This question was previously asked in

SSC JE EE Previous Paper 3 (Held on: 27 Jan 2018 Evening)

Option 3 : 1

General Awareness Subject Test 1

7367

20 Questions
20 Marks
20 Mins

Given that, self-inductances \({L_1} = {L_2} = 40\;mH\)

Mutual inductance (M) = 40 mH

Coupling factor \(k = \frac{M}{{\sqrt {{L_1}{L_2}} }} = \frac{{40 \times {{10}^{ - 3}}}}{{\sqrt {40 \times {{10}^{ - 3}} \times 40 \times {{10}^{ - 3}}} }} = 1\)