How Do You Calculate Apparent and Reactive Power in AC Circuits?

[mattyh3]

Yes, and that's the correct result for the impedance. It's a complex number with real and imaginary parts.

o great i thought i was wrong lol... so now i just need to take off the 10 ohm resistance from the real value of 19.40828158597505 to get z= 9.408 ohms

 

[gneill]

o great i thought i was wrong lol... so now i just need to take off the 10 ohm resistance from the real value of 19.40828158597505 to get z= 9.408 ohms

Yes, that's the idea.

 

[mattyh3]

Yes, that's the idea.

thanks for the help just need to find how to do complex calcs on my calculator now

 

[Big Jock]

May I ask gneill why we subtract 10 ohms as I can't find anything in my course about this? Just a quick question really so I know why this happens...

 

[gneill]

May I ask gneill why we subtract 10 ohms as I can't find anything in my course about this? Just a quick question really so I know why this happens...

If you look at the image attached to the first post you'll see that the load consists of some unknown impedance Z in series with a 10 Ω resistor. The given information was used to calculated the total impedance of the circuit, so Z + 10 Ω. By subtracting the 10 Ω we're left with Z alone.

 

[Big Jock]

Z = (10 Ohm)/.72 = 13.889 Ohms (@ arc cos .72) = 13.889 Ohms |_43.95' = 10 + j9.639
Would this also be an acceptable answer or are the ones listed a better route gneill?

 

[gneill]

Z = (10 Ohm)/.72 = 13.889 Ohms (@ arc cos .72) = 13.889 Ohms |_43.95' = 10 + j9.639
Would this also be an acceptable answer or are the ones listed a better route gneill?

The unknown Z adds to the 10 Ω to make up the entire impedance of the load. Z itself will have real and reactive components. 10 Ω is not the total real resistance in the circuit. So no, I'm afraid that your suggested approach will not work.

There are other possible approaches. For example, one might start by assuming that the total resistance is R and so the total real power dissipated would be P = (120V*pf)²/R. Solve for R. Then use that R and the pf to determine the reactive impedance. You still need to subtract the 10 Ω that's external to the Z to find the real part of Z.

 

[Electest]

Thought I would bump this thread as there doesn't seem to be a clear answer to question (4)

I looked at the problem like this:

We know the power dissipated = 375 watts. This will be dissipated as heat by the resistive elements of the circuit. Now we know that part of the total resistance is 10 ohms, as indicated by R on the diagram, but the unknown impedance shown as Z on the diagram will/may contain some resistance. (Only true power is dissipated, reactance does not consume (dissipate) any power)

So using Vr

Rt = (Vs*p.f)^2/375

= (120*0.72)^2/375

= 19.90656 ohms

Rz = Rt - R

= 19.90656 - 10

= 9.90656 ohms

9.90656 ohms is surely only resistive part of Z (not the answer they require) as this was part of the dissipated power so would be defined as Rz?

Would we not divide Rz by the known p.f to obtain our answer Z?

Thanks again

 

[gneill]

The power factor applies to the total impedance, not just the bit inside the "Z". Your Rz is not the total real part of the overall impedance of the circuit. Rt fits that bill. Z contains some reactance in addition to some resistance, and by the circuit diagram it would also be the total reactance for the circuit.

Now, the pf is the cosine of the angle between the hypotenuse of the impedance triangle and the real component. You're looking for the imaginary component. A little trig would do to find the reactance X (at least its magnitude). As for its sign, you're told that the power factor is lagging. That should give you a clue to whether the reactance is +jX or -jX in the overall impedance.

 

[Electest]

Thanks for the reply

See now I'm think I'm getting confused in what they are asking for!

I have the following figures

R=19.90656
Z=27.648
X=19.187
Theta= 43.95 degrees
P.F 0.72
Vs= 120v
Vr= 86.4v
Lagging = inductive

They only ask for Z

So what do they want?

Your help is much appreciated

 

[gneill]

Z is an impedance with both real and imaginary parts. It is comprised of a resistance (real) and a reactance (imaginary). The real part has been determined as Rz = Rt - 10 Ohms. You have the magnitude of the reactance as X = 19.187 Ohms. You also say that the load is inductive, so you should be able to "construct" Z from Rz and X...

 

[Electest]

Ok so,

Z = SQRT (R^2+X^2)

= SQRT (9.9065^2+19.187^2)

= SQRT (98.14+368.141)

= 21.596 ohmsHow's that? Any closer

Regards

 

[gneill]

No, you want the impedance in complex form, not just its magnitude.

Z = R + sjX

where s is +1 or -1 according to whether the load is inductive or capacitive.

 

[Electest]

I asked the tutor if the answer was to be in complex form after reading the original post this week and they said no. So how else would it be shown?

I also think I made a mistake. Should I be using a recalculated value of X = 9.55 from Rz*tan(43.95)

ThereforeZ = SQRT (R^2+X^2)

= SQRT (9.9065^2+9.55^2)

= SQRT (98.14+91.2)

= 13.76 ohms

Thanks

 

[gneill]

The question asks for the impedance. Impedance is a complex value (or a phasor, if that's the import of the chapter under study). A real number is NOT an impedance unless the impedance is a pure resistance.

The reactance can be calculated from the power factor and the circuit's total resistance. You can't use the partial resistance of a single component.

The reactance of the circuit is not 9.55 Ohms. It's larger than that. Check the phase angle that would result with a total resistance of 19.91 Ohms and a total reactance of 9.55 Ohms. Does it square with your power factor?

 

[Electest]

So my figures in post 42 were correct after all. I can see that now after thinking about it.

 

[Electest]

So my figures could be shown in complex form as

Z= 9.9065+19.187j

I will ask the tutor again stating what you have told me. Want to understand this properly after all.

Many thanks to you for your input and time.

Regards

 

[gneill]

So my figures could be shown in complex form as

Z= 9.9065+19.187j

Yes, that looks good. You could also specify it in polar form (magnitude + angle).

I will ask the tutor again stating what you have told me. Want to understand this properly after all.

That is the ultimate goal

Many thanks to you for your input and time.

No worries. Always happy to help!

 

[Electest]

So in polar form

Z=|Z|L φ

where |Z| represents the magnitude of the impedance and φ represents the phase angle.

Ztotal = (27.65 L +44 degrees) Ohms

Z = (21.52 L +62.58 degrees) Ohms

 

[gneill]

So in polar form

Z=|Z|L φ

where |Z| represents the magnitude of the impedance and φ represents the phase angle.

Ztotal = (27.65 L +44 degrees) Ohms

Z = (21.52 L +62.58 degrees) Ohms

The numbers and units (Ohms) are good, but the "L" is not. The angle alone tells you that the reactance lies along the +j axis and so the impedance has an inductive character.

 

[Electest]

Sorry this is how the learning material shows how to write it. It's not so much an L but a symbol that denotes an angle or am I missing the point.

Ztotal = 27.65 +44 degrees Ohms

Z = 21.52 +62.58 degrees Ohms

Regards

 

[gneill]

Sorry this is how the learning material shows how to write it. It's not so much an L but a symbol that denotes an angle or am I missing the point.

Ztotal = 27.65 +44 degrees Ohms

Z = 21.52 +62.58 degrees Ohms

Regards

Ah! Okay, understood. If you go to the Advanced editing panel to compose your post then you will find an actual angle symbol ∠ available in the Quick Symbols menu.

 

[Electest]

Ok brilliant.

So is my answer now correct for the question (4) or is there anything else to do?

Thanks

 

[gneill]

Ok brilliant.

So is my answer now correct for the question (4) or is there anything else to do?

Thanks

You're good to go with either representation of the impedance (they are after all just two representations of the same complex value).

 

[Electest]

Thanks again

So my answer should show the Z values calculated not the Z totals. I've got it now :-)

 

[Lightning9]

Hi, sorry to resurrect this thread but I'm confused by your working out. I understand your reason for using complex numbers but when I try and confirm the voltage drops they don't add up to the supply voltage 120V?

or am I missing something?
Thanks

 

[gneill]

Hi, sorry to resurrect this thread but I'm confused by your working out. I understand your reason for using complex numbers but when I try and confirm the voltage drops they don't add up to the supply voltage 120V?

or am I missing something?
Thanks

There's no way to tell if you're missing something if you don't show us what you have...

Show your work!

 

[Lightning9]

I was just using the equation Vs = √ V_R² + V_L²

But I now see you can't because of the complex number.

 

[woodyzzz]

For the circuit given in the power factor is 0.72 lagging and
the power dissipated is 375 W.

View attachment 42861

Determine the:
(1) apparent power
(2) reactive power
(3) the magnitude of the current flowing in the circuit
(4) the value of the impedance Z and state whether circuit is inductive or
capacitive.

(1) Apparent power = True power / Power Factor (S=P/pf)

= 375 / 0.72

= 520.8333 VA

(2) Reactive Power = SQRT (apparent power^2) - (True Power^2)

= SQRT (520.8333^2)-(375^2)

= SQRT (130642.3264)

= 361.4448 VAR

(3) Magnitude of Current

Power= Voltage*Current *Power factor (P=V*I*pf)

375 = 120*I*0.72
I = P/(V*pf)
I = 375/(120*0.72)
I = 4.34 A

(4) Total ohms = Voltage/Current
= 120/4.34
=27.6498 ohms (minus 10)
z = 17.65 ohms

The power factor is lagging within the circuit therefore the the virvuit is inductive
Am i on the right lines here? or is there a better way of calculating the above results

Hello Charger and Oneill

I am currently going through the same coursework and just wanted to share my answers and thoughts- hoping for some feedback so that I can be sure that what my understanding is, is correct

Now...there are numerous equations that involve P, R and I the first one that jumps out is the P=VI or P=I^2R P = dissipated power V= supply voltage I = current

now my thinking is you can't use the P=VI as this would only give you the true power value, which in turn would not give you the correct I value

I then ventured onto P= I^2R - but then i thought this would only apply to a purely resistive circuit, which this obviously isn't as the pf is lagging 0.72

I then decided on (s) = Vs*I s= apparent power (which includes the total circuit voltage) Vs= voltage supply

I = 520.83/ 120 = 4.34 A

is this the correct way of thinking ?

 

[gneill]

I then decided on (s) = Vs*I s= apparent power (which includes the total circuit voltage) Vs= voltage supply

I = 520.83/ 120 = 4.34 A

Yes, that will work.