Electro Magnetic Induction And Alternating Currents Question 581
Question: Energy stored in a coil of self inductance 40mH carrying a steady current of 2 A is [Kerala (Engg.) 2001]
Options:
A) 0.8 J
B) 8 J
C) 0.08 J
D) 80 J
Show Answer
Answer:
Correct Answer: C
Solution:
$ U=\frac{1}{2}Li^{2}\Rightarrow U=\frac{1}{2}\times 40\times {{10}^{-3}}\times {{(2)}^{2}}=0.08J $
