# 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.

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.

### Right Answer is: 72 J

#### SOLUTION

Since the mutual flux of the coil is in the same direction, therefore, the coils are Cumulatively Coupled.

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

Where,

L1 = Self-inductance of inductor 1

L2 = Self-inductance of inductor 2

M = Mutual inductance

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,

$\begin{array}{l} E = \dfrac{1}{2}{L_T}{I^2}\\ = \dfrac{1}{2} \times 9 \times {4^2} \end{array}$

E = 72 J

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