A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a resistance \( R_1 \) in series with the coil. If the resistance \( R_1 \) is replaced by a resistance \( R_2 \) in series with the coil, the voltmeter can measure upto \( \displaystyle\frac{V}{3} \) volts. Find the resistance \( R_3 \) in terms of \( R_1 \) and \( R_2 \) required to be connected to convert it into a voltmeter that can read upto 3V volts:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4R_1 - 3R_2 \)
A galvanometer is converted into a voltmeter by a large resistance in series with its coil. Full-scale deflection always needs the same current \( I_g \), so the range of the voltmeter is \( I_g \) times the total resistance of the coil and the series resistance.
\( \displaystyle I_g\,(G + R_1) = V \)
\( \displaystyle I_g\,(G + R_2) = \frac{V}{3} \)
Dividing the first equation by the second:
\( \displaystyle \frac{G + R_1}{G + R_2} = 3 \;\Rightarrow\; G + R_1 = 3G + 3R_2 \;\Rightarrow\; 2G = R_1 - 3R_2 \)
For a range of \( 3V \):
\( \displaystyle I_g\,(G + R_3) = 3V = 3\,I_g\,(G + R_1) \)
\( \displaystyle R_3 = 2G + 3R_1 \)
Putting \( 2G = R_1 - 3R_2 \):
\( \displaystyle R_3 = R_1 - 3R_2 + 3R_1 = 4R_1 - 3R_2 \)