Question

An inductor, L, of 2 H and a resistor, R, of 1000Ω are in series across an a.c. voltage of 10 V, frequency 60 Hz. Calculate the current in r.m.s. flowing,

Answer 1

According to the Ohm's law, the current amplitude is:

$$I = \dfrac{U}{Z}$$

where $U = 10V$ is the volage amplitude and $Z$ is the total impedance of the circuit.

The impedance of the series resistor and inductor is:

$$Z = \sqrt{R^2 + (2\pi fL)^2}$$

where $f = 60Hz$ is the frequency of a.c. voltage.

Substituting the numerical values, get:

$$I = \dfrac{U}{\sqrt{R^2 + (2\pi fL)^2}} = \dfrac{10}{\sqrt{1000^2 + (2\pi\cdot 60\cdot 2)^2}} \approx 0.00798A$$

The r.m.s. of a sinusoidal current is:

$$I_{rms} = \dfrac{I}{\sqrt{2}}\approx 0.00564A$$

Answer. 0.00564 A = 5.64 mA.