A series LCR circuit has $\mathrm{R}=200 \Omega, \mathrm{L}=663 \mathrm{mH}$ and $\mathrm{C}=26 \cdot 5 \mu…

A series LCR circuit has $\mathrm{R}=200 \Omega, \mathrm{L}=663 \mathrm{mH}$ and $\mathrm{C}=26 \cdot 5 \mu \mathrm{F}$. The applied alternating voltage has an amplitude of $50 \mathrm{~V}$ and a frequency of $60 \mathrm{~Hz}$ so that $\mathrm{X}_{\mathrm{L}}=250 \Omega$ and $\mathrm{Xc}=100 \Omega$. The peak current is
  1. $0.33 \mathrm{~A}$
  2. $0 \cdot 20 \mathrm{~A}$
  3. $0.50 \mathrm{~A}$
  4. $0 \cdot 25 \mathrm{~A}$

Solution

$\mathrm{R}=200 \Omega \quad \mathrm{L}=663 \mathrm{mH} \quad \mathrm{C}=26.5 \mu \mathrm{F}$ $\mathrm{V}_{0}=50 \mathrm{~V}, \mathrm{f}=60 \mathrm{~Hz}$ $\mathrm{X}_{\mathrm{L}}=250 \Omega, \mathrm{X}_{\mathrm{C}}=100 \Omega$ $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{1}-\mathrm{X}_{\mathrm{C}}\right)^{2}}=\sqrt{40000+22500}$ $=\sqrt{62500}=250$ $\therefore \mathrm{i}_{0}=\frac{\mathrm{V}_{0}}{\mathrm{Z}}=\frac{50}{250}=\frac{1}{5}=0.2 \mathrm{~A}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

Practice more AC Circuits questions on Aicharya