d
\(\text { Here, } L=20\, \mathrm{mH}=20 \times 10^{-3}\, \mathrm{H}\)

\(C=50\, \mu \mathrm{F}=50 \times 10^{-6}\, \mathrm{F}\)

\(R=40\, \Omega, V=10\, \sin \,340\, t=V_{0} \sin \omega t\)

\(\omega=340 \,\mathrm{rad} \,\mathrm{s}^{-1}, V_{0}=10\, \mathrm{V}\)

\(X_{L}=\omega L=340 \times 20 \times 10^{-3}=6.8\, \Omega\)

\(X_{C}=\frac{1}{\omega C}=\frac{1}{340 \times 50 \times 10^{-6}}=\frac{10^{4}}{34 \times 5}=58.82 \,\Omega\)

\(Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}=\sqrt{(40)^{2}+(58.82-6.8)^{2}}\)

\(=\sqrt{(40)^{2}+(52.02)^{2}}=65.62\, \Omega\)

The peak current in the circuit is

\(I_{0}=\frac{V_{0}}{Z}=\frac{10}{65.62} \,\mathrm{A}, \cos \phi=\frac{R}{Z}=\left(\frac{40}{65.62}\right)\)

Power loss in \(A.C.\) circuit,

\({=V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi=\frac{1}{2} V_{0} I_{0} \cos \phi}\)

\({=\frac{1}{2} \times 10 \times \frac{10}{65.62} \times \frac{40}{65.62}=0.46 \,\mathrm{W}}\)