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