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QJEEM22Ph4fa8: JEE Main 2022 Physics - Class 12 JEEM22Ph4fa8 Solution: A 110 V, 50 Hz, AC source is connected in the circuit (as shown …

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

JEE Main 2022 (Online) 26th June Morning Shift Physics - Alternating Current Question 79 English

Step-by-Step Verified Solution

1Z=(1XL−1XC)2\frac{1}{Z}=\sqrt{\left(\frac{1}{X_L}-\frac{1}{X_C}\right)^2}

At resonance, XL=XC&Z→∞\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \& \mathrm{Z} \rightarrow \infty

∴Ztotal circuit →∞\therefore \mathrm{Z}_{\text {total circuit }} \rightarrow \infty

i.e, I=0\mathrm{I}=0

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