Question

energy stored in a coil of self inductance 40mH carrying a steady current of 2A is

ans:80J

ans should be 0.08J or 80 J

Answer 1

Dear Student,

◆ Answer -

U = 0.08 J

● Explanation -

# Given -

L = 40 mH = 40×10^-3 H

I = 2 A

# Solution -

Energy stored in a coil is given by -

U = 1/2 LI²

Substitute values,

U = 1/2 × 40×10^-3 × 2²

U = 80×10^-3 J

U = 0.08 J

Hence, energy stored in a coil is 0.08 J.

Thanks dear. Hope this helps you...

Answer 2

Answer:

L = 40 mh = 40 × 10 ^3

Energy stored = 1/2 L ( I ) 2

= 1/2 × 40×10^3 × 2×2

= 0.08 j Answer.......

Explanation:

this will help you...