In the circuit shown, when key $\mathrm{K}_1$ is closed, then the ammeter reads $\mathrm{I}_0$ whether $\mathrm{K}_2$ is open or closed. But when $K_1$ is open the ammeter reads $I_0 / 2$, when $K_2$ is closed:

Assuming that ammeter resistance is much less than $R_2$ , the values of r and $R_1$ is ohms are:

0, 50

When $K_1$ is closed, $R_1$ is short-circuited

When $K_2$ is open, $i_0=\frac{E}{r+R_2}=\frac{E}{r+100}$          (i)

When $K_2$ is closed, $i_0=\frac{1}{2}\left[\frac{E}{r+50}\right]$                 (ii)

From these two equations, we get $r=0$

When $K_1$ is open and $K_2$ is closed. $\frac{i_0}{2}=\frac{E}{2\left(R_1+50\right)}$           (iii)

From eqs. (i) and (iii), we have $R_1=50 \Omega$