Two inductors of inductance 4 II and 8 II are arrange in series and a current of 10 A is passed through them. What is the energy stored in them?

Analyze the question: Two inductors (4 H and 8 H, where "II" is a typo for "H") are connected in series with a current of 10 A. We need the total energy stored.

Concept:

• Inductors in series add: L_total = L₁ + L₂
• Energy stored in an inductor: E = ½LI²

Solution:

1. Total inductance: L = 4 + 8 = 12 H

2. Energy stored: E = ½ × L × I² E = ½ × 12 × (10)² E = ½ × 12 × 100 E = 600 J

(Check on the other options: 133 J would come from treating the inductors as if in parallel, since L = 32/12 ≈ 2.67 H gives ½ × 2.67 × 100 ≈ 133 J. That is the trap, but the question says series.)

Tip: Inductors combine like resistors: series means add, parallel means use the reciprocal formula. Capacitors are the opposite. Always check "series" or "parallel" in the question before you calculate.