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
ENERGY IN A INDUCTOR = 1/2 X[ CURRENT ]2 X [ INDUCTANCE ] = 1/2 X 2 X 2 X 0.04
= 0.08 J
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
ENERGY IN A INDUCTOR = 1/2 X[ CURRENT ]2 X [ INDUCTANCE ] = 1/2 X 2 X 2 X 0.04
= 0.08 J