Electro Magnetic Induction And Alternating Currents Question 558
Question: The energy stored in a 50 mH inductor carrying a current of 4 A will be [MP PET 1999]
Options:
A) 0.4 J
B) 4.0 J
C) 0.8 J
D) 0.04 J
Show Answer
Answer:
Correct Answer: A
Solution:
$ U=\frac{1}{2}Li^{2}=\frac{1}{2}\times (50\times {{10}^{-3}})\times {{(4)}^{2}} $ $ =400\times {{10}^{-3}}=0.4J $
BETA
