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 a resistance $R_2$ is connected in series with it, then it came the $\frac{v}{3}$ 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
Let Ig be the current through galvanometer at full deflection.
$ V = I_g(R_g+R_1)$ .........(1)
$ \frac{V}{3} = I_g(R_g+R_2)$ .......(2)
Dividing the two we will get
$ 3 = \frac{R_g+R_1}{R_g+R_2}$
$ 2R_g = R_1 - 3R_2$
$ 3V = I_g(R_g+R_3)$ .......(3)
Dividing (1) and (3)
$ \frac{1}{3} = \frac{R_g+R_1}{R_g+R_3} \Rightarrow 2R_g = R_3 -3R_1$
$\Rightarrow R_1- 3R_2 = R_3 -3R_1 \Rightarrow R_3 = 4R_1 - 3R_2$