d
(d) Current will be maximum in the condition of resonance so \({i_{\max }} = \frac{V}{R} = \frac{V}{{10}}A\)

Energy stored in the coil \({W_L} = \frac{1}{2}Li_{\max }^2\)\( = \frac{1}{2}L{\left( {\frac{E}{{10}}} \right)^2}\)

\( = \frac{1}{2} \times {10^{ - 3}}\left( {\frac{{{E^2}}}{{100}}} \right)\)\( = \frac{1}{2} \times {10^{ - 5}}{E^2}\;joule\)

\(\therefore \) Energy stored in the capacitor

\({W_C} = \frac{1}{2}C{E^2} = \frac{1}{2} \times 2 \times {10^{ - 6}}{E^2} = {10^{ - 6}}{E^2}\;joule\)

\(\therefore \frac{{{W_C}}}{{{W_L}}} = \frac{1}{5}\)