Issue with impedance/power correction

  • Thread starter Machinia
  • Start date
  • #1
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Main Question or Discussion Point

Say I have a circuit like this: (R+L+R||L)

-----20ohm---150mH------------
|................................|..........|
Vs.............................80mH 120ohm
|................................|..........|
---------------------------------

Vs = 120V @ 60Hz

So I decide I want to add a capacitor across the source of 28.35uF to correct the power factor to 1. Then, because I'm bored, I want to recalculate the current through the circuit to prove to myself that the current hasn't changed, only the phase angle.

Therefore I do something like -XCj || ((R+XL1) + R||XL2)

And end up with an answer of about 292 + 0j ohms.

This gives me a current much smaller than before I added the capacitor, meaning my real power has changed which isn't supposed to happen. Can someone shed some light on this problem?
 

Answers and Replies

  • #2
vk6kro
Science Advisor
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Are you sure about that capacitor?

I did a simulation and a capacitor of about 50 µF seems to get the current more in phase with the supply voltage than the one you suggest.

I haven't calculated it though.
 
  • #3
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I'm pretty certain it's the right value.
 
  • #4
vk6kro
Science Advisor
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The current in the branch with the resistors and inductors in it can't be changed by adding a capacitor across the mains supply. The mains is assumed to have zero internal resistance.

The reduction in total current is due to the capacitor's current being out of phase with the inductive branch.

So, the capacitor should only be cancelling out the reactive component of the inductive branch's impedance.
 
  • #5
1,762
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Machinia, I get the same values you do, yet when I plug those values into SPICE, I do not get voltage and current in phase, instead I get current and voltage 180 deg out of phase. Still I believe your answer is correct.

R1 + XL1 = 20 + j56.55
R2 || XL2 = 7.13 + j28.37
Z = 27.13 + j84.97

Convert Z(series) to Z(parallel)
Rp = (Rs^2 + Xs^2)/Rs = 292.92
Xp = (Rs^2 + Xs^2)/Xs = 93.58

Find Capacitance
Cp = 1/(2 * pi * f * 93.58) = 28.34 uF
 
Last edited:
  • #6
Isn't it like simple KCL problem? If you are connect a capacitor across the source that is in parallel, then the current will be diverted in to two diffrent paths. ? :smile:

Also, My circuit maker simulation says that power factor won't be unity after adding 28.35 uF capacitor across the load. :confused:
 
Last edited:
  • #7
9
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I ran the circuit through SPICE also and adding/removing the capacitor did nothing to the phase angles...

But I'll ignore that for the moment.

Instead I'm concerned with the fact that I can't recalculate my real power anymore. Without the capacitor, the current through the circuit = 1.35A @ -72.3. So the real power is:

(1.35A)^2 * 20ohm + (other current)^2 * 120ohm = 120V * 1.35A cos(-72.3) = 49W

Now add the capacitor, and the current changes to:

120V / (292 + 0j) = 1.35A cos(-72.3) = 410mA (The reactive power of the L and C cancel each other)

The real power is still 120V * 410mA = 49W, but I can't get that same value by summing the power dissipated by each resistor. That was how I was initially approaching the problem rather than just using P=V*I. Something is going wrong for me and I don't know what it is.
 

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