
Hence, both capacitors are charged up to same magnitude of charge. So, $C_{1} V_{1}=C_{2} V_{2}$
Time constant.= $R C$
From graph,
$C_{2}$ is taking more time to charge
So, $R_{2} C_{2}>R_{1} C_{1} \ldots$ (ii)
From (i) and(ii) we can't say anything about $V_{1}$ and $V_{2}$.
$V_{1}$ can be greater than $V_{2}$ ot it can be less than $V_{2}$


