Alternating Current

JEE Main level, numerical answer, medium

18 of 190

Question 18 of 190

A series LCR circuit consists of R=80ΩR=80Ω. XL=100Ω{X}_{L}=100Ω, and XC=40Ω{X}_{C}=40Ω. The input voltage is 2500cos⁡(100πt)V2500\cos \left(100\pi t\right)V. The amplitude of current, in the circuit, is _____ AA.

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Answer: 25

Comparing input voltage with standard expression, V=V0cos⁡(ωt)V={V}_{0}\cos \left(\omega t\right), we get ω=100π\omega =100\pi.
Now, impedance Z=R2+(XL−XC)2Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}
=802+(100−40)2=\sqrt{{80}^{2}+{\left(100-40\right)}^{2}}
=100Ω=100\Omega
Therefore, peak current
i0=V0Z=2500100A=25A{i}_{0}=\dfrac{{V}_{0}}{Z}=\dfrac{2500}{100}A=25A

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