Parallel RC circuit in AC system?

Question

What value of resistance should be placed in parallel with a 50 microfarad capacitor in order to have a total power factor angle of -30 degrees on a 60 Hz AC system?

Answer 1

This problem requires a little bit of trigonometry and some understanding of complex impedances.

Impedance of the capacitor is:  Zc = 1/(j* 2 * pi * 60 * 50e-6)

Zc = 0 - j 53.052 ohms

Since we want to find R that will result in the net impedance having an angle of -30°, we can use the tangent function to determine this.

tan(30) = imaginary component / real component.

However, this is a parallel circuit question, so it is better to shift from impedances to admittances, which are the reciprocal of impedances.

Y = G + jB = 1/Z = 1/(R + jX)

So, the imaginary component is the capacitor admittance, Bc

Bc = 1/(53.052 ohms) = 2*pi*60 * 50e-6 = 0.018850 mhos (or Siemens)

Then,

G = Bc / tan(30°) = 0.032648 mhos

Yequivalent = 0.032648 + j 0.18850 mhos

When G is inverted, we get R = 1/G = 30.629 ohms.

Note that the equivalent impedance is:

Zeq = 1/Yeq = 22.972 - j13.263 ohms

= 26.526 @ -30° ohms