Two coils having self-inductance of 3 H and 2 H, respectively, have mutual inductance of 2 H. They are connected in series and carry a current of 4 A. Calculate the energy of the magnetic field when the self and mutual fluxes are in the same direction.

Concept:

Series Inductors:

When inductors are connected together in series so that the magnetic field of one links with the other, the effect of mutual inductance either increases or decreases the total inductance depending upon the amount of magnetic coupling. 

Mutually connected series inductors can be classed as either “Aiding” or “Opposing” the total inductance. If the magnetic flux produced by the current flows through the coils in the same direction then the coils are said to be Cumulatively Coupled. If the current flows through the coils in opposite directions then the coils are said to be Differentially Coupled as shown below:

Cumulatively Coupled:

Total self-inductance, L_T = L₁ + L₂ + 2 M

Where, 

L₁ = Self-inductance of inductor 1

L₂ = Self-inductance of inductor 2

M = Mutual inductance

Differentially Coupled:

Total self-inductance,

LT = L₁ + L₂ - 2 M

Energy store in Inductor in form of the magnetic field is given as,

\(E = \frac{1}{2} L_T I^2\)

Where I is the series current.

Calculation:

Given,
L₁ = 3 H
L₂ = 2 H
M = 2 H
I = 4 A 

Total inductance,

LT = L1 + L2 + 2 M = 3 + 2 + (2 x 2) = 9 H

Energy of magnetic field,

\(E=\frac{1}{2}L_TI^2=\frac{1}{2}\times9\times4^2=72J\)