A 110V, 50Hz, AC source is connected in the circuit (as shown in...

Let current in the inductance is ${i}_{L}$, through the capacitor is ${i}_{C}$ and through resistance is ${i}_{R}$ at any instance as shown in the figure.

According to the Kirchoff's junction law, ${i}_{R} = {i}_{L} + {i}_{C}$.

At resonance ${X}_{C} = {X}_{L}$ hence ${i}_{L} = {i}_{C}$. Also, current through the capacitor and inductor have phase difference of $\pi rad$. Therefore, they will form destructive interference and hence the current through the resistance will be $0$.