(a) Inductive reactance $X_L=2 \pi f L$ $X_L=2 \pi(50) 80 \times 10^{-3}=25.12 \Omega$
Capacitive reactance $X_C=\frac{1}{2 \pi f C}$
$$
X_C=\frac{1}{2 \times 3.14 \times 50 \times 60 \times 10^{-6}}=53.05 \Omega
$$
Impedance $=X_C-X_L=53.05-25.12=27.93 \Omega$
rms value of current $I_v=\frac{E_v}{Z}=\frac{230}{27.93}$
$=8.235 \mathrm{~A}$
Peak Value $I_0=I_{\mathrm{v}} \sqrt{2}=11.644 \mathrm{~A}$
(b) Potential drop across $L, V_L=I_L X_L=206.68 \mathrm{~V}$ Potential drop across $C$, $V_C=I_V \times X_C=436.87 \mathrm{~V}$
(c) Average power transferred to inductor is zero, because of phase difference $\pi / 2$.
$$
\mathrm{P}=E_v I_v \cos \phi . \quad \phi=\pi / 2, \therefore \mathrm{P}=0
$$
(d) Average power transferred to capacitor is also zero, because of phase difference $\pi / 2$.
$$
\begin{aligned}
&\mathrm{P}=E_v I_v \cos \phi . \\
&\phi=\pi / 2, \therefore \mathrm{P}=0
\end{aligned}
$$
(c) Total power absorbed by the circuit
$$
P_{\text {Total }}=P_L+P_C=0
$$