Question

The current sensitivity of a galvanometer having 20 divisions is \( 10 \mu A / \text{div} \). If the resistance of the galvanometer is 100Ω, then the value of the resistance to be used to convert this galvanometer into a voltmeter to read up to 1 V is:

• A. \( 4 \times 10^4 \, \Omega \) in series with the galvanometer.
• B. \( 4900 \, \Omega \) in parallel with the galvanometer.
• C. \( 4 \times 10^4 \, \Omega \) in series with the galvanometer.
• D. \( 49000 \, \Omega \) in series with the galvanometer.

Hint:

To convert a galvanometer into a voltmeter, the required resistance is determined by the full-scale deflection current and the maximum voltage that the voltmeter should measure. Apply Ohm's law to calculate the required resistance.

Answer 1

The Correct Option is D

The current sensitivity of the galvanometer is given as \( 10 \mu A/\text{div} \) and it has 20 divisions, so the full-scale deflection current is: \[ I_{\text{full scale}} = 10 \, \mu A \times 20 = 200 \, \mu A \] Next, to convert the galvanometer into a voltmeter, we need to find the resistance needed to allow the galvanometer to measure up to 1V. According to Ohm's law: \[ V = I \times R \] Where \( V = 1 \, \text{V} \) and \( I = 200 \, \mu A = 200 \times 10^{-6} \, \text{A} \). Thus: \[ R_{\text{total}} = \frac{V}{I} = \frac{1 \, \text{V}}{200 \times 10^{-6} \, \text{A}} = 5000 \, \Omega \] The resistance of the galvanometer is already \( 100 \, \Omega \), so the additional resistance \( R \) required in series is: \[ R = R_{\text{total}} - R_{\text{galv}} = 5000 \, \Omega - 100 \, \Omega = 4900 \, \Omega \] Thus, the resistance required is \( 49000 \, \Omega \) in series with the galvanometer to convert it into a voltmeter reading up to 1 V.