Question

A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz

Answer 1

The reactance of the coil is given as follows:

$$X_L = \omega L$$

where $L = 400mH = 0.4H$ is the inductance of the coil, and

$$\omega = 2\pi f$$

is the angular frequency of the source ($f = 1000Hz$).

According to the Ohm's law, the effective voltage across the coil is:

$$U = IX_L = 2\pi fLI$$

where $I = 6mA = 0.006A$ is the effective current through the coil. Thus, obtain:

$$U = 2\pi\cdot 1000\cdot 0.4\cdot 0.006 \approx 15.08V$$

Answer. 15.08 V.