AC Circuit Power/impedance question

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Discussion Overview

The discussion revolves around calculating various parameters of an AC circuit, including apparent power, reactive power, current, and impedance. Participants are attempting to solve a homework problem related to power factor and circuit characteristics, with a focus on understanding the relationships between these quantities.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the apparent power as 520.83 VA and the reactive power as 361.44 VAR based on the given power factor and true power.
  • Another participant points out that impedance should include both magnitude and angle, suggesting that the initial calculations may not fully address this requirement.
  • There is a discussion about whether the impedance should be calculated by subtracting the resistance from the total impedance, with some participants agreeing and others challenging this approach.
  • Participants debate the correct method to find the resistive and reactive components of impedance, with confusion arising over the definitions and calculations involved.
  • Some participants express uncertainty about the methods being discussed, particularly regarding the separation of resistance and reactance in the context of impedance.
  • One participant suggests using the formula for voltage across the resistor but receives feedback that this approach is not meaningful in the context of the discussion.

Areas of Agreement / Disagreement

There is no consensus on the correct method to calculate impedance and its components. Multiple competing views remain regarding how to approach the problem, particularly in separating the resistive and reactive parts of impedance.

Contextual Notes

Participants express confusion over the definitions and calculations related to impedance, suggesting that there may be a reliance on different educational materials that present varying approaches to the topic.

james123

Homework Statement


For the circuit given in FIGURE 1 the power factor is 0.72 lagging and the power dissipated is 375 W.

12-jpg.jpg


Determine the:

  1. (i) apparent power
  2. (ii) reactive power
  1. (iii) the magnitude of the current flowing in the circuit
  2. (iv) the value of the impedance Z and state whether circuit is inductive or capacitive.

The Attempt at a Solution



(i)Power factor=true power/apparent power
So, Apparent power =true power/power factor
=375/0.72= 520.83 VA

(ii) Reactive power= √Apparent power^2-truepower^2
=√520.83^2-375^2
=361.44 VAR

(iii) P=VI
So, I=P/V
=520.83/120
=4.34 A

(iiii) Impedance= Vs/I
=120/4.34
=27.658 Ω

The circuit is inductive because it has a lagging power factor.Can anybody confirm if I am along the right lines with this/ where I've messed up?
Any help is appreciated!
Many thanks.
 
Last edited by a moderator:
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P is, by convention, the symbol for real power. The symbol for apparent power is S.

If it asks for impedance, you need to give not just magnitude but also angle.

Otherwise, it seems okay.
 
If the 375W is the total power dissipated in the circuit,
james123 said:
Power factor=true power/apparent power
So, Apparent power =true power/power factor
=375/0.72= 520.83 VA
..this is correct.
james123 said:
iii) P=VI
So, I=P/V
=520.83/120
=4.34 A
Correct.
james123 said:
Impedance= Vs/I
=120/4.34
=27.658 Ω
No. That would the "total" impedance seen by the source, considering the 10 ohm resistance. You need to find the value of Z only.
 
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Thanks for the replies!

So would Z=Impedance-Resistance
=27.658-10=17.658 Ω ??

And phase angle:

Cosθ=R/Z
=10/17.658
=0.5663

θ=Cos^-1*0.5663
=55.5

So Z=17.658∠55.5°

Does this look better?
 
james123 said:
So would Z=Impedance-Resistance
=27.658-10=17.658 Ω
No.
First compute the resistance and reactance in Z using active and reactive power values. You know the current.
 
Is the reactance the total resistance *power factor ??

So, 27.658*0.72=19.92 ??

I really don't where I'm going from here, I've just followed the ways my learning materials have shown me
 
james123 said:
Is the reactance the total resistance **impedance** power factor ??
Yes, that would be an easier way to go. No need of active and reactive power.

So what is the resistance in Z?
 
10 Ω ?
 
james123 said:
Is the reactance the total resistance *power factor ??

So, 27.658*0.72=19.92 ??

I really don't where I'm going from here, I've just followed the ways my learning materials have shown me
cnh1995 said:
Yes, that would be an easier way to go. No need of active and reactive power.

So what is the resistance in Z?
Oh, I misread that. This actually gives you the total resistance. What is the resistance in Z then?
Similarly find the reactance in Z. You know Z and phi.
 
  • #10
Sorry but I don't know what method you're getting at?

The formula I have for impedance is Z=Vs/I

And to get the phase angle I have Cosθ*R/Z

Are these wrong?
 
  • #11
james123 said:
Sorry but I don't know what method you're getting at?

The formula I have for impedance is Z=Vs/I

And to get the phase angle I have Cosθ*R/Z

Are these wrong?
That gives the "total impedance" including the 10 ohm resistance, which is not same as Z in your diagram. You need to find only Z.
james123 said:
And to get the phase angle I have Cosθ*R/Z
It should be cosθ=Rtotal/Ztotal and Ztotal is not equal to Z.

You need to separate the resistive component of Z from the total resistance. What you calculated in #6 gives the "total resistance". So what is the resistive component of Z?
 
  • #12
Do I need to work out the resistance across the resistor?

So, Vr=IR
Vr=4.34*10=43.4 V ??
 
Last edited by a moderator:
  • #13
james123 said:
So, Vr=IR
Vr=4.34*10=43.4 Ω ??
That is meaningless.

First of all, understand how the components are connected.

You have an "impedance" Z in series with a 10 ohm "resistance". That Z contains a resistive component, say r, and an inductive component, say X.
So, you need to find r and X.

You have calculated the "total resistance" in the circuit in #6. How do you get r from that?
 
  • #14
Im sorry I really don't understand what you're getting at, my materials always give impedance as a single value in ohms.

Only thing I can think of is subtracting the 10Ω resistor from the total impedance giving 27.658-10=17.658 Ω
 
  • #15
james123 said:
my materials always give impedance as a single value in ohms.
And that value is the phasor sum of the resistive and reactive components of the impedance. You can't add (or subtract) impedance with resistance like that.

One last hint (I'm going to bed:sleep:):
Say you have two impedances Z1=R1+jX1 and Z2=R2+jX2,
then you can write
Z1+Z2=(R1+R2)+j(X1+X2).

If anything is still unclear, maybe you should look up some stuff on impedance.
 
  • #16
Okay, well thanks for the help mate, it's always appreciated. Think you're right, I'll go away and do some more revision.
 
  • #17
james123 said:
So would Z=Impedance-Resistance
=27.658-10=17.658 Ω ??
Impedance is like a vector, so you have to first break it into its horizontal and vertical components before doing any calculations.
 

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