JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 27)
An inductor of $$0.5 ~\mathrm{mH}$$, a capacitor of $$20 ~\mu \mathrm{F}$$ and resistance of $$20 ~\Omega$$ are connected in series with a $$220 \mathrm{~V}$$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $$\sqrt{x}$$ A. The value of $$x$$ is ___________
Answer
242
Explanation
$$
X_L=X_C
$$
So, $\mathrm{Z}=\mathrm{R}=20 \Omega$
$$ \begin{aligned} & \mathrm{i}_{\mathrm{rms}}=\frac{220}{20}=11 \\\\ & \mathrm{i}_{\max }=11 \sqrt{2}=\sqrt{242} \end{aligned} $$
So, $\mathrm{Z}=\mathrm{R}=20 \Omega$
$$ \begin{aligned} & \mathrm{i}_{\mathrm{rms}}=\frac{220}{20}=11 \\\\ & \mathrm{i}_{\max }=11 \sqrt{2}=\sqrt{242} \end{aligned} $$
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