Answer
(a) $0.146A$
(b) $0.1389A$
(c) $-0.1389A$
Work Step by Step
(a) We can find the required maximum current as
$I_{max}==\sqrt{2}I_{rms}$
$\implies I_{max}=1.414\times 0.103A$
$\implies I_{max}=0.146A$
(b) We know that
$V_{max}=\sqrt{2}V_{rms}$
$V_{max}=1.414\times 12V$
$V_{max}=17V$
Now $V=V_{max}sin(\theta-90^{\circ})$
$5.25=17sin(\theta-90^{\circ})$
This simplifies to:
$\theta=107.9^{\circ}$
The instantaneous current through the capacitor can be determined as
$I=I_{max} sin\theta$
$\implies I=0.146\times sin(107.9^{\circ})$
$\implies I=0.1389A$
(c) We know that
$V=V_{max}sin(\theta^{\prime}-180-90)$
$\implies 5.25=17sin(\theta{\prime}-180-90)$
$\implies \theta^{\prime}=17.9$
$\implies \theta^{\prime}=287.9^{\circ}$
Now $I=I_{max}sin\theta^{\prime}$
$\implies I=0.146\times sin287.9^{\circ}$
$\implies I=-0.1389A$
