# 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, L_{T} = L_{1} + L_{2} + 2 M

Where,

L_{1} = Self-inductance of inductor 1

L_{2} = Self-inductance of inductor 2

M = Mutual inductance

**Calculation:**

Given,

L_{1} = 3 H

L_{2} = 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**