Question

What inductance would be needed to store 1 KWh of energy in a coil carrying a 200 A current

• A. 100 H
• B. 180 H
• C. 200 H
• D. 450 H

Answer 1

Answer: (B)

180 H

Answer 2

U = 1KWH = 3.6 × 10⁶ J. By using $U = \frac{1}{2}Li^{2}$ ⇒ 3.6 × 10⁶ $= \frac{1}{2} \times L \times (200)^{2}$ ⇒ L = 180 H