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

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 $\pi \mathrm{rad}$. Therefore, they will form destructive interference and hence the current through the resistance will be $0$.