Physics - EMI/AC Question with Solution | TestHub

Since $f=900<f_{res}=1000$,

$$X_C>X_L$$

therefore, the circuit becomes capacitive, and the current leads the EMF.

At resonance, the impedance is minimum.

Not only at resonance, but under all conditions, the voltages across $L$ and $C$ differ in phase by $180^{\circ }$ when they are connected in series.