A series circuit has a resistor R=30 Ω, an inductor L=80 mH and a capacitor C=25 muF, driven by a source of rms voltage 100 V and angular frequency…

At $\omega=1000\ \mathrm{rad\,s^{-1}}$: $X_L=\omega L=80\ \Omega$ and $X_C=\frac1{\omega C}=\frac{1}{1000\times25\times10^{-6}}=40\ \Omega$. The impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{900+1600}=50\ \Omega$, so $I_{rms}=2$ A and $P=I^2R=4\times30=120$ W. At resonance $X_L=X_C$, so $Z=R=30\ \Omega$, $I_{rms}=\frac{100}{30}$ A and $P_{res}=\frac{V^2}{R}=\frac{10^4}{30}=333.3$ W. The ratio is $\frac{333.3}{120}=\frac{Z^2}{R^2}=\frac{2500}{900}=2.78$. The same ratio can be obtained directly as $P_{res}/P=(V^2/R)/(I^2R)$. Resonance at $\omega_0=1/\sqrt{LC}\approx707\ \mathrm{rad\,s^{-1}}$ lies below the given frequency, which is reasonable since the circuit is inductive at $1000\ \mathrm{rad\,s^{-1}}$ and the reactances do not cancel there, leaving a net reactance of $40\ \Omega$ that limits the current and wastes none of the supply as heat, which is why the dissipated power is smaller than at resonance.