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:

$4R_1-3R_2$

Let I_g 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$

$\Rightarrow R_1- 3R_2 = R_3 -3R_1 \Rightarrow R_3 = 4R_1 - 3R_2$