A series circuit has R=10 Ω, L=0.10 H and C=10 muF, connected to a variable-frequency source of rms voltage 100 V.

Resonance occurs at $\omega_0=1/\sqrt{LC}=1/\sqrt{0.10\times10^{-5}}=1000\ \mathrm{rad\,s^{-1}}$. The quality factor is $Q=\omega_0L/R=100/10=10$; A is true. At resonance the impedance equals $R$, so $I=100/10=10\ \mathrm{A}$ and the capacitor voltage is $I/(\omega_0C)=10\times100=1000\ \mathrm{V}$, ten times the source voltage (the factor $Q$); B is false. The power is $I^2R=100\times10=1000\ \mathrm{W}$, not $500\ \mathrm{W}$; C is false. At $1100\ \mathrm{rad\,s^{-1}}$, $X_L=110\ \Omega$ and $X_C=1/(1100\times10^{-5})=90.9\ \Omega$, so the net reactance is inductive, the current lags the voltage, and D is false. Only statement A holds, so the answer is option A alone. The key physical point is that at resonance the reactive voltages are $Q$ times the source voltage but cancel each other in the total.