Thevenin's Theorem: Solving Homework Statement on Load Current

[magician]

Homework Statement

FIGURE 1 shows a 50 Ω load being fed from two voltage sources via
their associated reactances. Determine the current i flowing in the load by:

(a) applying Thévenin’stheorem
(b) applying the superposition theorem
(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)

2. Homework Equations [/B]

The Attempt at a Solution

Could someone please help me get this started? The 'hand outs' I have been given are awful, and do not explain how to incorporate the reactance and p.f.

If all of the loads were simply a resistance, I can see this being fairly easy. Although I could be wrong! Any help is appreciated.

Many thanks

 

[gneill]

Hint: Convert the load to a complex impedance. Then treat everything like resistors but use complex arithmetic.

 

[magician]

Thank you for your reply.

Would this be using;

$Z = \sqrt{R^2+X^2}$

When R = 50 Ohms
And utilising the p.f. of 0.7 to get X?

 

[gneill]

Thank you for your reply.

Would this be using;

$Z = \sqrt{R^2+X^2}$

When R = 50 Ohms
And utilising the p.f. of 0.7 to get X?

Not quite. The 50 Ohms is the magnitude of the load impedance. The power factor tells you what the angle of the impedance is (for a complex number in magnitude / angle form). Remember that the power factor happens to be the cosine of the angle.

 

[magician]

Thanks again gneill,

Cosine of 0.7 = 45.57

arctan(51.01 / 50) = 45.57

Therfore: Z = 50 + j51.01

 

[NascentOxygen]

Thanks again gneill,

Cosine of 0.7 = 45.57

arctan(51.01 / 50) = 45.57

Therfore: Z = 50 + j51.01

No. You are given Z=50 ohms, not R=50 ohms.

 

[gneill]

With the angle and the magnitude you have the load impedance in polar form. Thus
$$Z_L = 50 \Omega ~~\angle 45.57°$$
You can convert this to rectangular form if you need it that way.

By the way, before you dive into the calculations for the Thevenin equivalent, take a close look at the definitions of the voltage supplies. While they both have the same frequencies they won't have the same phase angle (why is that do you think?).

 

[magician]

In rectangular form, would it be written as;

$Z_L = 35 \Omega + j35.71$

?

Then would I;

Disregard the right portion of the circuit and and find the total impedance of the coil and load. Find the equivalent Voltage across the open circuit terminals (disregarding the j6 coil?)

Really confusing! Apologies

 

[gneill]

To apply Thevenin here you want to disregard the load for now and concern yourself with just the source network. That means finding the open circuit voltage and equivalent impedance of the source network.

 

[NascentOxygen]

In rectangular form, would it be written as;

$Z_L = 35 \Omega + j35.71$

?

Then would I;

Disregard the right portion of the circuit and and find the total impedance of the coil and load. Find the equivalent Voltage across the open circuit terminals (disregarding the j6 coil?)

Really confusing! Apologies

With the load removed, you are left with two equal-frequency voltage sources connected by a pair of inductances. Find the voltage at the junction of the inductors and that's going to be your Thévenin voltage.

 

[magician]

Hi guys - Sorry for the late reply, but I went back to the drawing board as I just wasn't understanding it. And to be honest - I'm still lost.

From the notes I have:

"First find the equivalent resistance. To do this we remove the load..." OK That bit makes sense.

"Replace all sources with their internal resistance.." OK - Couldn't be simpler!

"Next, find the Thevenin equivalent voltage, that is the voltage across the open-circuit terminals.." What open-circuit terminals?

 

[gneill]

"Next, find the Thevenin equivalent voltage, that is the voltage across the open-circuit terminals.." What open-circuit terminals?

The open terminals that are where the load was connected before you removed it in a prior step.

 

[magician]

But, then the circuit would be in 2 parts, split down the middle.

$Edited$

Oh dear. I believe I was taking it far too literally. Ignore the above statement!

Would the next step then be;

To work out the current by adding the resistance in series. Then reconnecting the load, and work out the current flowing through the load, again in a series formation??

 

[NascentOxygen]

The load stays right out of the picture while you are determining the Thévenin voltage.

You are at the stage of trying to determine the voltage at the junction of the inductances.

 

[gneill]

If you're concerned about how the circuit looks when the load is removed, redraw it so it looks more familiar:

Once again I encourage you to look at the specifications of V1 and V2. See post #7.

 

[magician]

Thanks again to you both for the above post's. It's been a great help so far.

So, I believe I am at the point of calculating the equivalent impedance.

Which I have calculated as;

Z_t = (j4 * j6) / (j4 + j6)
Z_t = -24 / j10
Z_t = j2.4

Then I need to find the equivalent voltage.

So the current in the circuit is;

I = (V1 - V2) / (j4 + j6)

The terminal voltage is 'V2' plus the volt drop across the j6 coil... Isn't it?

...Before I carry on, I'd like to make sure I'm on the right track!

gneill - I have no idea why the phase angles would be different, and what the significance of that is. Completely dumbfounded thus far...*sigh*

 

[gneill]

gneill - I have no idea why the phase angles would be different, and what the significance of that is. Completely dumbfounded thus far...*sigh*

Take a close look at the definitions of those voltages. If you're going to use phasors to do the calculations you want to make sure that both of them are based on the same trig function, either sine or cosine. That or carry the full functions through the math and use trig identities to sort out adding or subtracting them. It's usually easier to just make them both either sines or cosines right from the start.

 

[magician]

OK.

So cosine is phase shifted by 90°.

V2 = √2 * 415 Cos(100π t - 90)

 

[NascentOxygen]

I would have replaced sin by cos(wt+90°), since sine leads cosine by 90°

EDIT: and I would have been wrong

What you wrote is correct.

 

[peppa]

Can anyone confirm how the current was achieved i.e.

V1 (sqr2x415cos(100pi t) - V2 (sqr2x415cos(100pi t-90)/ j4 + j6

as I am not sure where to start concerning the voltages?

 

[NascentOxygen]

Can anyone confirm how the current was achieved i.e.

V1 (sqr2x415cos(100pi t) - V2 (sqr2x415cos(100pi t-90)/ j4 + j6

as I am not sure where to start concerning the voltages?

That expression is desperately in need of some more brackets! Here, borrow some spare ones I keep handy for such a purpose: ❲❳❲❳❲❳❲❳[/size]

The current flowing between the pair of sources is simply...

Current = voltage difference ÷ impedance

 

[gneill]

Can anyone confirm how the current was achieved i.e.

V1 (sqr2x415cos(100pi t) - V2 (sqr2x415cos(100pi t-90)/ j4 + j6

as I am not sure where to start concerning the voltages?

Write the voltages as phasor quantities. Then the math is just complex arithmetic.

 

[peppa]

Thank you for your response

I'm not sure how to convert these voltages to phasor quantities would it be like this

V1 = (415(cos(0)+jsin(0)) - V2 (415(cos(90)+jsin(90))

Though the square root of 2 x 415 is rms volts?

 

[gneill]

Thank you for your response

I'm not sure how to convert these voltages to phasor quantities would it be like this

V1 = (415(cos(0)+jsin(0)) - V2 (415(cos(90)+jsin(90))

Don't combine V1 and V2 right away. Just write them as separate phasor quantities to begin with. The problem wants you to apply different methods to solve for the load current, so you'll need all the component values to be separate.

Though the square root of 2 x 415 is rms volts?

RMS is PEAK divided by √2. The expressions given for the voltage functions should be interpreted as peak voltages. The fact that they contain a √2 makes it easy to divide by √2 to find their RMS magnitude ;)

I suggest that you let V₁ provide the reference angle for the phasors, so you can just write V₁ as: 415 V ∠ 0° , or in complex form, 415 + j0 V.

You've got the right idea about how to convert the time functions to complex phasor values.

 

[peppa]

Thanks gneil

Sorry about the lateness in reply though i am still struggling to confirm my answer

If V1 = 415+j0v (415(cos(0)+(jsin(0))
V2 = 0 +j415 (415(cos(90)+(jsin(90))

Then subtract them (V1-V2) in complex form or keep it in polar? I'm struggling with this as i can't seem to find any examples of phase angles at 0 degrees and your point about keeping component values separate has confused me

I've been reading other post s on polar and rectangle form including extracts on allaboutcircuits.com but can't seem to find a solution for the voltage difference.

415∠0/7.211∠56.31?

j4+j6 = 7.211(cos56.31+jsin56.31) = 7.211∠56.31

 

[gneill]

Thanks gneil

Sorry about the lateness in reply though i am still struggling to confirm my answer

If V1 = 415+j0v (415(cos(0)+(jsin(0))
V2 = 0 +j415 (415(cos(90)+(jsin(90))

Then subtract them (V1-V2) in complex form or keep it in polar? I'm struggling with this as i can't seem to find any examples of phase angles at 0 degrees and your point about keeping component values separate has confused me

I'm not sure why you want to subtract them. What's your plan?

Since you're looking for the Thevenin equivalent, my instinct would be to find Vth using nodal analysis. Take a look at the circuit rearrangement I presented in post #15.

I suggest that you want to keep the components separate because you have other methods to apply (problem parts b and c) that will require them to be separate.

I've been reading other post s on polar and rectangle form including extracts on allaboutcircuits.com but can't seem to find a solution for the voltage difference.

They are both just complex numbers. Apply the standard rules of complex arithmetic. Addition and subtraction are carried out in rectangular form, adding or subtracting the real and imaginary components separately.

415∠0/7.211∠56.31?

j4+j6 = 7.211(cos56.31+jsin56.31) = 7.211∠56.31

Not sure what you're doing there. Clearly j4 + j6 = j10 no?

 

[peppa]

My work folders give no examples of what they ask in the assessments which make them glorified paper weights. I have used this website for most of my learning resources but feel my understanding of certain areas is a little scattered. My thought process for this question was as follows

for a)

Zt = j4xj6/j4+j6 = j2.4 ohms

I then need to find the l voltage (Vt) which is the terminal voltage plus the volt drop across the j6 ohm

Vt = V2 + I (V1-V2/j4+j6) x j6 = ? V

Then i can draw the thevenin equivalent circuit and combine it with the ZL ( I = V/35Ω+j35.71) giving the current through the load?

Im stuck pretty early on but i will try nodal anaylsis if that is how i can fine Vt?

 

[gneill]

My work folders give no examples of what they ask in the assessments which make them glorified paper weights. I have used this website for most of my learning resources but feel my understanding of certain areas is a little scattered. My thought process for this question was as follows

for a)

Zt = j4xj6/j4+j6 = j2.4 ohms

Good.

I then need to find the l voltage (Vt) which is the terminal voltage plus the volt drop across the j6 ohm

Vt = V2 + I (V1-V2/j4+j6) x j6 = ? V

I'm not sure what the variable l really represents in the above equation. But it looks like you have the basics of a valid approach: With the load removed find the current in the remaining loop and then sum the voltages along a path from ground to the terminal node (where the inductors meet).

I think nodal analysis is the more obvious approach though, since you can do everything at once by solving one equation.

Then i can draw the thevenin equivalent circuit and combine it with the ZL ( I = V/35Ω+j35.71) giving the current through the load?

Sure.

Im stuck pretty early on but i will try nodal anaylsis if that is how i can fine Vt?

Nodal analysis will certainly work.

 

[peppa]

Thankyou gneil

'l' was a mistype sorry

I will try nodal analysis and post my results

 

[magician]

OK.

This has been an absolute nightmare for the last month or so! This is what I have so far...

Thevenin equivalent reactance;

1/Z_th = 24j/10 = 2.4j

v1 = sqrt2 * 415sin(100pi.t +90)
v2 = sqrt2 * 415sin(100pi.t)

converting to rms value:

v1 = 415∠90
v2 = 415∠0

circuit eq now;

-v1 + j4 * i + j6 * i + v2 = 0

substituting v1 & v2 I then get;

i = 58.69∠45

Thevenin equivalent voltage is

Vth = 6j * i + v2
Vth = 299.26∠56.31

The load connected is:

Z_L = 50 ∠-arcos 0.7
Z_L = 50∠-45.572

Vth = i * 2.4j + i * 50∠-45.572
Thus;

i = 6.193∠99.89A

Reforming the current;

i = sqrt2 * 6.193sin(100pi.t + 99.89)A

And relax... And pray...

PS. I obviously have a lot more of the working out inbetween most of that! But hopefully you can tell me I have followed the right path.

Many thanks

 

[gneill]

The load power factor is said to be lagging which implies that the load impedance should have an inductive component (current lags voltage). That in turn implies that the angle associated with that impedance should be positive. So fix up your load impedance.

 

[magician]

Ah, damn it! That makes sense when you put it like that!

OK. I will re calculate using;

Z_L=50∠45.572°Ω

Thank you

 

[Ebies]

Thevenin equivalent reactance;

1/Z_th = 24j/10 = 2.4j

Quote from post #30

How do you calculate Thevenin's equivalent reactance when there is no frequency supplied, I understand how to calculate reactance using the formula Reactance=2*pi*f*H(Henry), but I am kind of lost as to how to calculate it in this instance, I have to agree with Peppa with the work books supplied being as good as glorified paperweights!

 

[gneill]

Quote from post #30

How do you calculate Thevenin's equivalent reactance when there is no frequency supplied, I understand how to calculate reactance using the formula Reactance=2*pi*f*H(Henry), but I am kind of lost as to how to calculate it in this instance, I have to agree with Peppa with the work books supplied being as good as glorified paperweights!

The individual impedances of the components are already given. Take a close look at the circuit diagram that was provided in the first post.

 

[Ebies]

am I correct then at saying j4 and j6 are the impedances respectively? this equation is giving me a hard time haha they only show examples of thevenins theorem using resistors which I totally get but then they throw the odd ball like this and I am completely lost like the other chaps on here...

oh and you multiply j4&j6 to get j24?

 

[gneill]

am I correct then at saying j4 and j6 are the impedances respectively?

Yes. Those are the impedances of the two inductors.

this equation is giving me a hard time haha they only show examples of thevenins theorem using resistors which I totally get but then they throw the odd ball like this and I am completely lost like the other chaps on here...

oh and you multiply j4&j6 to get j24?

The 'j' represents the square root of negative one. That is, the impedances are imaginary values (and are complex in general). So the arithmetic you do with them must follow the rules of complex arithmetic.

 

[magician]

Hey Ebies,

As gneill has already said, "The 'j' represents the square root of negative one."

To help you get started, below is the calculations I used;

1/Z_th = 1/j4 + 1/j6
1/Z_th = (4+6)/j24
1/Z_th = 10/j24

Therefore;
Z_th = j24/10
Z_th = j2.4

Now following the procedure I did above, [post 30] you'll be there in no time, aslong as you take your time with the complex arithmetic.

 

[Ebies]

thanks gneil and magician, your help is much appreciated... I want to try and crack this one out before Wednesday morning as I am heading offshore and won't be able to do any of this whilst I am away and trying to come back to a half completed equation after a few weeks at sea is pants as you forget half of what you have done before you left...

 

[magician]

thanks gneil and magician, your help is much appreciated... I want to try and crack this one out before Wednesday morning as I am heading offshore and won't be able to do any of this whilst I am away and trying to come back to a half completed equation after a few weeks at sea is pants as you forget half of what you have done before you left...

Completly understand Ebies.

Put up what you've got.

Ive had mine marked and it is correct

 

[Electest]

Hi Guys

I'm really trying to get my head round this and would appreciate some pointers. I understand obtaining the Thevenins equivalent impedance (Zt) but am struggling obtaining The Thevenins equivalent Voltage (Vt).

Looking at the circuit drawing the j4Ω & j6Ω inductances act as a potential divider between the 2 voltage sources, do they not?

The voltage across the j6Ω is therefore given by $$\left (v1-v2 \right ) \frac{6}{4+6}$$

where:

v1 = 415∠90 v or 0+j415 v
v2 = 415∠0 v or 415+j0 v

Then I could minus the voltage across the j6Ω from the v2 Source to get my answer?

Thanks in advance

 

[Ebies]

Hi guys, just got back from my off shore trip, been a short one to say the least. I attempted this during my time away and to be completely honest I am still at a loss,

I understand calculating the Zth using product over sum rule,

I have used the p.f. of 0.7 lagging to calculate theta=45.572996 deg
I understand that v1 = sqrt2 * 415sin(100pi.t +90) and v2 = sqrt2 * 415sin(100pi.t) respectively

what I do not get however is:

- How do you get from v1 = sqrt2 * 415sin(100pi.t +90) and v2 = sqrt2 * 415sin(100pi.t) to either Polar or Rectangular form? I an change from Polar to Rect and vice versa but not completely sure how to get there from what's given by V1 and V2 so just a nudge in the right direction here will help...

thank you kindly

 

[gneill]

If you have a time signal of the form $A sin(\omega t + \phi)$ or $A cos(\omega t + \phi)$ then you can use the phasor form $A ∠ \phi$. So then you have the polar form of the signal in the frequency domain. You should already know how to convert a polar form to rectangular form.

Note that it is critical that when you convert a phasor result back to the time domain that you employ the same trig function that you started with otherwise you'll introduce a spurious phase difference (the phase difference between a sine and cosine function).

 

[Ebies]

thanks for that gneill

 

[Ebies]

out of interest can I make the following statement and would it ring true for imaginary numbers? this would give me my Thevein Voltage when the circuit has been redrawn as originally posted but without the Load of 50Ω

the voltage drop across j6Ω would be 1.5x more than the voltage drop across j4Ω and as such write it as:

Vj6Ω= (V2-V1)x(j6/j10)...
I know this will work with normal numbers for both voltage supplies and resistances but not sure if it will work when you use imaginary numbers...

 

[gneill]

I think it should work if the numbers are purely real or purely imaginary (as in this case). Else there will be "interaction" between the real and imaginary components, and the phase of the result will change.

 

[Ebies]

not completely sure it works this way when complex arithmic is involved... I Vth=352.14∠-45 and clearly it is not the same as Vth in Magician's earlier post... could however be down to me miscalculating... Can anyone please confirm my working out on this...

(V2-V1)*(j6/j10)
=[(415+j0)+(-0-j415)]*(0.6)
=[(415-j0)+(0-j415)]*(0.6)
=415-j415*(0.6)
=249-j249

which in turn gives me: Vth=352.14∠-45

Sorry I keep on posting on this post but I am trying to calculate this equation and need to ensure I understand it...

 

[gneill]

You've calculated the potential across the j6 Ω inductor with an assumed polarity of - on its left and + on its right end (since you've assumed current flow to be counterclockwise by your choice of "V2 - V1"), but you've not yet found the Thevenin voltage. For that you need to take into account the fact that V2 is between that inductor and the reference node (ground).

 

[Ebies]

so does that mean I need to calculate the p.d over the j4Ω and then sum the two resultant p.d's together? I am a bit lost here now... can you point me in the right direction here please gneill?

 

[gneill]

You are looking for the open terminal voltage where the load sits. That is, the Thevenin voltage. To do that you should take a "KVL walk" from one terminal to the other, summing the potential changes. You've found the potential change across the j6 Ohm impedance, so you could choose the path that includes it (in blue below):

 

[Ebies]

so in essence I stick to the blue path to calc j4Ω as well yes? or do I follow the red path for the second part?