Question

Two inductor coils L1 = 18 mH and L2 = 2 mH are perfectly coupled. If a current of 2 A is reversed in first coil within 2 millisecond, then the EMF induced in the second coil will be

Answer 1

Answer:

12 volt

Explanation:

L=18*2^1/2=6

e=LdI/dt

  =6*4/2

  =12volt

Answer 2

Given: Two inductors $L_{1}$ and $L_{2}$ of 18 mH and 2 mH.

Current of 2 A is reversed in first coil within 2 millisecond.

To Find:

EMF induced in the second coil.

Solution:

Inductance of first coil = 18 mH

Change in current ΔI = 2 A

Change in time Δt = 2 mS

We Know that,

Induced emf = L.ΔI /Δt

$e_{1}$ = $L_{1}$.ΔI /Δt

$e_{1}$ = 18 ×2/2

$e_{1}$ = 18 Volts

Since,  first coil is perfectly coupled with second coil

∴ EMF induced in second coil will be same as that in First coil.

EMF induced in the second coil = 18 Volts

Hence, EMF induced in the second coil is 18 volts.