Question

An inductor of 200 \(\mu\)H is connected to an alternating source of frequency 50 Hz. Then the inductive reactance is:

• A. \(6.28 \times 10^{-2}~\Omega\)
• B. \(1.64 \times 10^{-2}~\Omega\)
• C. \(3.14 \times 10^{-2}~\Omega\)
• D. \(4.82 \times 10^{-2}~\Omega\)

Hint:

Inductive reactance depends linearly on frequency: \(X_L = 2 \pi f L\). Convert microhenry to henry before calculation.

Answer 1

The Correct Option is A

Step 1: Reactance formula.
X_L = 2π f L.

Step 2: Unit conversion and substitution.
L = 200 μH = 200×10⁻⁶ H, f = 50 Hz. X_L = 2π(50)(200×10⁻⁶) = 6.28×10⁻² Ω.

Step 3: Conclusion.
The inductive reactance is 6.28×10⁻² Ω.