JEE MAIN - Physics (2022 - 26th June Morning Shift - No. 26)
A 110 V, 50 Hz, 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
Explanation
$$
\frac{1}{Z}=\sqrt{\left(\frac{1}{X_L}-\frac{1}{X_C}\right)^2}
$$
At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \& \mathrm{Z} \rightarrow \infty$
$\therefore \mathrm{Z}_{\text {total circuit }} \rightarrow \infty$
i.e, $\mathrm{I}=0$
At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \& \mathrm{Z} \rightarrow \infty$
$\therefore \mathrm{Z}_{\text {total circuit }} \rightarrow \infty$
i.e, $\mathrm{I}=0$
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