d
\(\text { Given }: i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \text { ampere }\)

Compare it with \(i=i_{0} \sin (\omega t),\) we get

\(i_{0}=\frac{1}{\sqrt{2}} \mathrm{A}\)

Given \(: e=\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right)\) volt

Compare it with, we get

\(e_{0}=\frac{1}{\sqrt{2}} V, \phi=\frac{\pi}{3}\)

\(\therefore i_{\mathrm{rms}}=\frac{i_{0}}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \mathrm{A}=\frac{1}{2} \mathrm{A}\)

\(e_{\mathrm{rms}}=\frac{e_{0}}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \mathrm{V}=\frac{1}{2} \mathrm{V}\)

Average power consumed in the circuit,

\(P = {i_{rms}}\,{e_{ems}}\,\cos \,\phi \)

\(=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \cos \frac{\pi}{3}=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{8} \,\mathrm{W}\)