Two inductors, each having self-inductance of 1 H, are connected in parallel in such a way that their fluxes act in the same direction. If the mutual inductance is 0.5 H, then find the equivalent inductance of the parallel connection.

Question

Question

This question was previously asked in

HPCL Engineer Electrical 01 Nov 2022 Official Paper

Answer 1

The correct answer is option 2);0.75 H

Concept:

The equivalent inductance of series aiding connection is:

L_eq = L₁ + L₂ + 2M.

The equivalent inductance of series opposing connection is:

L_eq = L₁ + L₂ – 2M.

The total inductance, LT for two parallel aiding inductors is given as

L_T = \(L_1 L_2 - M^2\over L_1+L_2- 2M\)

Calculation:

Given

Two inductors, having self-inductance of 1 H

M = 0.5

LT = \(L_1 L_2 - M^2\over L_1+L_2- 2M\)

= \(1-0.25 \over 1\)

= 0.75 H