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 :

Solution

Physics sir · 4 steps

At full deflection the whole 100 V drives just 5 mA, so find the total resistance first, then take away 50 Ω.

1. Series connection. The resistor R goes in series with the galvanometer, so the same current flows through both. At the full 100 V reading, that current must be the galvanometer's maximum, I_g = 5 mA = 5 × 10⁻³ A.

   V = I_g (G + R)

2. Total resistance. G + R = V/I_g = 100/(5 × 10⁻³) = 20000 Ω.
3. Take away G. The galvanometer already supplies 50 Ω of this, so R = 20000 - 50 = 19950 Ω.
4. So. A series resistor of 19950 Ω is needed, which is option (d).

The idea to keep. For a voltmeter, V/I_g is the total resistance of the whole meter. The galvanometer is part of that total, so subtract G to get the series resistor.

Answer from the NTA final answer key. Solution by PrepWiser.