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

[charger9198]

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

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

 

[gneill]

I find your calculation of total ohms (part 4) to be a bit dubious. Where you've made the assumption that you can just subtract the resistor value from the impedance magnitude to leave the impedance Z of the "mystery component" is also suspect. Impedance will have real and imaginary parts.

I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: i = I (cos(\phi) - j\;sin(\phi)), and then find the total (complex) impedance of the circuit:

Z_{tot} = \frac{120 V}{i} = (19.91 + j 19.19) \Omega

Then subtract the given resistor from the real part of that to leave the impedance Z.

 

[charger9198]

Ok thanks gneill

so ill use

ϕ=acos(power factor)
=acos (.72)
= 0.76699

then;

i = 4.34(cos(0.76699)-jsin(0.76699))
i = 4.339566
Zt=120/4.339566
=27.28723 (minus 10)
=17.28723

 

[gneill]

Ok thanks gneill

so ill use

ϕ=acos(power factor)
=acos (.72)
= 0.76699

then;

i = 4.34(cos(0.76699)-jsin(0.76699))
i = 4.339566 <---- This should be a complex value (real + imaginary components)
Zt=120/4.339566
=27.28723 (minus 10)
=17.28723

The complex current should retain real and imaginary components so that the impedance value obtained will also have real and imaginary components. These correspond to the resistance and reactance of the circuit.

 

[charger9198]

Think I've got it;

i=4.34(cos(0.76699)-jsin(0.76699)

i = 4.34(0.720003-0.693971)
i = (3.12481-3.01183j)

So

Z=120/(3.12481-3.01183j)

Z=19.9079 + 19.1882j

Minus ten from real gives

9.9079 ohms

This better? Are my answers ok for the other parts of the question?

 

[gneill]

Yes, that looks better. So the "mystery component" has both resistance and inductance.

The other answers look fine to me; nicely calculated.

 

[charger9198]

Thanks for you help as always :)

 

[gneill]

Thanks for you help as always :)

You're very welcome!

 

[stemurdo]

gneill said - I'd be more inclined to use the current magnitude and its known phase angle obtained from the power factor to construct the complex current: , and then find the total (complex) impedance of the circuit:



Charger said - ϕ=acos(power factor)
=acos (.72)
= 0.76699

I work the phase angle out at 43.95, where does this 0.76699 come from? I'm having a real problem getting my head round this 'imaginary' thing.

 

[gneill]

Charger said - ϕ=acos(power factor)
=acos (.72)
= 0.76699

I work the phase angle out at 43.95, where does this 0.76699 come from? I'm having a real problem getting my head round this 'imaginary' thing.

0.76699 is the angle in radians. You can work in degrees or radians as long as you're consistent.

 

[stemurdo]

ok so i use cos-1 x power factor, gives me a phase angle of 43.95 degrees.

i=I(cos(43.95)-jsin(43.95))
=4.34(cos(43.95)-jsin(43.95))
=4.34(0.7199-j0.694)
=(3.1244-j3.012)

is this right so far? this is where I'm getting a bit lost and not sure of the next step

 

[stemurdo]

am i right in saying that the jvalue is a negative value?
so it would actually be (3.1244 - (-3.012))
giving 120/6.134 = 19.555 for the real value? then how do you get the second jvalue?

 

[gneill]

am i right in saying that the jvalue is a negative value?
so it would actually be (3.1244 - (-3.012))
giving 120/6.134 = 19.555 for the real value? then how do you get the second jvalue?

You don't add the real and imaginary components of a complex number, they are separate entities. Take a look at post #5 in this thread. The impedance will have real and imaginary components, too.

 

[stemurdo]

sorry gneill, i can only ask for patience here. i appreciate the time.
i'm obviously missing something here, or just not grasping the concept at all!
if you don't add or subtract the jvalue then what purpose does it serve? how do you divide 120 by 3.124 and come up with 19.9 or whatever? my brain has packed up and gone home. :-)

 

[gneill]

sorry gneill, i can only ask for patience here. i appreciate the time.
i'm obviously missing something here, or just not grasping the concept at all!
if you don't add or subtract the jvalue then what purpose does it serve? how do you divide 120 by 3.124 and come up with 19.9 or whatever? my brain has packed up and gone home. :-)

You should review the manipulation of complex numbers; how to add, subtract, multiply, and divide them. http://www.allaboutcircuits.com/vol_2/chpt_2/6.html to get you started.

If you enter "Complex number arithmetic" into a Google search you'll get more information, and even links to online complex number calculators.

 

[terrytibbs]

Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?

 

[gneill]

Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?

27.63 Ω represents the magnitude of the impedance. Impedance is a complex value comprised of a real (resistive) and an imaginary (reactive) part. In this sense, yes, the magnitude will be greater than the size of the individual components: $|Z| = \sqrt{R^2 + X^2}$ . So for X nonzero, |Z| > R.

 

[mattyh3]

Think I've got it;

i=4.34(cos(0.76699)-jsin(0.76699)

i = 4.34(0.720003-0.693971)
i = (3.12481-3.01183j)

So

Z=120/(3.12481-3.01183j)

Z=19.9079 + 19.1882j

Minus ten from real gives

9.9079 ohms

This better? Are my answers ok for the other parts of the question?

i don't understand how get from Z=120/(3.12481-3.01183j) to Z=19.9079 + 19.1882j

can anyone explain this please if possible followed the link at the end but can understand from that how it get to Z=19.9079 + 19.1882j

 

[gneill]

i don't understand how get from Z=120/(3.12481-3.01183j) to Z=19.9079 + 19.1882j

can anyone explain this please if possible followed the link at the end but can understand from that how it get to Z=19.9079 + 19.1882j

It's basic complex arithmetic. You can convert the denominator to polar form (magnitude + angle) then do the division, convert the result back to rectangular form. Or, clear the imaginary term from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator.

A web search on "complex arithmetic" will likely turn up a tutorial.

 

[mattyh3]

It's basic complex arithmetic. You can convert the denominator to polar form (magnitude + angle) then do the division, convert the result back to rectangular form. Or, clear the imaginary term from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator.

A web search on "complex arithmetic" will likely turn up a tutorial.

you see none of this is in the lesson book for this course the only way i can find is z=v/i

there is polar form but nothing like what's on this page.. i still can't understand what your trying to tell me

just tried a complex arithmetic calculator online and i don't get them answers

 

[gneill]

you see none of this is in the lesson book for this course the only way i can find is z=v/i

there is polar form but nothing like what's on this page.. i still can't understand what your trying to tell me

just tried a complex arithmetic calculator online and i don't get them answers

Your course may assume that you've studied complex arithmetic in another course; It's generally taught in a mathematics course. If you haven't taken such a class, then you should find an online tutorial on complex numbers.

 

[mattyh3]

Should the impedance be Vs/I which is 27.63 ohms. If there is an Inductance or Capacitance in the circuit the Z will be higher than R?

well i thought it would be z=v/i but looks different on here and inductance becuase p.f is lagging

 

[mattyh3]

Your course may assume that you've studied complex arithmetic in another course; It's generally taught in a mathematics course. If you haven't taken such a class, then you should find an online tutorial on complex numbers.

i did a maths course before been aloud on this and there is nothing in there on this or how to do it

 

[gneill]

well i thought it would be z=v/i but looks different on here and inductance becuase p.f is lagging

Z = V/I is correct, assuming that V and I are complex values (well, here V is assumed to be the reference for the phasor angles, so V's imaginary component will be zero making V a real value). The I is determined from the given information using the power factor to determine the apparent current in complex form.

 

[mattyh3]

still can't find anything of any use that i can understand where the 120 goes and you end up with 19.9079+19.1882j

 

[gneill]

still can't find anything of any use that i can understand where the 120 goes and you end up with 19.9079+19.1882j

It's just division. A real number, 120, is being divided by the complex number 3.12481-3.01183j .

 

[mattyh3]

It's just division. A real number, 120, is being divided by the complex number 3.12481-3.01183j .

if i divide them into 120 it get 38.402+39.843j i don't understand any other way of doing it as am not great with maths

 

[gneill]

if i divide them into 120 it get 38.402+39.843j i don't understand any other way of doing it as am not great with maths

You'll have to find a complex number / complex arithmetic tutorial to learn about it. Or make use of an online complex arithmetic calculator (although that won't help you in tests or exams!).

Here's an example online calculator (found with via google with a couple of seconds effort):

http://www.1728.org/compnumb.htm

 

[mattyh3]

You'll have to find a complex number / complex arithmetic tutorial to learn about it. Or make use of an online complex arithmetic calculator (although that won't help you in tests or exams!).

Here's an example online calculator (found with via google with a couple of seconds effort):

http://www.1728.org/compnumb.htm

thanks for the help but i can't even use the calculator it gives me weird answers nothing like what's on here

this is what i get 19.408277013161786 + i*19.70643553618951

 

[gneill]

thanks for the help but i can't even use the calculator it gives me weird answers nothing like what's on here

this is what i get 19.408277013161786 + i*19.70643553618951

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

 

[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.