Question

An alternating pd 220V amplitude and frequncy 50Hz is applied to a series circuit containing R=400 ohms, L=0.35H and C= 18 micro farad (a) what is Z? (b) what is the current amplitude ,is it lagging the supply pd (c) what is the power factor?

Answer 1

a)

The impedance of the circuit is

$Z=\sqrt{R^2+(2\pi fL-\frac{1}{2 \pi f C})^2}$

$=\sqrt{(400)^2+(2 \pi (50)(0.35)-\frac{1}{2 \pi (50) (18 \times 10^{-6})})^2}$

=$405.5 \Omega$

b)

The current amplitude is

$I=\frac{V}{Z}=\frac{220}{405.5}=0.54 A$

c)

The power factor is

$=\frac{R}{Z}=\frac{400}{405.5}=0.986$