MHT CET · Physics · Alternating Current
In the circuit shown in the figure, a.c. source gives voltage \(\mathrm{V}=20 \cos (2000 \mathrm{t})\). Impedance and r.m.s current respectively will be

- A \(10 \Omega, \sqrt{2} \mathrm{~A}\)
- B \(5 \Omega, 2 \mathrm{~A}\)
- C \(15 \Omega, \frac{2 \sqrt{2}}{3} \mathrm{~A}\)
- D \(5 \Omega, 1 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(15 \Omega, \frac{2 \sqrt{2}}{3} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{V}=20 \cos 2000 \mathrm{t} \therefore \omega=2000 \mathrm{rad} / \mathrm{s} \\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2000 \times 5 \times 10^{-3}=10 \Omega \\
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega \\
& \therefore \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \text { which gives resonance. } \\
& \mathrm{Z}=\mathrm{R}=10 \Omega+5 \Omega=15 \Omega
\end{aligned}
\)
The peak current, \(\mathrm{I}_0=\frac{\mathrm{V}_0}{\mathrm{R}}=\frac{20}{15}=\frac{4}{3} \mathrm{~A}\) R.M.S. current, \(I_{r m s}=\frac{I_0}{\sqrt{2}}=\frac{2 \sqrt{2}}{3} \mathrm{~A}\)
\begin{aligned}
& \mathrm{V}=20 \cos 2000 \mathrm{t} \therefore \omega=2000 \mathrm{rad} / \mathrm{s} \\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2000 \times 5 \times 10^{-3}=10 \Omega \\
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2000 \times 50 \times 10^{-6}}=10 \Omega \\
& \therefore \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \text { which gives resonance. } \\
& \mathrm{Z}=\mathrm{R}=10 \Omega+5 \Omega=15 \Omega
\end{aligned}
\)
The peak current, \(\mathrm{I}_0=\frac{\mathrm{V}_0}{\mathrm{R}}=\frac{20}{15}=\frac{4}{3} \mathrm{~A}\) R.M.S. current, \(I_{r m s}=\frac{I_0}{\sqrt{2}}=\frac{2 \sqrt{2}}{3} \mathrm{~A}\)
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