JAMB - Physics (2011 - No. 33)
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?
600 J
50 J
133 J
250 J
Penjelasan
Inductance (L) in series: 4 + 8 = 12
Therefore Energy stored =
W = \(\frac{1}{2} LI^2\)
= \(\frac{1}{2} \times 12 \times 10^2\)
= 600J