A steady current of 2A flows in a coil of emf 12V for 0.4s. A back emf of 3V was induced during this period. The stored energy in the loop that can be utilized is
-
A.
9.6J -
B.
2.4J -
C.
12.0J -
D.
7.2J
Correct Answer: Option B
Explanation
Energy stored = Ivt = current x back e.m.f x time
= 2 x 3 x 0.4
= 2.4J