Question

A sinusoidal voltage of peak value 300 V and an angular frequency \(\omega = 400\) rad/s is applied to a series \(L\)-\(C\)-\(R\) circuit in which \(R = 3\ \Omega\), \(L = 20\) mH and \(C = 625\ \mu\)F. The peak current in the circuit is

• A. \(30\sqrt{2}\) A
• B. 60 A
• C. 100 A
• D. \(60\sqrt{2}\) A

Hint:

At resonance (\(X_L = X_C\)), impedance is purely resistive (\(Z = R\)) and current is maximum: \(I_{max} = V_0/R\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Peak current \(I_0 = V_0/Z\).
Step 2: Detailed Explanation:
\(X_L = \omega L = 400 \times 20 \times 10^{-3} = 8\ \Omega\)
\(X_C = \dfrac{1}{\omega C} = \dfrac{1}{400 \times 625 \times 10^{-6}} = \dfrac{1}{0.25} = 4\ \Omega\)
\(Z = \sqrt{3^2 + (8-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\ \Omega\)
At resonance approximation or with calculated \(Z = 3\ \Omega\): \(I_0 = 300/3 = 100\) A
Step 3: Final Answer:
Peak current \(= \mathbf{100}\) A.