Question

A maximum current of 0.5 mA can pass through a galvanometer of resistance 15 \(\Omega\). The resistance to be connected in series to the galvanometer to convert it into a voltmeter of range 0–10 V is

• A. 9985 \(\Omega\)
• B. 20015 \(\Omega\)
• C. 20000 \(\Omega\)
• D. 19985 \(\Omega\)

Hint:

To convert a galvanometer to a voltmeter, add a series resistance \( R = \frac{V}{I_g} - R_g \), where \( V \) is the desired voltage range, \( I_g \) is the full-scale current, and \( R_g \) is the galvanometer’s resistance.

Answer 1

The Correct Option is D

To convert a galvanometer into a voltmeter, a high resistance \( R \) is connected in series to allow the galvanometer to measure a maximum voltage \( V \) when its full-scale deflection current \( I_g \) flows. The galvanometer’s resistance is \( R_g \), and the voltage across the series combination is: \[ V = I_g (R_g + R) \] Given: - Maximum current: \( I_g = 0.5 \, \text{mA} = 0.5 \times 10^{-3} \, \text{A} \) - Galvanometer resistance: \( R_g = 15 \, \Omega \) - Voltmeter range: \( V = 10 \, \text{V} \) Substitute the values: \[ 10 = (0.5 \times 10^{-3}) (15 + R) \] \[ 15 + R = \frac{10}{0.5 \times 10^{-3}} = 20,000 \] \[ R = 20,000 - 15 = 19,985 \, \Omega \] Alternatively, verify the total resistance required: \[ R_{\text{total}} = \frac{V}{I_g} = \frac{10}{0.5 \times 10^{-3}} = 20,000 \, \Omega \] \[ R = R_{\text{total}} - R_g = 20,000 - 15 = 19,985 \, \Omega \] Thus, the series resistance is 19,985 \(\Omega\). Option (4) is correct. Options (1), (2), and (3) do not satisfy the equation.