# Current values after power factor correction

I am a student tasked with correcting the power factor of an RC circuit to unity with a parallel inductor... Vrms = 240V, 50Hz R = 12ohm C = 220uF. I have calculated the inductor value by calculating the reactive power. I simulated the circuit and the voltage and current are now in phase so I am happy it is correct.

Before the power factor was corrected the current waveform had a peak of 18A and a negative peak of -18A.

As can be seen in the image the current waveform now has a peak of 25.42A and a negative peak of 2.37A, with an rms value of 16.07A.

I just don't understand how these values are correct for the current. Can anybody explain why they are.

• Andy's right here its the initial phase condition but here is a caveat on Remanence. See here the comparison tinyurl.com/yzdktbh8 I'm going back to refinishing my hardwood floor. Commented Mar 17, 2021 at 13:23
• I just wanted to compliment you on your thought process and reaction. You solved the problem, but something seemed not right and you were perplexed. So you asked on here to see if someone could explain it, and they did, and now you understand much better. This is all very good. Good job. You have the right philosophy and instinct and attitude to be a good engineer. When little things are out of place, whenever possible, track it down and figure it out to make sure it will not become a bigger problem later. Commented Mar 21, 2021 at 20:16

Instead of beginning the transient analysis with a sinewave at 0°, try driving it with a cosine wave that begins at 90°. In this way, you automatically cause the inductor current to begin rising through zero and you won't need to wait for the circuit to settle down: -

Of course this then puts emphasis on the capacitor conducting current and you might find a happy medium around 45°. Try it and see.

• Due to transformer and motor Remanence this won't be actual as this creates an initial condition in real grids and ignored in this simulation. So in transients you want to reclose in the same phase that the current was broken on grids. ABB has systems to do this and protect high power magnetics from satuation. Commented Mar 17, 2021 at 13:20
• Starting the simulation with the phase at 90° gave me the values I expected and I understand why you suggested it. Thanks very much. Commented Mar 17, 2021 at 13:32

Well, in order to achieve $$\\cos\varphi=1\$$ we need the reactive power to be equal to $$\0\space\text{VAR}\$$. The reactive power in your initial circuit is given by:

$$\text{Q}=\frac{240^2}{\sqrt{12^2+\left(\left(2\pi\cdot50\cdot220\cdot10^{-6}\right)^{-1}\right)^2}}\cdot\sin\left(\frac{3\pi}{2}+\arctan\left(\frac{12}{2\pi\cdot50\cdot220\cdot10^{-6}}\right)\right)=$$ $$-\frac{1742400 \pi ^2}{\sqrt{\left(36000000+121 \pi ^2\right) \left(15625+1089 \pi ^2\right)}}\approx-17.6486\space\text{VAR}\tag1$$

Using a parallel inductor, we get (and I will let you show that that is the case):

$$\varphi=\begin{cases} \arctan\left(\omega\text{CR}\right)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{if}\space\space\omega\text{L}-\frac{1}{\omega\text{C}}=0\\ \\ \arctan\left(\omega\text{CR}\right)-\arctan\left(\frac{\omega\text{L}-\frac{1}{\omega\text{C}}}{\text{R}}\right)\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{if}\space\space\omega\text{L}-\frac{1}{\omega\text{C}}>0\\ \\ \arctan\left(\omega\text{CR}\right)-\left(\frac{3\pi}{2}+\arctan\left(\frac{\text{R}}{\frac{1}{\omega\text{C}}-\omega\text{L}}\right)\right)\space\space\space\space\space\space\space\space\text{if}\space\space\omega\text{L}-\frac{1}{\omega\text{C}}<0 \end{cases}\tag2$$

The general formula for the reactive power in your circuit with parallel inductor is:

$$\text{Q}_\text{new}=\frac{\text{V}_{\text{in|rms}}^2}{\omega\text{L}}\cdot\sqrt{\frac{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}{\text{R}^2+\left(\frac{1}{\omega\text{C}}\right)^2}}\cdot\sin\left(\varphi\right)\tag3$$

Now, it is not hard to show that there is no real solution, for $$\\text{L}\$$, such that:

$$\frac{\text{V}_{\text{in|rms}}^2}{\omega\text{L}}\cdot\sqrt{\frac{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}{\text{R}^2+\left(\frac{1}{\omega\text{C}}\right)^2}}=0\tag4$$

So, we need to take a look at when:

$$\sin\left(\varphi\right)=0\tag5$$

When $$\0\le\varphi<2\pi\$$. And that gives:

$$\varphi=0\space\vee\space\varphi=\pi\tag6$$

Now, using $$\(2)\$$ and $$\(6)\$$ you can show that:

$$\text{L}=\text{CR}^2+\frac{1}{\omega^2\text{C}}\tag7$$