MHT CET · Physics · Alternating Current
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
- A \(0.33 \mathrm{~A}\)
- B \(0 \cdot 20 \mathrm{~A}\)
- C \(0.50 \mathrm{~A}\)
- D \(0 \cdot 25 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(0 \cdot 20 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{R}=200 \Omega \quad \mathrm{L}=663 \mathrm{mH} \quad \mathrm{C}=26.5 \mu \mathrm{F}\)
\(\mathrm{V}_{0}=50 \mathrm{~V}, \mathrm{f}=60 \mathrm{~Hz}\)
\(\mathrm{X}_{\mathrm{L}}=250 \Omega, \mathrm{X}_{\mathrm{C}}=100 \Omega\)
\(\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{1}-\mathrm{X}_{\mathrm{C}}\right)^{2}}=\sqrt{40000+22500}\)
\(=\sqrt{62500}=250\)
\(\therefore \mathrm{i}_{0}=\frac{\mathrm{V}_{0}}{\mathrm{Z}}=\frac{50}{250}=\frac{1}{5}=0.2 \mathrm{~A}\)
\(\mathrm{V}_{0}=50 \mathrm{~V}, \mathrm{f}=60 \mathrm{~Hz}\)
\(\mathrm{X}_{\mathrm{L}}=250 \Omega, \mathrm{X}_{\mathrm{C}}=100 \Omega\)
\(\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{1}-\mathrm{X}_{\mathrm{C}}\right)^{2}}=\sqrt{40000+22500}\)
\(=\sqrt{62500}=250\)
\(\therefore \mathrm{i}_{0}=\frac{\mathrm{V}_{0}}{\mathrm{Z}}=\frac{50}{250}=\frac{1}{5}=0.2 \mathrm{~A}\)
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