Question:
Find the peak current and resonant frequency of the following circuit (as shown in figure).
Find the peak current and resonant frequency of the following circuit (as shown in figure).
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The source has \(\omega = 100\) rad/s, which is not the resonant frequency, so find \(Z\) at this \(\omega\) for the peak current. The resonant frequency is \(f_0 = \frac{1}{2\pi\sqrt{LC}}\).
Updated On: Nov 4, 2026
- 2 A and 100 Hz
- 2 A and 50 Hz
- 0.2 A and 100 Hz
- 0.2 A and 50 Hz
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The Correct Option is D
Solution and Explanation
Figure: series L = 100 mH, C = 100 \(\mu\)F, R = 120 \(\Omega\) with source \(V = 30\sin 100t\), so \(V_0 = 30\) V and \(\omega = 100\) rad/s.
\(X_L = \omega L = 100 \times 0.1 = 10\ \Omega\).
\(X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\ \Omega\).
\(Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{120^2 + 90^2} = 150\ \Omega\).
Peak current \(I_0 = \frac{V_0}{Z} = \frac{30}{150} = 0.2\) A.
Resonant frequency \(f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 10^{-4}}} = \frac{1}{2\pi \times 3.16\times10^{-3}} \approx 50\) Hz.
Answer: (4) 0.2 A and 50 Hz
\(X_L = \omega L = 100 \times 0.1 = 10\ \Omega\).
\(X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-4}} = 100\ \Omega\).
\(Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{120^2 + 90^2} = 150\ \Omega\).
Peak current \(I_0 = \frac{V_0}{Z} = \frac{30}{150} = 0.2\) A.
Resonant frequency \(f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 10^{-4}}} = \frac{1}{2\pi \times 3.16\times10^{-3}} \approx 50\) Hz.
Answer: (4) 0.2 A and 50 Hz
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