In an electric circuit, an inductor of inductance 0.5 H and resistance 50 is connected to an alternating…

Analyze the question: A coil (inductor) with inductance L = 0.5 H and resistance R = 50 Ω is connected to an AC source of frequency f = 60 Hz. We need the impedance Z of this series R-L circuit.

Concept: The inductor opposes alternating current through its inductive reactance:

X_L = 2πfL

Resistance and reactance are 90° out of phase, so they add as vectors:

Z = √(R² + X_L²)

Solution:

1. Inductive reactance: X_L = 2 × π × 60 × 0.5 = 60π ≈ 188.5 Ω (using π = 3.142)

2. Impedance: Z = √(50² + 188.5²) Z = √(2500 + 35532) Z = √38032 Z ≈ 195.0 Ω

Checking the options:

• A (50 Ω) is just the resistance, ignoring the reactance.
• B (450.5 Ω) and D (1950.1 Ω) don't match the calculation.
• C (195.0 Ω) matches our result.

Exam tip: For any series R-L circuit, remember the steps: find X_L = 2πfL first, then use Z = √(R² + X_L²). Do not simply add R and X_L (that would give 238.5 Ω). Also, a quick check helps: since X_L (188.5 Ω) is much larger than R (50 Ω), Z should be only a little above X_L, so about 195 Ω makes sense.