A series circuit has a resistor of 30 Ω, an ideal inductor of 0.10 H and a capacitor of capacitance C. It is connected to a source of 100 V (rms) at…

The inductive reactance is $X_L=\omega L=400\times0.10=40\ \Omega$ and $X_C=80\ \Omega$, so $X_L-X_C=-40\ \Omega$. The impedance is $Z=\sqrt{30^2+40^2}=50\ \Omega$ and $I_{rms}=100/50=2.0\ \mathrm{A}$: (A) is true. Because $X_C>X_L$ the circuit is capacitive, $\tan\phi=(X_L-X_C)/R=-4/3$, so the current leads the voltage by $53^\circ$: (B) is false. The average power is $I^2R=4\times30=120\ \mathrm{W}$, the same as $VI\cos\phi=100\times2\times0.6$: (C) is true. The capacitance is $C=1/(\omega X_C)=31.25\ \mu\mathrm{F}$, so the resonant frequency is $\omega_0=1/\sqrt{LC}\approx566\ \mathrm{rad\,s^{-1}}$, not $400$: (D) is false. The phase angle can also be seen from the power factor $\cos\phi=R/Z=30/50=0.6$, which confirms the power $VI\cos\phi=120\ \mathrm{W}$ and shows that all of it is dissipated in the resistor, because the ideal inductor and capacitor absorb none on average.