection with 500 V, the current should remain the same (since the number of divisions is fixed at 100), and the total voltage across the meter should be 500 V. So, the total resistance is: \[ R_{\text{total}} = \frac{500}{0.02} = 25000\ \Omega \] The series resistance \(R_s\) should be: \[ R_s = R_{\text{total}} - R_{\text{internal}} = 25000 - 50 = 24950\ \Omega \] Thus, the required series resistance is: \[ \boxed{24950\ \Omega} \] Final Answer: 24950 \(\Omega\)