AC Circuit Power/impedance question

[james123]

Homework Statement

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

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.

 

[NascentOxygen]

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.

 

[cnh1995]

If the 375W is the total power dissipated in the circuit,

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

..this is correct.

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

Correct.

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.

 

[james123]

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?

 

[cnh1995]

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.

 

[james123]

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]

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?

 

[james123]

10 Ω ?

 

[cnh1995]

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

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.

 

[james123]

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?

 

[cnh1995]

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.

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

It should be cosθ=R_total/Z_total and Z_total 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?

 

[james123]

Do I need to work out the resistance across the resistor?

So, Vr=IR
Vr=4.34*10=43.4 V ??

 

[cnh1995]

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?

 

[james123]

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

 

[cnh1995]

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):
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.

 

[james123]

Okay, well thanks for the help mate, it's always appreciated. Think you're right, I'll go away and do some more revision.

 

[NascentOxygen]

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.