A galvanometer having a coil resistance 100 Ω gives a full scale deflection when a current of 1 mA is passed through it. What is the value of the resistance which can convert this galvanometer into a voltmeter giving full scale deflection for a potential difference of 10 V?

Solution

Physics sir · 4 steps

The coil already takes 100 ohms of the 10 kilo-ohm total, so the series resistor supplies only the rest.

1. Series resistor. To make a voltmeter, connect a large resistance R in series with the coil. At full scale the same current I_g = 1 mA flows through both the coil and R.

   V = I_g(G + R)

2. Total resistance. For 10 V at full scale, G + R = 10/(1 × 10⁻³) = 10 000 Ω.
3. Subtract the coil. R = 10 000 - 100 = 9900 Ω.
4. So. The series resistance is 9.9 kΩ, which is option (c).

The idea to keep. For a voltmeter, V/I_g gives the total resistance of the whole meter. Subtract the coil resistance G to find the series resistor you need.

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