When the current in the winding changes from 0 to 20 A in 0.2 s, an EMF

When the current in the winding changes from 0 to 20 A in 0.2 s, an EMF of self-induction of 60 V appears in it. Determine the inductance of the coil.

Initial data: I1 (initial current in the winding) = 0 A; I2 (final current rate) = 20 A; Δt (time of rise of the current in the winding) = 0.2 s; Eis (self-induction EMF generated in the coil) = 60 V.

The inductance of the coil can be expressed from the formula: Eis = L * ΔI / Δt = L * (I2 – I1) / Δt, whence L = Eis * Δt / (I2 – I1).

Let’s calculate: L = 60 * 0.2 / (20 – 0) = 0.6 G.

Answer: The inductance of the coil is 0.6 H.