When an inductive coil is connected to the U = 220 V DC power supply, the current is I = 20 A; when it is connected to the F = 50 Hz, u = 220 V AC power supply, the current is I = 20 a

When an inductive coil is connected to the U = 220 V DC power supply, the current is I = 20 A; when it is connected to the F = 50 Hz, u = 220 V AC power supply, the current is I = 20 a

Well, there's no electricity
The inductance coil hinders the AC action, and the general problem is that there is no current

When an inductive coil is powered by a 10V DC power supply, the measured current is 2.5A. When the power supply is 50 Hz, the measured current is 10A. Calculate the parameter r.l.of the coil

A known quantity, the voltage of 50 Hz power supply, is missing. Obviously, it is not 10 V. It is assumed that it is 220 v
DC resistance: r = 10 / 2.5 = 4 ohm
Total reactance: z = 220 / 10 = 22 ohm
Inductive reactance to 50 Hz: RL = root sign (22 * 22-4 * 4) = root sign 468 = 21.64 ohm

There is a coil whose resistance can be ignored. Connect it to a 220 V, 50 Hz AC power supply. The current through the coil is measured to be 2A. Calculate the inductance of the coil

Inductive reactance XL = 220 / 2 = 110 ohm, XL = 2 Π LF, l = 110 / (2 × 3.14 × 50) = 0.35

One coil is connected to u = 120V DC power supply, I = 20A; if it is connected to f = 50 Hz, u = 220 V AC power supply, I = 28.2a. Calculate the resistance R and inductance L of the coil

The calculation formula of resistance 6 ohm is L = (USN ^ 2) / L. the meaning of each letter: u represents the permeability of the medium in the coil, s represents the area of the coil, n represents the number of turns of the coil, and l represents the length of the coil[