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

The total resistance \(R\) of the voltmeter is calculated using the formula:

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

The parameters are defined as follows:

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

Upon substituting these values:

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

Therefore, the required resistance is \(19950 \, \Omega\), matching Option (3).