Verifying Complex Power Calculations

In summary, the conversation discussed the calculation of complex power delivered by a source using the formula S=VI*. There was some confusion about using the conjugate or not, but it was determined that using S=VI* gives the true active power dissipated in the circuit. There was also mention of a different formula, S=1/2*Vm*Im*, but it is unclear where this formula came from. The conversation also touched on the use of RMS values in complex power calculations.
  • #1
eehelp150
237
0

Homework Statement


Calculate the complex power delivered by the source
V = 12cos(wt) V

upload_2016-10-24_21-11-55.png


Homework Equations


V = IR

The Attempt at a Solution


1. I combined 8ohm resistor and 8j ohm inductor in parallel to get 4+4j ohms
2. I combined that with 4ohm resistor in series to get a Zth of 8+4j ohms
3. V = IR -> I = V/R = 12/(8+4j) amps
4. I = 6/5 - 3i/5

Complex power = V * I conjugate =
12 * (6/5 + 3i/5)
= 14.4 + 7.2i

Did I do this right?
 

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  • #2
eehelp150 said:
Complex power = V * I conjugate =
12 * (6/5 + 3i/5)
= 14.4 + 7.2i
Hmm..Your calculations are all correct but I am not sure about this one. When we say complex power "absorbed" by the load, we take the conjugate of the current. So for lagging loads, the complex power comes out to be P+jQ, with positive imaginary part, indicating absorption of reactive power. When the load is leading, the complex power "absorbed" by the load becomes P-jQ, where imaginary part is negative, indicating it is supplying reactive power. In your problem, the source is supplying reactive power and the load is absorbing that reactive power. Hence, IMO, complex power supplied by the source should be VI and not VI*. So it should be S=14.4-j7.2.
 
  • #3
cnh1995 said:
Hmm..Your calculations are all correct but I am not sure about this one. When we say complex power "absorbed" by the load, we take the conjugate of the current. So for lagging loads, the complex power comes out to be P+jQ, with positive imaginary part, indicating absorption of reactive power. When the load is leading, the complex power "absorbed" by the load becomes P-jQ, where imaginary part is negative, indicating it is supplying reactive power. In your problem, the source is supplying reactive power and load is absorbing that reactive power. Hence, IMO, complex power supplied by the source should be VI and not VI*. So it should be S=14.4-j7.2.
The current is I = 6/5 - 3i/5
Wouldn't the complex conjugate be 6/5 + 3i/5?
 
  • #4
eehelp150 said:
The current is I = 6/5 - 3i/5
Wouldn't the complex conjugate be 6/5 + 3i/5?
Yes but I think we should not take the conjugate here. The expression S=VI* gives the complex power "absorbed" by the load. The question is asking the complex power supplied by the source. Here, reactive power is being supplied by the source, hence, it should be negative.
 
  • #5
I'm pretty sure that the question author intended that the power delivered by the source is to be interpreted as being identical to the power "handled" by the load. I'd use the complex conjugate of the source current to find that power, and leave any interpretations of the imaginary component for later.
 
  • #6
cnh1995 said:
Yes but I think we should not take the conjugate here. The expression S=VI* gives the complex power "absorbed" by the load. The question is asking the complex power supplied by the source. Here, reactive power is being supplied by the source, hence, it should be negative.
So source delivers 14.4 - 7.2i and load absorbs 14.4 - 7.2i?
 
  • #7
You had it right in your first post.
 
  • #8
gneill said:
You had it right in your first post.
How would I know which (conjugate or not) to use on an exam?
 
  • #9
I think you should use the formula VI* as gneill said earlier. It holds for both sending and receiving ends. I think the interpretation of reactive power I wrote is incomplete (or wrong, maybe). I'll search and post if I find any explanation for this.
 
  • #10
cnh1995 said:
I think you should use the formula VI* as gneill said earlier. It holds for both sending and receiving ends. I think the interpretation of reactive power I wrote is incomplete (or wrong, maybe). I'll search and post if I find any explanation for this.
Well, after doing some math, it turns out that the conjugate part has nothing to do with reactive power but it affects the active power P. VI* gives the true active power P dissipated in the circuit. If VI is used, the value of P (real part of complex power) is different from the actual active power dissipated in the circuit. So, always use S=VI*. S=VI is not even a valid expression. My apologies for misguiding you with this equation.
 
  • #11
cnh1995 said:
Well, after doing some math, it turns out that the conjugate part has nothing to do with reactive power but it affects the active power P. VI* gives the true active power P dissipated in the circuit. If VI is used, the value of P (real part of complex power) is different from the actual active power dissipated in the circuit. So, always use S=VI*. S=VI is not even a valid expression. My apologies for misguiding you with this equation.
It's alright, I really appreciate your help!
 
  • #12
gneill said:
You had it right in your first post.
cnh1995 said:
Well, after doing some math, it turns out that the conjugate part has nothing to do with reactive power but it affects the active power P. VI* gives the true active power P dissipated in the circuit. If VI is used, the value of P (real part of complex power) is different from the actual active power dissipated in the circuit. So, always use S=VI*. S=VI is not even a valid expression. My apologies for misguiding you with this equation.

Apparently the correct formula was:
1/2 * S * I*

Any idea why?
 
  • #13
eehelp150 said:
Apparently the correct formula was:
1/2 * S * I*

Any idea why?
Where did you get this formula?
 
  • #14
cnh1995 said:
Where did you get this formula?
Solutions were posted. Solution for this problem is:
S = 1/2 * Vm * Im* = 7.2 + j3.6 VA

I looked through my notes and this was in it:
S = V * I* = 1/2 * Vm * Im<θv-θi = Vrms * Irms <θv-θi

Also in my notes:
Vrms = Vm/sqrt(2)
Irms = Im/sqrt(2)

I'm assuming the voltage source provided (12cos(wt)) was not in RMS, so in order to use S = V * I*, we have to convert it to RMS first? It's weird because my textbook uses the formula S = V * I* and not S = 1/2 * V * I*

I will ask the professor in person but do you have any idea when I'm supposed to used S = V * I* and when I'm supposed to use S = 1/2 V * I*?
 
  • #15
eehelp150 said:
S = 1/2 * Vm * Im*
Ah.. that's right. I missed that the voltage given is peak voltage and not rms. I read only the 12∠0° part and didn't notice V=12coswt written above the diagram, which indicates 12V is actually the peak voltage. So when the voltage equation is given in terms of peak voltage, use
eehelp150 said:
S = 1/2 * Vm * Im*
Or, convert it into rms and then use
S=VI*.
 
  • #16
cnh1995 said:
Ah.. that's right. I missed that the voltage given is peak voltage and not rms. I read only the 12∠0° part and didn't notice V=12coswt written above the diagram, which indicates 12V is actually the peak voltage. So when the voltage equation is given in terms of peak voltage, use

Or, convert it into rms and then use
S=VI*.
So if I were given only the diagram, (12<0 V), should I assume peak or RMS? When the function is given (12cos), should I always assume peak unless it specifically says RMS?
 
  • #17
eehelp150 said:
When the function is given (12cos)
It means the peak voltage is 12V. RMS value is never written in terms of a sinusoidal equation since it's the dc equivalent of the ac waveform. So whenever you see a sinusoidal equation, it gives the peak of the sinusoid. If the voltage is given directly as 12∠0°, it could be rms or peak. It is usually mentioned in the problem whether to read it as rms or peak. But in some of my books, it is assumed to be rms by default.
 
Last edited:
  • #18
cnh1995 said:
It means the peak voltage is 12V. RMS value is never written in terms of a sinusoidal equation since it's the dc equivalent of the ac waveform. So whenever you see a sinusoidal equation, it gives the peak of the sinusoid. If the voltage is given directly as 12∠0°, it could be rms or peak. It is usually mentioned in the problem whether to read it as rms or peak. But in some of my books, it is assumed to be rms by default.
So in other words, if I see the sinusoidal equation, I should always convert to RMS first.
 
  • #19
eehelp150 said:
So in other words, if I see the sinusoidal equation, I should always convert to RMS first.
If you intend to work with power, yes. Otherwise it's not really an issue, particularly if you're expected to supply an answer for currents or voltages in the same form as given (you'd just end up removing the same scaling factor at the end).
 

Related to Verifying Complex Power Calculations

1. What is complex power?

Complex power is a mathematical representation of the total power in an AC circuit, taking into account both the active power (real power) and the reactive power. It is expressed as a complex number, with the real part representing the active power and the imaginary part representing the reactive power.

2. Why is it important to verify complex power calculations?

Verifying complex power calculations ensures that the power measurements in an AC circuit are accurate and reliable. This is important for troubleshooting and maintaining the circuit, as well as for ensuring the safety and efficiency of the circuit.

3. How can I verify complex power calculations?

Complex power calculations can be verified through various methods, including using power meters, analyzing circuit diagrams, and performing mathematical calculations using the known values of voltage, current, and power factor.

4. What are some common errors in complex power calculations?

Some common errors in complex power calculations include incorrect measurement of voltage or current, incorrect use of power factor, and not taking into account both the real and reactive power components. It is important to double-check all input values and calculations to avoid these errors.

5. Is there a standard for verifying complex power calculations?

Yes, there is a standard for verifying complex power calculations called the IEEE Standard for Verification of Complex Power Calculations (IEEE Std 1459-2010). This standard provides guidelines and methods for verifying complex power calculations in AC circuits.

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