Here,

Inductance, 𝐿 = 80 mH

Capacitance, 𝐶 = 60 μF

rms voltage, 𝑉 = 230 V

Frequency, 𝑓 = 50 Hz

(a) $\omega =2\pi f=2\pi \times 50Hz=100\pi rad{s}^{-1}$

current amplitude,

$\begin{array}{l}{I}_m=\sqrt{2}\times \frac{V_{rms}}{Z}=\frac{V_{rms}\sqrt{2}}{X_L-X_C}=\frac{230\times \sqrt{2}}{(100\pi \times 80\times {10}^{-3}-\frac{1}{100\pi \times 60\times {10}^{-6}})}\\ =-11.6A\end{array}$

rms current,

$I_{rms}=\frac{I_m}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}=-8.24A$

(b) the rms value of potential drop across the inductor,

$V_{Lrms}=I_{rms}\times X_L=8.24\times 100\pi \times 80\times {10}^{-3}=207V$

rms voltage across the inductor,

$V_{Crms}=I_{rms}\times X_C=8.24\times \frac{1}{100\pi \times 60\times {10}^{-6}}=437V$

(c) As the phase difference between voltage and current through the inductor
is π/2, the average power transferred over a complete cycle by the source to
The inductor will always be zero.

$P_{l,avg}=0$

(d) For 𝐶, voltage lags by 𝜋𝜋/2. As the phase difference between voltage and
current through the capacitor is 𝜋𝜋/2, the average power transferred over a
complete the cycle by the source to inductor will always be zero.

$P_{l,avg}=0$

(e) As there is no average power transfer over a complete cycle by the source
to the circuit elements, so the average power absorbed by the circuit is
zero.