Engineering AC Circuit (RL) Homework: Finding the Series R-L Equivalent for Two Loads

[galaxy_twirl]

Homework Statement

In the circuit shown in Figure Q2(b), two loads A and B are connected to the source. The cable connecting the power supply to the load is modeled as series R and L of values R = 0.1Ω, L = 0.1 mH as shown in the figure.

The loads have the following specifications:
Load A: 230V, 50Hz, 1kW, pf=0.5 lagging.
Load B: 230V,50Hz, 1kW, pf=0.87 lagging

Find the series R-L equivalent of the two loads together.

Figure 2(b):

Homework Equations

Z_L = jωL
P_active = V_rms*I_rms*cos∅ (∅ means phi)
P_reactive = V_rms*I_rms*sin∅ (∅ means phi)

The Attempt at a Solution

Given the specifications for loads A and B, am I right to say that as long as I supply 230V, 50Hz to those 2 loads, I will get a power output of 1kW? Just wondering, does "Find the series R-L equivalent of the two loads" mean I have to find the impedances of Load A and B and convert them from parallel to series connection (just like 2 parallel batteries)?

Since the loads have lagging power factor, they are inductors.

As I am not given more information on the loads, I can only deduce L from P_active = V_rms*I_rms*cos∅, and v_L(t) = L(di_L(t)/dt)). Am I right to say this? However, I am stuck as I don't know V_rms and I_rms.

May I have some hints please?

Thank you. :)

 

[NascentOxygen]

Given the specifications for loads A and B, am I right to say that as long as I supply 230V, 50Hz to those 2 loads, I will get a power output of 1kW?

Yes, 1kW from each.

Just wondering, does "Find the series R-L equivalent of the two loads" mean I have to find the impedances of Load A and B and convert them from parallel to series connection (just like 2 parallel batteries)?

Convert from 2 inductors and 2 resistances to a single equivalent L-R.

 

[galaxy_twirl]

Convert from 2 inductors and 2 resistances to a single equivalent L-R.

Thank you. :) What do you mean by 2 inductors and 2 resistances? Just to confirm, Load A and B are not purely inductors right? (I am getting a bit suspicious because it dissipates active(real) power, as stated, "1kW".)

If that is the case, I will have Z_A = R + jX, where R is resistance and X is the inductance (or imaginary part), right?

 

[NascentOxygen]

Each load can be considered to comprise one R and one L, and the combination determines that load's power factor.

The R accounts for real power.

 

[galaxy_twirl]

I see. Thanks. I shall go and try it and let you know if I can get the answer, prolly a few hours later as I have to run now. :)

 

[galaxy_twirl]

I am still not getting what does "find the series R-L equivalent of the 2 loads together mean". Does it mean I find the total impedance and then do 1/Z_eff = 1/Z₁ + 1/X₂ ?

i.e.: For Load A, I do:

P_active = V_rms*I_rms*cos∅ (phi).

Hence, P_active = V_rms*(V_rms / Z)*cos∅, so substituting all values in, we have:

1000 = (230² / Z_A)*cos∅.

However, I was wondering, the active power will not be the same since the inductor and resistor took up some voltage before it reaches the node of Load A. Am I right to say this? And may I know if my working above is correct?

Thanks!

 

[NascentOxygen]

Yes, you can sum the reciprocals of the impedances, the loads are parallel. Your method looks like it should work.

Inclusion of the cable impedance gives the problem realism. You are correct in saying the loads will not receive their rated 230V, but that is not relevant to the specific task you are tackling here.

 

[galaxy_twirl]

Yes, you can sum the reciprocals of the impedances, the loads are parallel. Your method looks like it should work.

Inclusion of the cable impedance gives the problem realism. You are correct in saying the loads will not receive their rated 230V, but that is not relevant to the specific task you are tackling here.

I see. I didn't know the cable will have impedance. I suppose it is linked to resistance in the cable and passing an AC voltage changes the term resistance to 'impedance', right?

May I know why is it not a concern that the loads are not receiving 230V? I thought they will not give the max rated power output of 1kW if they are not supplied with 230V of p.d. across them.

Thank you. :)

 

[NascentOxygen]

The calculations you are doing concern load specifications, and these are specified at 230V.

A cable has resistance, also inductance, as well as capacitance to earth, so it has a characteristic impedance.

 

[galaxy_twirl]

The calculations you are doing concern load specifications, and these are specified at 230V.

I see. But if I don't supply 230V as in this case, won't I be wrong if I use 1kW as the power output? Sorry. I am a bit confused at the moment. :(

 

[gneill]

I see. But if I don't supply 230V as in this case, won't I be wrong if I use 1kW as the power output? Sorry. I am a bit confused at the moment. :(

The loads are defined by Specification:

The loads have the following specifications:
Load A: 230V, 50Hz, 1kW, pf=0.5 lagging.
Load B: 230V,50Hz, 1kW, pf=0.87 lagging

That means that these loads, if connected to an ideal 230 V, 50 Hz source without any intervening cable would draw the specified power at the specified phase angle. This gives you enough information to characterize these loads in terms of their impedances. The power cable is irrelevant for this problem.

Perhaps there are other sections to the problem which you've not mentioned yet where the cable impedance becomes important?

 

[NascentOxygen]

I see. But if I don't supply 230V as in this case, won't I be wrong if I use 1kW as the power output? Sorry. I am a bit confused at the moment. :(

In practice, the supply voltage is never going to be exactly what the load was nominally specified to operate from. This is a fact of life, and power engineers accept this. Sure, it does mean that the load probably won't be drawing exactly 1,000.0 watts of power whenever its voltage departs from 230.0V but loads are designed to be tolerant of voltage errors (and may compensate for it by various means if ever it's important). If precise calculations are needed then that can be done, but here you are basing your calculations on the nominal design voltage, 230V. So long as the cable impedance is within acceptable limits, the loads will work just fine on whatever the voltage ends up being.

 

[galaxy_twirl]

The loads are defined by Specification:
That means that these loads, if connected to an ideal 230 V, 50 Hz source without any intervening cable would draw the specified power at the specified phase angle. This gives you enough information to characterize these loads in terms of their impedances. The power cable is irrelevant for this problem.

Perhaps there are other sections to the problem which you've not mentioned yet where the cable impedance becomes important?

I see. Thanks for your explanation. :)

Perhaps there are other sections to the problem which you've not mentioned yet where the cable impedance becomes important?

Hmm. There are 2 small question parts related to this question. They are:

Part ii) Find the power factor as seen from the source.
Part iii) Find the value of the capacitor to be connected across the source so that the source voltage and the total current drawn from it are in phase with each other.

I think the questions above should not interfere with how I see/factor in the RL of the cable. :)

 

[galaxy_twirl]

In practice, the supply voltage is never going to be exactly what the load was nominally specified to operate from. This is a fact of life, and power engineers accept this. Sure, it does mean that the load probably won't be drawing exactly 1,000.0 watts of power whenever its voltage departs from 230.0V but loads are designed to be tolerant of voltage errors (and may compensate for it by various means if ever it's important). If precise calculations are needed then that can be done, but here you are basing your calculations on the nominal design voltage, 230V. So long as the cable impedance is within acceptable limits, the loads will work just fine on whatever the voltage ends up being.

I see. Alright. I shall try it out later. :) Thanks! I am just a bit puzzled about the voltage issue but other than that, I should be fine. :)

 

[gneill]

Hmm. There are 2 small question parts related to this question. They are:

Part ii) Find the power factor as seen from the source.
Part iii) Find the value of the capacitor to be connected across the source so that the source voltage and the total current drawn from it are in phase with each other.

I think the questions above should not interfere with how I see/factor in the RL of the cable. :)

Okay, these two questions are where the cable impedance comes into play. The source sees the cable impedance as part of its total load.

 

[galaxy_twirl]

Okay, these two questions are where the cable impedance comes into play. The source sees the cable impedance as part of its total load.

I see. Thanks! I will take a look again tomorrow and if I have any problems I will post here again. :)

 

[galaxy_twirl]

I got Z_A = 26.45 Ohms after working out the above. However, how do I express this in terms of Z = R + jX = R + jωL?

I realized my teacher somehow arrived at 13.23 Ohms for R. How and why did my teacher do that? *puzzled* :S

 

[gneill]

I got Z_A = 26.45 Ohms after working out the above. However, how do I express this in terms of Z = R + jX = R + jωL?

I realized my teacher somehow arrived at 13.23 Ohms for R. How and why did my teacher do that? *puzzled* :S

26.45 is the magnitude of the "A" impedance. What's its angle?

 

[galaxy_twirl]

26.45 is the magnitude of the "A" impedance. What's its angle?

From power factor = cos∅, cos∅=0.5, hence, by cos^-1 0.5 = 60°. Angle = 60°

 

[gneill]

From power factor = cos∅, cos∅=0.5, hence, by cos^-1 0.5 = 60°. Angle = 60°

Yup. So you have the polar form of that impedance. Convert to rectangular.

 

[galaxy_twirl]

Yup. So you have the polar form of that impedance. Convert to rectangular.

OH! I get the picture now. Oh dear. I think one really cannot do engineering without a strong Maths foundation. Sigh.

Thanks for pointing it out. :)

 

[NascentOxygen]

I got Z_A = 26.45 Ohms after working out the above. However, how do I express this in terms of Z = R + jX = R + jωL?

I realized my teacher somehow arrived at 13.23 Ohms for R. How and why did my teacher do that? *puzzled* :S

You said the power factor (which means cos ɸ) is 0.5 here,
so to find the real component of Z you can go 26.45Ω x 0.5

 

[galaxy_twirl]

You said the power factor (which means cos ɸ) is 0.5 here,
so to find the real component of Z you can go 26.45Ω x 0.5

Ah yes, I forgot for a moment about complex numbers, and hence I couldn't see that I need to convert polar back to rectangular form, which is:

x = rcosθ
y = rsinθ

Thanks for your help! :)

 

[NascentOxygen]

Finding the real component was straight-forward.

Q: how to determine sin ɸ without needing to determine ɸ itself?

 

[galaxy_twirl]

I am guessing, I have to use the Power Triangle, with Apparent Power, on the HYP, Active power as the base of the triangle and Reactive Power as the vertical of the triangle. The triangle is a right-angled triangle. Am I right? :)

 

[NascentOxygen]

I am guessing, I have to use the Power Triangle, with Apparent Power, on the HYP, Active power as the base of the triangle and Reactive Power as the vertical of the triangle. The triangle is a right-angled triangle. Am I right? :)

All very true, and that's what you will use ...

though doesn't touch on the answer I had in mind. oo)

 

[galaxy_twirl]

All very true, and that's what you will use ...though doesn't touch on the answer I had in mind. oo)

I see. Haha. :) You may want to share your answer in mind with me, perhaps I may learn something new. :D

 

[NascentOxygen]

I had in mind Euler's Theorem. Undoubtedly you have studied it in maths, but perhaps not yet related it to engineering. (pronounced OILERS)

 

[galaxy_twirl]

I had in mind Euler's Theorem. Undoubtedly you have studied it in maths, but perhaps not yet related it to engineering. (pronounced OILERS)

I see. I heard of Euler's formula when learned complex number at A levels, which is, any complex number of the form:

z = a + bi

can be expressed in the form z = re^iΘ. where r = √(a² + b²) and θ = tan^-1(b/a).

Is this the Euler's Theorem in your mind? :)

 

[NascentOxygen]

Oops, what I had in mind is a Pythagorean identity

Using that first one, you can find the magnitude of sine (x) if you already know cosine (x).

(I'm not sure why I decided to credit it to Euler, but Euler has enough to his credit without me heaping more still, even in error.)

 

[galaxy_twirl]

Oops, what I had in mind is a Pythagorean identity

View attachment 75436

Using that first one, you can find the magnitude of sine (x) if you already know cosine (x).

(I'm not sure why I decided to credit it to Euler, but Euler has enough to his credit without me heaping more still, even in error.)

I see. I like the figure you attached above. Haha. :) Thanks a lot for your help! :D Yupp, Euler is a great Mathematician I think. I have heard of his name in my Science Club in high school, so I guess he must be very famous and made lots of contribution to the field of Mathematics. :)