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
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
Energy stored = Ivt = current x back e.m.f x time
= 2 x 3 x 0.4
= 2.4J
Energy stored = Ivt = current x back e.m.f x time
= 2 x 3 x 0.4
= 2.4J