A coil takes 4 A from a 12 V DC source. When connected to a 12 V, 50 rad/s AC source, the current is 2.4
A. The inductance of the coil is:

First, consider the DC circuit. The coil takes 4 A from a 12 V DC source.In a DC circuit, the inductor acts as a short circuit (resistance is zero), so the current is limited only by the resistance R of the coil: VDC ​= IDC ​R 12 = 4 × R R = 12 / 4 ​= 3 Ω  Now, consider the AC circuit. The coil is connected to a 12 V, ω = 50 rad / s AC source, and the current is IAC ​ =2.4 A. In an AC circuit with a coil (inductor and resistor), the impedance Z determines the current: VAC = IACZ 12 = 2.4 × Z Z = 12 / 2.4 ​= 5 Ω The impedance Z of a series RL circuit is given by: where R is the resistance, L is the inductance, and ω is the angular frequency. We have Z = 5 Ω, R = 3 Ω, and ω = 50rad/s. Substitute these values into the impedance formula:   Square both sides: 25 = 9 + 2500L² 16 = 2500L² L² = 16/2500 Taking square root both sides: L = 4 / 50 L = 0.08 H