
(b) Charging current, $I = \frac{E}{R}\ e^{- \frac{t}{RC}}$
Taking log both sides,
$\ {Log}{I = \log{\left( \frac{E}{R} \right) - \frac{t}{RC}}}$
When R is doubled, slope of curve increases. Also at t = 0, the current will be less. Graph Q represents the best.