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

Answer: (C)

19950Ω

Answer 2

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).