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 r.m.s voltage across the inductor.

Answer 1

The impedance in the circuit

$$Z=\sqrt{R^2+(2\pi f L)^2}\\ =\sqrt{1000^2+(2\pi \times 60\times 2)^2}=1252\:\Omega$$

The r.m.s current in the circuit

$$I=\frac{V}{Z}=\frac{10}{1252}=0.008\:\rm A$$

The r.m.s voltage across the inductor

$$V_L=IX_L=I(2\pi f L)=0.008\times 2\pi \times 60\times 2=6\:\rm V$$