Answer
The total current through $R_1$ and $R_2$ together is more than the current through $R_1$ previously.
Work Step by Step
Let $\Delta V$ be the potential difference across the battery.
Initially, the potential difference across $R_1$ is $\Delta V$
After $R_2$ is added in parallel, the potential difference across each resistor is $\Delta V$
Initially, the current through $R_1$ is $~~i_1 = \frac{\Delta V}{R_1}$
After $R_2$ is added in parallel, the current through $R_1$ is still $~~i_1 = \frac{\Delta V}{R_1}$
After $R_2$ is added in parallel, the current through $R_2$ is $~~i_2 = \frac{\Delta V}{R_2}$
After $R_2$ is added in parallel, the total current is $(i_1+i_2)$ which is more than the previous current of $i_1$
The total current through $R_1$ and $R_2$ together is more than the current through $R_1$ previously.
