Question

A pure inductor of self inductance 500 mH is connected to an alternating voltage supply of \(100\sqrt{2}\pi \sin 100\pi t\) volt. The current in the circuit in ampere is

• A. \(\sqrt{2}\)
• B. 5
• C. \(5\sqrt{2}\)
• D. \(2\sqrt{2}\)
• E. 2

Hint:

In AC circuits, unless "peak" or "instantaneous" is mentioned, "current" or "voltage" always refers to the RMS value. Be careful with \(\pi\) in the numerator and denominator; they usually cancel out in textbook problems.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
In a purely inductive AC circuit, the current is limited by inductive reactance (\(X_L\)). The current lags the voltage by \(90^\circ\). When "current" is asked without qualification, we find the RMS value.

Step 2: Key Formula or Approach:
1. Inductive Reactance: \(X_L = \omega L\)
2. Peak current: \(I_0 = \frac{V_0}{X_L}\)
3. RMS current: \(I_{rms} = \frac{I_0}{\sqrt{2}}\)

Step 3: Detailed Explanation:
Given voltage equation: \(V = 100\sqrt{2}\pi \sin 100\pi t\)
Peak voltage \(V_0 = 100\sqrt{2}\pi\)
Angular frequency \(\omega = 100\pi\)
Inductance \(L = 500 \text{ mH} = 0.5 \text{ H}\)

Step 1: Calculate Inductive Reactance:
\[ X_L = 100\pi \times 0.5 = 50\pi \Omega \]

Step 2: Calculate Peak Current:
\[ I_0 = \frac{100\sqrt{2}\pi}{50\pi} = 2\sqrt{2} \text{ A} \]

Step 3: Calculate RMS Current:
\[ I_{rms} = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \text{ A} \]

Step 4: Final Answer:
The current in the circuit is 2 A.