# Is my solution correc(about complex power)

Discussion in 'Homework Help' started by bpmy123, Jun 25, 2015.

1. ### bpmy123 Thread Starter New Member

Jun 24, 2015
6
0

The circuit is in the sinusoidal steady state.Find the active power and reactive power of Z.

|Vrms|=10V |Z|=(3^2 + 1^2)^1/2
|Irms|= |Vrms| /(|Z|+1)
then we find complex power of Z =|Irms|^2 *Z
And we can get active power and reactive power.
Is this correct? If not,how to solve it?

2. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
No. Because the total impedance is not |Z| + 1Ω

Z is the series resistance (at least effectively) of a 3Ω impedance and a -j1Ω impedance. But you found did not just add the magnitudes of the components to find |Z|. So would you do so when you have an additional 1Ω in series?

3. ### bpmy123 Thread Starter New Member

Jun 24, 2015
6
0
OK,I tried another way to solve.
Vz=(3-j)/(1+3-j)*10
Iz=10/(1+3-j)
complex power Sz=Vz * Iz^*
Is this correct now???

4. ### bpmy123 Thread Starter New Member

Jun 24, 2015
6
0
Or I can directly use
Vz=[(3-j)/(1+3-j)]*10
Sz=(|Vz|^2)/Z to skip compute Iz

5. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
You need to start using units properly.

(3-j)/(1+3-j)*10 and 10/(1+3-j) are just numbers. They are not voltages or currents.

What you should have put was:

Vz=[(3-j)Ω/(1+3-j)Ω]*10V
Iz=10V/(1+3-j)Ω

This is far more than just pedantic nitpicking. As an engineer/technician you have the moral and legal responsibility to exercise due diligence in the work you perform. This includes taking all reasonable steps to check your work for errors. Tracking your units is perhaps the single most powerful and effective error-detecting tool available to the engineer. Most mistakes you make (not all) will mess up the units and allow you to detect them almost immediately, rather than at the end of several pages of work when the final answer is so absurd that you can't help but notice that it's wrong and rather than when you are a defendant in a wrongful death case because you failed to catch a design error in a product. If nothing else, as a student, it is in your best interest to track units because doing so will let you turn in work that has fewer errors AND that took you less time to complete because you didn't waste a bunch of time cranking out solutions to problems that were guaranteed to be wrong starting with your first or second line of setup.

Try it both ways and see if you get the same answer (hint: you won't).

You've overlooked a subtle, but critical, point.

6. ### bpmy123 Thread Starter New Member

Jun 24, 2015
6
0
You're right,I'll try to correctly put unit.
Back to the square one,I compute in both way and get the reactive powers with different signs.
I confused,which way is correct,or both wrong (if I put units).
Could you give me a straight answer?

7. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
No, because you've already been given the "straight answer" in your text and/or in your lecture and probably in examples worked out in both. That didn't do it. So being told the same "straight answer" by some stranger on the internet is unlikely to make a difference.

But by struggling with it and asking the questions and fighting with it yourself, with leading hints and suggestions from us, you are highly likely to end up focusing on the key concepts that will make this make sense to YOU and, consequently, allow you to understand (not memorize) those concepts for the long term.

If we accept as our starting point that

$
\bar{S} \; = \; \bar{V} \cdot \bar{I^*}
$

(and if you don't know WHY we have I* in that formula, then you need to go back and go through that material until you do)

What happens if you replace I with its equivalent in terms of V and Z?

8. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hi,

Just a quick note...

You can calculate power in at least three ways:
P=I^2*Z
P=V^2/Z
P=V*I

and you should get the same result (including signs) for any of these.

Your goal should be to compute a single complex number like:
a+b*j

no matter what you are trying to calculate. There are cases where either a or b or both will be zero, but that's not the case here just so you know. Also, if not zero then a or b can be plus or minus, but for one particular problem they will always come out to be the same no matter how you compute it, as long as you handle the complex math right of course.

It seems like you are going about it correctly, so it seems that you just need to practice using complex math a little more, like multiplying, dividing, etc., and probably finding the complex conjugate of a complex number.

If you keep track of the units it helps to determine if you got the right answer in the end. If the units dont work out right, then you probably did it wrong.

Last edited: Jun 26, 2015
9. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
Two of these are wrong.

These don't work for complex power (which is given the symbol 'S' since 'P' is used for real power).

$
\bar{S} \; \equiv \; P \; + \; jQ
$

where S (a vector) is complex power, P (a scalar) is real power, and Q (a scalar) is reactive power. This is by definition

P and Q are previously defined and, in terms of the phasor voltage and current, are:

$
P \; = \; \| \bar{V} \| \| \bar{I} \| \cos $$\angle \bar{V} - \angle \bar{I}$$
\;
Q \; = \; \| \bar{V} \| \| \bar{I} \| \sin $$\angle \bar{V} - \angle \bar{I}$$
$

We need to subtract the phase of the current from the phase of the voltage (as opposed to the other way around) because we have defined reactive power to be positive for inductive loads and, for inductive loads, voltage leads current.

If we just multiply the phasor voltage by the phasor current, we ADD the angles where as we need to subtract them. This is accomplished by taking the complex conjugate of the phasor whose angle we need to subtract which, in this case, is the current phasor. So

$
\bar{S} \; \equiv \; P \; + \; jQ \; = \; \bar{V} \cdot \bar{I^*}
$

Saying S = V*I does NOT work and is incorrect!

We can relate the phasor voltage and current by the impedance

$
\bar{V} \; = \; \bar{I} \cdot \bar{Z}
$

Substituting this into the formula for complex power gives us

$
\bar{S} \; = \; \bar{V} \cdot \bar{I^*}
\bar{S} \; = \; \bar{I} \cdot \bar{Z} \cdot \bar{I^*}
\bar{S} \; = \; $$\bar{I} \cdot \bar{I^*}$$ \bar{Z}
\bar{S} \; = \; {\| \bar{I} \|}^2 \cdot \bar{Z}
$

To write S in terms of V and Z we start by rearranging the relations for Ohm's Law

$
\bar{V} \; = \; \bar{I} \cdot \bar{Z}
\bar{I} \; = \; \frac{\bar{V}}{\bar{Z}}
$

Substituting this into the formula for complex power gives us

$
\bar{S} \; = \; \bar{V} \cdot $$\frac{\bar{V}}{\bar{Z}}$$^*
\bar{S} \; = \; \bar{V} \cdot \frac{\bar{V^*}}{\bar{Z^*}}
\bar{S} \; = \; \frac{\bar{V} \cdot \bar{V^*}}{\bar{Z^*}}
\bar{S} \; = \; \frac{{\| \bar{V} \|}^2}{\bar{Z^*}}
$

10. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hello again,

Yes you are right, it's bad form to write it like that when at least one is a vector.

For this problem i think this works when V is not a vector but I and Z are:
P=(V*I)*
P=(I^2*Z)*
P=(V^2/Z)*

where Z is the total impedance and * is the conjugate, or in your notation:
S=(V*I)*
S=(I^2*Z)*
S=(V^2/Z)*

In this way we are always working with complex numbers alone and the math that goes with them. No trig needed.

11. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
But V is a phasor and so it IS a vector!

You are using * for two different things. I'm assuming you mean:

P = (V x I)*
P = (I^2 x Z)*
P = (V^2/Z)*

Two of these are incorrect. If V is not a vector (I assume it is the magnitude of the voltage phasor), then your first equation says that P has the angle of the current irrespective of the angle of the voltage. But it's the DIFFERENCE between the voltage and current angles that matter!

Your second equation will have the wrong sign on the angle. Your third equation will be correct.

If V and I are phasors (which that ARE!) then it is important to realize that

(V·I)* is NOT the same as V·I*

The correct relationships are

S = V·I*
S = I² · Z
S = V² / Z*

where · is vector/scalar multiplication as appropriate, * is complex conjugate, and I² and V² are the square of the magnitudes of I and V respectively.

12. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hi again,

Maybe i didnt explain this correctly. I'll use S instead of P now, also as you did.

S=(V*I)*

Now here the second asterisk is the conjugate as you guessed, and the first is multiplication:
S=(V . I)* {rule: eval inside parens first, then take the conjugate}

Now first we have:
V=10
and this is at an angle of 0 degrees. Next we have:
Z=z+1=4-j

Now compute I as:
I=V/Z=10/(4-j)

and converting that to the form a+b*j we have:
I=40/17+10/17*j {rule: multiplication and division have equal priority, eval from left to right}

Now since V=10, we have:
V times I=10*I=400/17+100/17*j

Taking the conjugate, we get:
S=400/17-100/17*j {Result #1}

The result you posed was:
P=|V|*|I|*cos(angle(V)-angle(I))
Q=|V|*|I|*sin(angle(V)-angle(I))

|V|=10
|I|=10/sqrt(17)
|V|*|I|=100/sqrt(17)
angle(V)=0
angle(I)=atan(1/4)
cos(0-atan(1/4))=4/sqrt(17)
sin(0-atan(1/4))=-1/sqrt(17)
P=100/sqrt(17)*4/sqrt(17)=400/17
Q=100/sqrt(17)*-1/sqrt(17)=-100/17

so we have:
S=P+Q*j=400/17-100/17*j

or just:
S=400/17-100/17*j {Result #2}

Comparing:
Result1-Result2=0

because:
S=400/17-100/17*j {Result #1}
S=400/17-100/17*j {Result #2}

Also, for general impedance:
Z1=a+b*j
and V1 a real number, we have:
I1=V1/Z1=V1/(a+b*j)
V1*I1=V1^2/(a+b*j)
V1^2/Z1=V1^2/(a+b*j)
I1^2*Z1=V1^2/(a+b*j)
so we have a triple equality:
V1*I1=V1^2/Z1=I1^2*Z1

Also, did you mean to write instead of (V,I complex here):
START QUOTE
The correct relationships are
S = V·I*
S = I² · Z
S = V² / Z*
END QUOTE

S=V . I* {same}
S=|I|^2 . Z {different}
S=V . V* . Z/|Z|^2 {different}

where the little dot is multiplication and the asterisk is the conjugate.

Last edited: Jun 27, 2015
13. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
And this claim is simply wrong.

Okay. You can evaluate the inside first and then take the conjugate or you can distribute the conjugate and then multiply.

S = (V . I)* = (V*).(I*)

Both are the same and both yield the wrong result since the claim is simply wrong.

Let's work a simple example to show this.

Say V = 100V @ 30° and I = 10A @ 30°

V and I are in phase and so we have nothing but real power, meaning that S will be real.

If we use your rule, we have:

S = (V . I)*
S = [(100V @ 30°)(10A @ 30°)]*
S = [1000 W @ 60°]*
S = 1000 W @ -60°

Notice that this is exactly the same result as

S = (V . I)*
S = (V*).(I*)
S = (100V @ 30°)*.(10A @ 30°)*
S = (100V @ -30°).(10A @ -30°)
S = 1000 W @ -60°

The correct formula, again, is

S = V.I*
S = (100V @ 30°).(10A @ 30°)*
S = (100V @ 30°).(10A @ -30°)
S = 1000 W

14. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hello again,

You seem to be misinterpreting my posts. I had said already that V is NOT a vector in the formula:
S=(V . I)*
In this formula V must be a scalar. No way around this, so showing an example where V is a vector is not going to help. The original problem has V=10 at angle of 0 degrees, which can be taken to be a scalar, and that is very typical of real life situations because the line voltage is often taken to be at zero degrees.

The more universal formula is of course:
S=V . I*

but i thought i made that clear, and we both got the same results for this one

But, try your 2nd and 3rd formulas on the same example you posted or better yet with different angles for V and I, then try my 2nd and 3rd formulas in the previous post. I did make a typo though and had to correct the 3rd formula so be sure to see that again first.

Since we agree on S=V.I* we can use that as a reference.
For general impedance vector:
Z=a+b*j

and general voltage vector:
V=c+d*j

we have:
I=(c+d*j)/(a+b*j)

Now V.I* is:
V*conjugate(I)=(c+d*j)*(c-d*j)/(a-b*j)=(d^2+c^2)/(a-j*b)

Now compute |I|^2*Z:
|I|^2=(d^2+c^2)/(b^2+a^2)
|I|^2.Z=(d^2+c^2)*(a+b*j)/(b^2+a^2)
factor that:
|I|^2.Z=(d^2+c^2)/(a-j*b)

so we know that:
V.I*=|I|^2.Z

which means my second formula looks good, and the third formula is:
V.V*.Z/|Z|^2

which comes out the same as V.I* above.

I^2*Z=(c+d*j)^2/(a+j*b)

is not the same as V.I* which we both agree is correct. To loose the center term we need:
(c+d*j)*(c-d*j)

in the numerator rather than (c+d*j)^2.

And just to reiterate one more time, the formulas that include quantities inside the parens with the outside taking the conjugate, they are only intended for use with a real voltage like we find all around like 120vac 60Hz for example. This is a common case and the formulas are simple.
I do agree however, that V.I* is the best one, because that includes any case, vector or not.

A small point is that the complex power is also sometimes represented by a bold P:
P=P+Q*j

I might have to play with that bold character thing a little
Wow the bold char thing worked, cant believe it
I'll have to get used to doing this more often to make things more clear. I guess you have mastered this already.

Last edited: Jun 27, 2015
15. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
If V is a scalar, then what does V* mean above?

16. ### MrAl Distinguished Member

Jun 17, 2014
2,554
515
Hello again,

First, the third formula can be simplified:
S=|V|^2/Z*

In the formula:
S=V.I*

V is a complex number like a+b*j.

In the formula:
S=(V.I)*

V is a real number a+0*j.

If we distribute the conjugate mark we see:
S=(V*.I*)

but the conjugate of a real number is just the real number again, so we have:
S=(V.I*)=V.I*

and we know that works for both V and I complex so it must work for V real and I complex.

Let me see if i can use that bold case thing again...

Formula A:
S=V.I*

Formula B:
S=(V.I)*