Answer
a) $V_{max}=9.0V$
b) $V=-6.7V$
c) $V=6.7V$
Work Step by Step
(a) We know that
$X_c=\frac{1}{\omega C}$
We plug in the known values to obtain:
$X_c=\frac{1}{2\pi(30.0)(22\mu F)}=60.29\Omega$
Now $V_{max}=X_cI_{rms}$
$V_{max}=60.29(0.15)$
$V_{max}=9.0V$
(b) We know that
$I=I_{max}sin\theta$
$\theta=sin^{-1}\frac{I}{I_{max}}=sin^{-1}\frac{0.10}{0.15}=41.8^{\circ}$
Now $V=V_{max}sin(\theta -90^{\circ})$
$V=9.0sin(41.8^{\circ}-90^{\circ})=-6.7V$
(c) We know that
$V=9.0sin(138-90)=6.7V$
