Answer:
Option B,C
Explanation:
$Z= \sqrt{R^{2}+X_{C}^{2}}=\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}$
In case (b) capacitance C will be more. Therefore, impedance Z will be less.
Hence, current will be more.
$\therefore$ Option(b) is correct
Further $V_{C}=\sqrt{V^{2}-V_{R}^{2}}$
$\sqrt{V^{2}-(IR)^{2}}$
In case (b), since current l is more
Therefore , $V_{C}$ will be less
$\therefore$ Option (c) is correct
$\therefore$ Correct options are (b) and (c)