Complex Power

Thread Starter

RoKr93

Joined Jun 17, 2013
24
Hopefully this is the last you'll hear from me regarding this course. ;)

I think I went about this the wrong way:


I don't have ω, which makes for an ugly answer. In addition, I know that S = V_eff x I_eff*, and I didn't use effective values, which means my result is probably wrong. But it'd be hard to find the effective value of the voltage without knowing ω so I could find the magnitude. The current source is throwing me off- if it were just another impedance element, it'd be a simple matter to just combine impedances and whatnot, but I'm not sure what to do in this situation.
 

t_n_k

Joined Mar 6, 2009
5,448
You don't need the angular frequency. The inductance is given in ohms not Henry.
 

Thread Starter

RoKr93

Joined Jun 17, 2013
24
Oh jeez...

I think I've been studying for too long. Good catch; thank you.
 

WBahn

Joined Mar 31, 2012
24,581
Remember when I tried to stress the importance of tracking your units?

Well, you chose to ignore me and almost immediately you paid the price.

You have an given inductive impedance of j3Ω. In your first equation you put jωL. But then in your second equation you just grab the number 3 from the given impedance and throw it at the equation. Where as had you done it correctly you would have had j(j3Ω)ω which is -3Ωω and, either immediately or eventually would have seen that the units don't work out. But, instead, you insist on thinking that "wasting" time with all this units nonsense is pointless and proceed to waste a lot more time and end up with an answer that you have no idea whether it is wrong or not, when the fact that it was wrong was evident in the very second line.
 

xbt

Joined Aug 2, 2013
3
A trick or bad question... ?
Are we assuming perfect sources here (ie zero internal impedance) ?
If so then surely as the question is worded then the 0.5∠90∘amps source won't absorb any complex power ?
On the other hand is the 2Ω resistor is part of the 0.5∠90∘amps source ? (it is the only resistive element in the circuit), then Vs = 0.5∠90∘A x 2Ω = 1∠90∘volts and it would absorb V*V/R watts i.e. 1/2 = 0.5 watt. (question doesn't mention d.c. power as might be absorbed from 5V dc source. )

It occurs to me that the circuit might have been drawn incorrectly, such that the question intended 2Ω to be the internal resistance of the 5V d.c. source and the j3Ω inductive reactance to be the internal impedance of the a.c. source.
 

WBahn

Joined Mar 31, 2012
24,581
A trick or bad question... ?
Why do you say that?
Are we assuming perfect sources here (ie zero internal impedance) ?
There's no basis to assume anything else.
If so then surely as the question is worded then the 0.5∠90∘amps source won't absorb any complex power ?
Why do you say that?
On the other hand is the 2Ω resistor is part of the 0.5∠90∘amps source ? (it is the only resistive element in the circuit), then Vs = 0.5∠90∘A x 2Ω = 1∠90∘volts and it would absorb V*V/R watts i.e. 1/2 = 0.5 watt. (question doesn't mention d.c. power as might be absorbed from 5V dc source. )
So you are saying that, if the 2Ω is treated as the internal source of the current supply that it will ABSORB 0.5W of power independent of how it is used in a circuit?
It occurs to me that the circuit might have been drawn incorrectly, such that the question intended 2Ω to be the internal resistance of the 5V d.c. source and the j3Ω inductive reactance to be the internal impedance of the a.c. source.
What 5V DC source?

There is an AC voltage source that has a phase angle of 0°.

The only ambiguous thing about the problem is whether the voltage and current values on the sources are amplitude or RMS.


Just treat the current source as a black box load that is specified as having a current of 0.5A at and angle of +90°, analyze the circuit (there's only a single unknown node voltage) and multiply the voltage across the black box by the current through the black box and you have the power absorbed by the black box.
 
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