A moving coil galvanometer has resistance 50 \Omega and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 k \Omega resistance. The maximum voltage, that can be measured using this voltmeter, will be close to :

Current Electricity PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q61 with Solution

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Correct answer: D

Resistance of galvanometer: $G = 50\,\Omega$
Full-scale deflection current: $I_g = 4\text{ mA} = 4 \times 10^{-3}\,\text{A}$
Series resistance used to make voltmeter: $R = 5\,\text{k}\Omega = 5000\,\Omega$

To make a voltmeter from a galvanometer, a high resistance is connected in series. So total resistance of the voltmeter is

R_{\text{total}} = G + R = 50 + 5000 = 5050\,\Omega

At full-scale deflection, current through the voltmeter is $I_g$ . Hence maximum voltage is

V_{\max} = I_g\,R_{\text{total}}

Substituting values,

V_{\max} = 4 \times 10^{-3} \times 5050

V_{\max} = 20.2\,\text{V}

20.2\,\text{V} \approx 20\,\text{V}

So the correct option is D.

Stored correct answer: D

This matches our derived answer.