Electro Magnetic Induction And Alternating Currents Question 594
Question: A coil resistance 20W and inductance 5H is connected with a 100V battery. Energy stored in the coil will be [MP PMT 2003]
Options:
A) 41.5 J
B) 62.50 J
C) 125 J
D) 250 J
Show Answer
Answer:
Correct Answer: B
Solution:
$ U=\frac{1}{2}Li^{2}=\frac{1}{2}L{{( \frac{E}{R} )}^{2}}=\frac{1}{2}\times 5\times {{( \frac{100}{20} )}^{2}}=62.50J $
