# complex power supplied

Discussion in 'Homework Help' started by tunasengok, Feb 18, 2012.

Nov 19, 2011
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2. ### epsilonjon Member

Feb 15, 2011
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First step I would take would be to use the complex power supplied by the voltage source (and the known value of the voltage source itself) to calculate the current through it. What do you get if you do this? You can then use Kirchoff's current law on one of the nodes of the parallel impedances to calculate the current through L. It's then straightforward to calculate L itself.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It would be surprising that the source produces leading power factor. It might be worth checking the complex phasor angle is negative rather than positive.

Last edited: Feb 19, 2012
4. ### nishu_r Member

Jun 2, 2012
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Can anyone explain how i can go about solving this problem?

well using the nodal analysis if i find the current through inductor, what then? how to obtain value of L.

thanks

5. ### jegues Well-Known Member

Sep 13, 2010
735
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All the complex power supplied to the circuit must be absorbed by the two inductors in the circuit.

Figure out how much complex power the known inductor is absorbing and what remains must be absorbed by the unknown inductor L.

From here you should be able to deduce the value of L.

6. ### WBahn Moderator

Mar 31, 2012
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It's hard to say since there are two sources in the circuit. But it's definitely one of those sanity checks that should be performed. It may be correct, but sufficient analysis should be done to establish that it really is correct and it really does make sense that it works out the way that it does.

7. ### WBahn Moderator

Mar 31, 2012
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Demons lie along that path.

First, there are two sources, so you have to determine how much reactive power both are supplying. Second, the two sources are probably not supplying reactive power in phase with each other. Third, even if there were only one source, knowing how much reactive power is associated with one inductor and the source is not sufficient to determine the reactive power in the other inductor. Think of a resonant circuit. At resonance, the supply is not providing any reactive power yet both the capacitor and the inductor have reactive power associated with them. Off resonance, you should be able to come up with a frequency in which the magnitude of the reactive power in the source, the cap, and the coil are all equal. Now, the laws of physics still hold and so carefully computing the reactive powers being very careful to take the phase relationships into account should work.

8. ### WBahn Moderator

Mar 31, 2012
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If you want to know the value of a resistor, you need to know BOTH the voltage across it and the current through it at the same moment in time. The same is true for a capacitor or an inductor. In the time domain, you need the value of one and the derivative of the other at the same moment in time. In the phasor domain, the complex impedance takes care of this for you and, like the resistor, you just need to know BOTH the phasor voltage across it and the phasor current through it, combined with the impedance for an inductor.

A good place to start is from what you need and work toward what you know. In this case, we need both the (phasor) voltage and current associated with L. Okay, what is the most closely related information that, if we had it, would allow us to calculate that? How about the voltage across and the current through the RL branch containing the unknown L? Now treat these as two separate sub problems: How can I get the voltage across the branch and how can I get the current through the branch.

A different approach is to simply work the following problem symbolically: Given the circuit with an inductor of unspecified value L, determine the complex power, in terms of L, that is supplied by the voltage source? Once you have that, then you just set it equal to the given complex power and solve for L.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You are absolutely right. Goes to show one shouldn't assume anything without some careful thinking. It's actually a great problem.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Per the recommended caution regarding my assumption I've attempted to solve the problem.

My approach was

1. Choose a suitable reference point on the circuit.
2. Assume the complex power from the voltage source is correct - I assume the source delivers a leading power factor.
3. Calculate the current from the voltage source.
4. Calculate the drop in the series 14Ω + 7/8H section [I'll call it Vd]
5. Calculate the resulting voltage across the current source [I'll call it Vp]
6. Find the total current Ip into the parallel branch [20Ω||(12+jXL)] - Current from step 2 above + 2A
7. Calculate the branch impedance Zp=Vp/Ip
8. Equate the value Zp to the parallel section comprising [20Ω||(12+jXL)] to determine XL and hence L.
Problem is I end up with Zp=16.475-j5.293 Ω which is incompatible with the unknown branch equivalent of [20Ω||(12+jXL)]. So I'm stumped if there is a solution.

11. ### WBahn Moderator

Mar 31, 2012
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I solved it as well and came up with a complex inductance, namely (-1.262-j1.591)H. Clearly that is not possible. I haven't actually checked, but I'm pretty sure my answer and yours don't agree. I know it is very possible I've made a mistake.

Taking your answer, it would appear that if the "inductor" were actually a 2.16mF capacitor, it would come close to satisfying your solution.

I'm gonna play with it some more. I'm thinking that generating a plot of the power factor of the voltage source as a function of the unknown reactance should shed quite a bit of light on things.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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If a cap is used instead of an inductor the effective impedance would have the form

$Z_p=\frac{7.68E3 + 20X_C^2}{1024+X_C^2}-j\frac{400X_C}{1024+X_C^2}$

I then have to satisfy two conditions with my actual Zp=16.475-j5.293 Ω

namely the real part

$\frac{7.68E3 + 20X_C^2}{1024+X_C^2}=16.475$

and the imaginary part

$\frac{400X_C}{1024+X_C^2}=5.293$

For the real part I get Xc=51.061Ω

For the imaginary part I get Xc= 57.88Ω or 17.69Ω

So it's clear even this approach wouldn't give a solution either. I can't equate the 3 elements comprising the required Zp [L replaced by C] with the given value constraints - notwithstanding the open choice of capacitance.

If I removed the 3 existing elements in parallel with the current source & placed a series impedance Zs=16.475-j5.293 Ω in parallel with the current source then the result would be consistent.

Last edited: Jun 23, 2012
13. ### WBahn Moderator

Mar 31, 2012
18,085
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Agreed. I said a cap would be close, not exact. Using a 2.16F cap would yield a branch impedance (including the 20Ω resistor) of (17.67-j5.293)Ω. I suppose an arguably closer match could be had with a slightly larger cap to spread the error between the two.

I've got the equations worked out (assuming no errors) for the apparent power in the voltage source as a function of device reactance of that component (I assumed it was a purely reactive element). So I can plot the trajectories for both capacitive and inductive components. Since it is parametric, I can tweak the other components to see if any of them bring about a match.

14. ### WBahn Moderator

Mar 31, 2012
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Here is a plot of the complex power in the supply as a function of capacitive reactance (blue) or inductive reactance (red) for the load. The black plots are for the targeted magnitude and phase angle that was requested in the problem.

As can be seen, a capacitive solution exists that is near the target. But if the sign of the angle asked for on the complex power is flipped, then the power curve doesn't come anywhere close to it.

I think putting together a spreadsheet to parametrically plot something like this is a very useful exercise. It let's you change all of the parameters, such as the source value or any of the component values, and see how that affects the power curve as the load reactance is swept.

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15. ### nishu_r Member

Jun 2, 2012
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Sir, in the step where you calculate current from the voltage source then finding the voltage drop across the impedance $(14 + j7)$Ω, don't we have to take in account the current from the current source?

16. ### WBahn Moderator

Mar 31, 2012
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How would you take it into account?

Since the voltage source, 14Ω resistor, and the 0.875H inductor are in series, whatever current flows in one IS the current flowing in all three. It doesn't matter how you arrive at that current (as long as you do it correctly). So, given the voltage sources voltage and the complex power it is delivering, you can calculate the current. Once you have that current, you have the current in the other two components as well.

Does this mean that the current in the current source has no affect on the current in the voltage source? Of course not. But the effect it has is taken into account in the complex power that was given (assuming it is correct, which all indications are that it is not). If you change the current in the current source, you would have a different complex power for the voltage source, but whatever it turns out to be, you can still use it to calculate the current in the voltage source.

17. ### nishu_r Member

Jun 2, 2012
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So it means that the current we arrive at using the complex power delivered by Voltage source in this circuit is actually the sum of currents from the current source and the current due to the voltage source, i.e it implies? am i right sir.

So the same voltage source for the same impedance configuration would deliver a different complex power in the absence of the current source?

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I think you've missed the point.

The problem itself is based on an incorrect assumption. This is an old thread which started back in February. My guess is that the OP either incorrectly transcribed the problem or the original source was flawed. Either way, the OP has not returned with follow-up information about the problem so we are forever clueless with respect to that matter.

Perhaps WBahn would be kind enough to pose a similar problem for you to solve - with the condition that a valid solution is possible. You could attempt to solve said problem and the discussion surrounding a correct solution would better inform you of the steps & assumptions involved.

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19. ### nishu_r Member

Jun 2, 2012
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where did i go wrong in my statement sir? , and yes a similar problem would be of lot help since i am not able to comprehend the mistake in my understanding. thanks for your time.

20. ### WBahn Moderator

Mar 31, 2012
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Yes, on all points. The voltage source does not live in isolation. It is part of a circuit and the power it provides (or absorbs) is a result of the total interaction of that source with the entire circuit. Change the circuit and you change the interaction.

Consider a circuit consisting of two batteries with the negative terminals connected together and the positive terminals tied to the opposite ends of a resistor. For discussion sake, let's say that one battery, the one we want to know the power about, is 10V and the resistor is 10kΩ. Now let's say that the other battery is 9V. How much power is delivered by the 10V battery? It would be [(10V-9V)/10kΩ]10V = 1mW. Now what if the other battery was 15V? It would be [(10V-15V)/10kΩ]10V = -5mW, the minus sign indicating that the battery is actually absorbing power. What if the other battery were removed and the connection made to ground? Then it would be 10mW. What if the other battery were -10V? Then it would be 20mW.

See how that works?

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