# When a constant voltage of 15 V was applied to the coil, the current in it was 0.5 A.

**When a constant voltage of 15 V was applied to the coil, the current in it was 0.5 A. When the same alternating voltage was applied with a frequency of 50 Hz, the current was reduced by 40%. What is the inductance of the coil?**

We find the active resistance of the coil using Ohm’s law, we get:

Ra = U / I = 15 / 0.5 = 30 Ohm

Since its strength decreased by 40% to alternating current, the total resistance increased by 40%, then it will become equal to:

R = Ra + 40 * Ra / 100 = 1.4 * Ra = 42 Ohm.

The total resistance of the circuit is reflected by the formula:

R ^ 2 = Ra ^ 2 + Rr ^ 2.

Then:

Rr ^ 2 = R ^ 2 – Ra ^ = 42 ^ 2 – 30 ^ 2 = 1764 – 900 = 864 = 29.4.

We use the formula for inductive reactance:

Rl = ωL;

L = Rl / ω = Rl / 2πf = 29.4 / 2π * 50 = 0.09 H

Answer: 90 mH.