Question

A galvanometer has a resistance of 50 Ω and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance

• A. 5975Ω
• B. 20050Ω
• C. 19950Ω
• D. 19500Ω

Answer 1

The Correct Option is C

We use the formula for the total resistance \(R\) of the voltmeter:

\[ R = \frac{V}{I} - R_G \]

Where:

• \(V = 100 \, \text{V}\) is the maximum voltage.
• \(I = 5 \times 10^{-3} \, \text{A}\) is the maximum current.
• \(R_G = 50 \, \Omega\) is the resistance of the galvanometer.

Substituting the values:

\[ R = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19950 \, \Omega \]

Thus, the resistance required is \(19950 \, \Omega\), which corresponds to Option (3).