Question 100 of 190

A $110V,50Hz,AC$ source is connected in the circuit (as shown in figure). The current through the resistance $55\Omega$, at resonance in the circuit, will be _____$A$.

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Answer: 0

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 $0$.