# Power factor correction in a simple phasor circuit

I've been having a bit of trouble with a relatively simple power factor correction problem involving a simple circuit in the frequency domain ($f=50\,\text{Hz},\omega=2\pi f=100\pi\,\text{rad/s}$).

From what I can tell, my first step is to determine the complex power $S=V_{eff}I_{eff}^*=\frac{V_{eff}V_{eff}^*}{Z_{eff}^*}=\frac{|V_{eff}|^2}{Z_{eff}^*}$ consumed by the circuit in its current state (without the corrective capacitor), in which case I figured the total impedance should be $Z=-j10\,\Omega+\dfrac1{\dfrac1{20\,\Omega}+\dfrac1{j20\,\Omega+10\,\Omega}}\approx11.435\angle{-19.653^\circ}\,\Omega$.

However, this strikes me as quite odd; $\arg S=\arg Z<0$ suggests that the power is lagging and therefore (intuitively) the power factor cannot be 'corrected' with the use of another capacitor (negative reactance), requiring instead an inductor (positive reactance).

In fact, when trying to solve this problem, I found two results -- both giving positive reactances $X_C=-j\frac1{\omega C}$ and thus 'negative' capacitances. This is nonsensical, no? Have I computed the total impedance incorrectly?

Thanks.

• What about an inductor ;) – JonRB Dec 7 '14 at 21:45
• Yes, of course an inductor would work to correct the power factor :-) This would explain the positive reactance. The problem itself however speaks of a capacitor, which is my point of confusion. – oldrinb Dec 7 '14 at 21:48

Your total impedance is correct. As you can see cos(-19.653)=0.94 which is below 0.95.

If you calculate the total impedance when the 10 Ohms resistor is shorted you'll find 10<0, i.e. PF=1.

Therefore you can increase the PF by placing a small impedance accross the 10 Ohms resistor, and then the equivalent impedance modulus of both components will be below 10 Ohms (and below the small impedance). To get PF=1 you would place a really small reactance, whether it is an inductor or a capacitor.

To get PF=0.95 you can also succeed either with an inductor or with a capacitor.

Let's solve this using a capacitor. One way to do it is to calculate the input impedance (seen from the source).

If I is the current flowing from the source, the source voltage can be written as

$$V=-j10·I + \frac { I·(j20+Zx) } {20 +j20+Zx }·20$$

The right term is the voltage in resistor 20 obtained by multiplying the current in this resistor (current divisor formula) by 20. And Zx is parallel between 10 Ohms and unknown reactance $Zx=\frac {-jXc·10} {10-jXc}$.

Therefore the Z, input impedance is

$$Z=-j10 + \frac { (j20+\frac {-jXc·10} {10-jXc}) } {20 +j20+\frac {-jXc·10} {10-jXc} }·20$$

Using algebra (or some symbolic math package) you can get real part and imaginary part of Z.

Dividing Imaginary part and real part, and simplifying, you get: $$\frac {\Im (Z)}{\Re{(Z)}}=-2.5·\frac {Xc^2} {7Xc^2-40Xc+400}$$

This division must be equal to $$-\tan(\arccos(0.95)\approx -0.329$$

Solving for Xc, you get 2 solutions, taking the positive one, $Xc=8.821 \Omega$. The capacitance is $C \approx 361 \mu F$

• Thank you for the response. How would the power factor of the supply be 1 with the 10 Ohm shorted? I'm not following. – oldrinb Dec 7 '14 at 22:12
• If you try $Z=-j10 + \frac 1 { \frac 1 {20} + \frac 1 {j20}}\ = 10 \Omega$. PF=cos(0)=1. – Roger C. Dec 7 '14 at 22:17
• And by placing a capacitive reactance of 0.01 Ohms, for example, in parallel with 10 Ohms resistor you're close to short it. If you look at the equivalent $\frac 1 { \frac 1 {10} + \frac 1 {-j0.01}}\approx -0.01 j$. – Roger C. Dec 7 '14 at 22:24
• Okay, returning to your answer it appears I made a critical error earlier. As it stands now, I've computed the power $S$ consumed by the circuit as it stands now and then assumed the supplied power after adding the capacitor would just be $S'=S-j\omega C|V|^2$ where $V$ is the voltage across the across the $10\,\Omega$ resistor and thus also the voltage across the capacitor. Does this approach seem correct to you? – oldrinb Dec 7 '14 at 23:37
• If current changes in a branch voltages inside do change too. I've updated the answer to show one way to solve this problem. – Roger C. Dec 8 '14 at 1:04