Alternating Current

JEE Main level, numerical answer, medium

81 of 190

Question 81 of 190

An inductor of 0.5mH0.5mH, a capacitor of 20μF20\mu F and resistance of 20Ω20\Omega are connected in series with a 220V220V ac source. If the current is in phase with the emf, the amplitude of current of the circuit is xA\sqrt{x}A. The value of xx is-

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Answer: 242

Here, emf and current are in phase, so the circuit is either pure resistive circuit or case of resonance,
XL=XC{X}_{L}={X}_{C}
Thus, the impedance of circuit is Z=R2+(XL−XC)2=R=20ΩZ=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}=R=20\Omega
Now, rms value of current is irms=22020=11A{i}_{rms}=\dfrac{220}{20}=11A
Using the relation between peak and rms value of current, we have peak value i0=irms2=112=242A{i}_{0=}{i}_{rms}\sqrt{2}=11\sqrt{2}=\sqrt{242}A.
Hence, the value of x=242x=242.

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