A coil of inductance 0.12 H and resistance 4$\Omega$, is connected across a 240V, 50Hz supply. Calculate the current through it. [$\pi$ = 3.142]

Analyze the question: A coil has both inductance and resistance, so it's an R-L series AC circuit. The current is found from I = V/Z, where Z is the impedance.

Concept:

• Inductive reactance: X_L = 2πfL
• Impedance: Z = √(R² + X_L²)
• Current: I = V/Z

Solution:

1. Inductive reactance: X_L = 2 × 3.142 × 50 × 0.12 X_L = 37.704 Ω

2. Impedance: Z = √(4² + 37.704²) Z = √(16 + 1421.57) Z = √1437.57 Z ≈ 37.91 Ω

3. Current: I = V/Z = 240 / 37.91 ≈ 6.33 A

Answer: I ≈ 6.3 A, which is option A.

Why the other options are wrong:

• C (37.2 A) and B (33.3 A) come from mistakes like using the wrong formula or dividing by the wrong quantity.
• D (40 A) is not obtainable from the given values.
• Using only R (240/4 = 60 A) or only X_L (240/37.7 ≈ 6.37 A) would be incomplete. Here X_L is much larger than R, so the answer is close to 6.4 A, but the full impedance gives 6.3 A.

Tip: For any coil with resistance on AC, never use R alone or X_L alone. Compute X_L = 2πfL first, then Z = √(R² + X_L²), then I = V/Z. If X_L is much larger than R, Z is only slightly larger than X_L, which helps you check your answer quickly.