Thevenin's Theorem: Solving Homework Statement on Load Current

[gneill]

so n essence I stick to the blue path to calc j4Ω as well yes?

You've already taken into account the effect of the j4Ω inductor when you found the potential drop across the j6Ω one. Your calculation effectively found the total current due to the sources in series with the inductors and multiplied that by the j6Ω impedance (a voltage divider scenario).

Either path will yield the same result, thanks to KVL holding around the loop.

I think my preferred approach would have been to just use nodal analysis at the junction of the inductors, so one equation with one unknown (Vth) to solve.

 

[Ebies]

This is as far as I have managed o get and now I am stuck with the equation. not sure how to deal with and equation that has a mix of unknowns and imaginary numbers in it... also attached is my diagram that I am using so the equation makes sense... I spose if I can learn where I am going wrong now I won't make the same mistake again... by the way X is the node between the two inductors or in other words thevenins voltage

my equation thus far:

I1 + I2 +I3 =0

Therefore: (V1 - X) / j4 + (V2 - X) / j6 + (0 - X) / 50∠45.6 (which is the load and lag angle)

(0 + j415 - X) / j4 + (415 + j0) / j6 + (0 - X) / 67.7 + j42.3 (Load and angle converted to Rectangular form)

this is where I am stuck now, hence not trying nodal analysis as I was not 100% sure about it...

 

[Ebies]

I understand division with complex numbers but not when there's an unknown involved... not sure ha to do with the unknown to get it to the other side of the equation...

 

[gneill]

∠This is as far as I have managed o get and now I am stuck with the equation. not sure how to deal with and equation that has a mix of unknowns and imaginary numbers in it... also attached is my diagram that I am using so the equation makes sense... I spose if I can learn where I am going wrong now I won't make the same mistake again... by the way X is the node between the two inductors or in other words thevenins voltage

my equation thus far:

I1 + I2 +I3 =0

Therefore: (V1 - X) / j4 + (V2 - X) / j6 + (0 - X) / 50∠45.6 (which is the load and lag angle)

Okay, the Thevenin voltage is the unloaded output voltage of the network. That means you want to remove the load impedance, so the last term should be removed. Set the remaining expression to equal zero. Solve for X.

(0 + j415 - X) / j4 + (415 + j0) / j6 + (0 - X) / 67.7 + j42.3 (Load and angle converted to Rectangular form)

this is where I am stuck now, hence not trying nodal analysis as I was not 100% sure about it...

With the load term out of the picture you should find an expression for X that involves just some complex arithmetic.

 

[Ebies]

thanks mate, your right the arithmetic is complex haha

 

[Ebies]

last question... j0 * j = 0 or j^2 ??

 

[gneill]

Anything multiplied by zero is zero.

 

[Ebies]

Anything multiplied by zero is zero.

I didn't know if I should treat it as zero sorry. bone question...

 

[Ebies]

I finally worked out Vth gneill... thanks for the help and guidance along the way

 

[Ebies]

gneill, can I ask this? why do we use the rms value for the calculations? should 415 not be multiplied with sqrt 2? when working with thev theorem you want the max values right?

 

[gneill]

gneill, can I ask this? why do we use the rms value for the calculations? should 415 not be multiplied with sqrt 2? when working with thev theorem you want the max values right?

You can use either peak or rms values if you're just looking for voltages or currents. You can write out the time domain functions for the results easily enough with or without the root 2 . If you're looking to calculate power though, it's easier to start with the rms values for the voltages and currents.

 

[Ebies]

thank you very much buddy.. you are a star by far... bloody genius I tell you!

 

[Ebies]

ok so I used my thev equivalent voltage and the total impedance (which was j4 & j6 combined and then added with the load of 50Ω @ 0.7p.f.) to calculate the current I with Vth/Zth and came up with a completely different answer to what it should be thus making me think I made a mistake somewhere... I get i=5.8∠8.9°

 

[gneill]

The magnitude is about right. The angle's off though. What are your (rectangular format) values for V_th, Z_th, and Z_L?

 

[Ebies]

Vth=166+j249 , Zth=35+j38.10714214 and ZL=35+j35.70714214

I have checked my calculations for calculating the current and its correct so one of my actual values must be up the duffer...

 

[gneill]

Vth=166+j249 , Zth=35+j38.10714214 and ZL=35+j35.70714214

Zth can't have any real component as it consists of the reduction of two imaginary values (inductance impedances). Should be something closer to 3j Ohms.

Your ZL looks fine, some resistance and a positive imaginary value implying inductance, which is what you'd expect for a load with a lagging power factor.

Your Vth looks a bit odd to me, if you started with v1 being 415V and v2 as 415V ∠ -90°. Looks like the real and imaginary component magnitudes have been swapped.

 

[Ebies]

I had V1=415∠90 and V2=415∠0... thanks for the hand there again gneill... with regards to Zth - I see what you mean by Zth, sorry I labelled em different in the equation... my Zth=j2.4

That explains my phase shift as I put myself out a few degrees by not reading the question properly... thanks again gneill

 

[Ebies]

ok so going off what gneill suggested last night I get a new revised figure for Vth=247.02-j164.69... what I do not understand however is why using a cos trig function or sine trig function yield different results...? ie the real and imaginary parts being swapped...

 

[gneill]

ok so going off what gneill suggested last night I get a new revised figure for Vth=247.02-j164.69... what I do not understand however is why using a cos trig function or sine trig function yield different results...? ie the real and imaginary parts being swapped...

The difference is in the phase of the complex number. The magnitude should be the same either way. A 90° phase difference is equivalent to rotating your phasor through 90°, which effectively swaps the real and imaginary components.

 

[Ebies]

yes you are correct the magnitude is very much the same sin rig funct (299∠56) and cos trif funct (296∠-33). makes sense now thanks...

 

[Ebies]

was thinking last night whilst in bed trying to fall asleep, the reason my current was incorrect was probably due to me adding the reactance and load and treating it as one component whereas I should have kept them as two separate components worked out the total circuit current then deducted the reactance current drawn from circuit current which would leave me with the load current

 

[gneill]

Well, if you've reduced the network that feeds the load to a Thevenin equivalent, then it is correct to sum the Rth and RL for the total impedance. Then $I = Vth/(Rth + RL)$. That will yield the load current (because there's only one current in a series circuit).

 

[Ebies]

ah gotcha... does this seem more realistic than last nights current then... 5.74∠-81.17 or 5.71cos(100pi*t-81.17) as I used the cos trig funct as per your post #66

 

[gneill]

Yes, that looks good. There is probably some accuracy lost in the second decimal places due to rounding of intermediate values, but it looks like you've reach the right result.

 

[Ebies]

when I do the actual calculation I do not round this was purely for speed purposes to see what result it would yield... thanks a million for your guidance it is very much appreciated i understand this a bit more than before and its not as hard as first perceived... you are a star...!

 

[nautalus]

Hi can someone explain why my answers are different I am well confused

 

[gneill]

Hi can someone explain why my answers are different I am well confused

Check your calculation of the load impedance. What's the magnitude of the impedance that you calculated?

 

[Gremlin]

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).

This question has been killing me for weeks. I thought I was gaining a good understand and this question just blew me away, so I had to look for assistance on the net - I'm a bit relieved that I'm not the only one that's struggled with it.

gneill, I looked at your post #17 (below) and went away to try and make the voltages based on the same trig function and the t time aspect of it completely knackered me, as I had no idea how to get rid of it. Then I see your post above - and I don't think I have come across that in my course notes - and it almost looks too easy. You can just lose $sin(\omega t)$ or $cos(\omega t)$ and that's OK?

Initially we're told that v1 = √2 x 415 x (cos (100 x pi x t)) Volts

and we're then saying that:

v1 = 415 ∠ 0 Volts and subsequently that v2 = 415 ∠ 90 Volts as (sin (x) = cos (90 - x)) and 90-0 = 90.

That can't be right, surely?

I'm confident - if i can get to that point - that I can convert polar to rectangular form and vice versa, but could you just explain what you mean when you say "... polar form of the signal in the frequency domain". This has done me in a bit. And once again you just can't get away from this question.

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.

 

[gneill]

This question has been killing me for weeks. I thought I was gaining a good understand and this question just blew me away, so I had to look for assistance on the net - I'm a bit relieved that I'm not the only one that's struggled with it.

gneill, I looked at your post #17 (below) and went away to try and make the voltages based on the same trig function and the t time aspect of it completely knackered me, as I had no idea how to get rid of it. Then I see your post above - and I don't think I have come across that in my course notes - and it almost looks too easy. You can just lose $sin(\omega t)$ or $cos(\omega t)$ and that's OK?

Yes, that's okay. This is the beauty of using phasors to analyze AC circuits. By Euler's formula:

$A e^{j(ωt+ q)} = A cos(ωt + q) + j A sin(ωt + q)$

and a phasor is a projection on an axis of the resulting rotating vector. For example, $Re(Ae^{jωt})$ would be the projection on the x-axis and would be identical to Acos(ωt).

When a circuit is driven by sources with a single frequency, at steady state all the voltages and currents in the circuit can be represent by phasors and the important property is that all these rotating vectors will maintain fixed magnitudes and angular relationships between each other. So the convention is to simply carry the rotation (a time domain property) along as implied, and deal with just the magnitudes and angular relationships. When drawn as a diagram, the phasors look like fixed vectors and the rotation is implied.

This is doable because

$Ae^{j(ωt+ Φ)} = A e^{jΦ} e^{jωt}$

We call $A e^{jΦ}$ the phasor representation for purposes of analysis and the $e^{jωt}$ part is the time varying bit implied to be common to all of the phasors. Note that $A e^{jΦ}$ can be represented as a complex constant or a constant vector.

Initially we're told that v1 = √2 x 415 x (cos (100 x pi x t)) Volts

and we're then saying that:

v1 = 415 ∠ 0 Volts and subsequently that v2 = 415 ∠ 90 Volts as (sin (x) = cos (90 - x)) and 90-0 = 90.

That can't be right, surely?

Sure. Time domain sources specified as cosine or sine functions usually imply that their magnitudes are given as peak values. Dropping the $\sqrt{2}$ turns them into RMS values suitable for doing power calculations.

I'm confident - if i can get to that point - that I can convert polar to rectangular form and vice versa, but could you just explain what you mean when you say "... polar form of the signal in the frequency domain". This has done me in a bit. And once again you just can't get away from this question.

415 ∠ 90 would be an example of a the polar form of the signal in the frequency (phasor) domain. In rectangular form it would be 0 + j 415 .

 

[Gremlin]

Thanks, I'll have another look at it this evening.

 

[Gremlin]

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

converting to rms value:

v1 = 415∠90
v2 = 415∠0

I am not far off throwing myself out of the window with this. I can get j2.4 - that's straight forward - but the above is killing me.

Originally it's:

v1 = √2 x 415 x cos(100 x π x t) Volts
v2 = √2 x 415 x sin(100 x π x t) Volts

I understand that cos (x) = sin (x + 90), that's sound. So:

v1 = √2 x 415 x sin(100 x π x t + 90) Volts
v2 = √2 x 415 x sin(100 x π x t) Volts

Ok. Then i would set t=0 and that would give us:

v1 = √2 x 415 x sin(90)
v2 = √2 x 415 x sin(0)

Now at this point i refer to my own notes i can see "V = V∠θ (r.m.s) or V = √2 x V ∠ (peak)" and i assume that's relevant. Although at this point i just hit a brick wall. People are just losing √2 and i have no idea why or how. Can anyone help me please?

 

[gneill]

I am not far off throwing myself out of the window with this. I can get j2.4 - that's straight forward - but the above is killing me.

Originally it's:

v1 = √2 x 415 x cos(100 x π x t) Volts
v2 = √2 x 415 x sin(100 x π x t) Volts

I understand that cos (x) = sin (x + 90), that's sound. So:

v1 = √2 x 415 x sin(100 x π x t + 90) Volts
v2 = √2 x 415 x sin(100 x π x t) Volts

Ok. Then i would set t=0 and that would give us:

v1 = √2 x 415 x sin(90)
v2 = √2 x 415 x sin(0)

Not quite sure why you're setting t to zero, unless its to highlight the phase constants?

Anyways, there's no point in playing with t since you're about to abandon the time domain entirely and move to the frequency domain (or I suppose one might call it the Laplace domain) by switching to phasor notation.

Now at this point i refer to my own notes i can see "V = V∠θ (r.m.s) or V = √2 x V ∠ (peak)" and i assume that's relevant. Although at this point i just hit a brick wall. People are just losing √2 and i have no idea why or how. Can anyone help me please?

A few things to note:

When sources are given in the time domain they are usually specified in terms of peak value. So unless otherwise indicated, a source that's defined by a sine or cosine function of time will have its magnitude specified as a peak value.

To convert between peak and RMS for sinusoids you divide the given magnitude by √2. To convert from RMS to peak you multiply the magnitude by √2.

Algebraically the √2 is just a scaling constant, so for voltage and current values you can take it out or put it back anywhere along the line of your calculations to convert from one to the other.

If you are doing power calculations, unless you're looking for instantaneous power values in the time domain you want to use RMS values for the voltages and currents. Using RMS leads to the proper "real world" power values for steady state operation of the given circuit.

In the present problem the author conveniently specified the time domain voltage sources with the √2 in plain site, allowing one to simply trim it off and have nice round numbers for the RMS values.

At the end of the problem if you need to convert from the phasor domain back to the time domain then you are welcome to tack the √2 back onto the magnitude when you write the sine or cosine functions of time.

 

[The_daddy_2012]

I've read and gone over this feed so many times and I'm still confused. I can find the thevenin reactance, that's easy. But this is what I've got and I have no idea why it's so different to everyone else's in regards to the current out.

j4 + j6 = j10 ohms

V1= 415angle90 = j415

Therefore, to find the current I1 and thus find the thevenin voltage:

j415/j10 can't divide by a j number so multiply by conjugate.

j415/j10 x -j10/-j10 = 4150/100 = 41.5A so I1 apparently = 41.5A

Then multiply I1 by the resistance in parallel with the load/exposed terminals, as far as I can see that's both of them.

41.5 x j10 = j415 and I'm back exactly where I started.

This is how the hand out says to do it. Then do the same for the other voltage source and add the two voltages together to get Vth.

Where have I gone wrong?

 

[The_daddy_2012]

Following on I have finished the calculation lowing ahead as I thought.

Maybe it's not j10 I multiply by, but that would make v1 multiplied by j6 giving j249V as the thevenin voltage of v1

Then working through the same method but multiplying by j4 this time as it's now the one in parallel I get

V2th = 166V

Then adding V1th and V2th together.

= 166+j249V

Confused

 

[magician]

Following on I have finished the calculation lowing ahead as I thought.

Maybe it's not j10 I multiply by, but that would make v1 multiplied by j6 giving j249V as the thevenin voltage of v1

Then working through the same method but multiplying by j4 this time as it's now the one in parallel I get

V2th = 166V

Then adding V1th and V2th together.

= 166+j249V

Confused

Remember. The PF is lagging so the circuit equation (for calculating the current) will begin like this:

CIRCUIT EQUATION;

-V1 + j4 * I + j6 * I + V2 =0

Thus;

-415<90° + j4 * I + j6 * I + 415<0° = 0

Continue with the above until you get your result for current, I.

Then you can quite easily find the Thevevin equivalent Voltage, Vth

 

[The_daddy_2012]

Hi magician,

Thank you for the quick response.

What I'm getting from that is a current of

I = 41.5 - j41.5

But I'm now confused as to where I am. Is this the overall current of the of the Thevenin equivalent circuit? How do I find the thevenin voltage from here?

Cheers

 

[magician]

Then if you have the current, continue to find the Thevevin equivalent voltage.

Once you have that, then you should be able to sketch a very simplified drawing. From that calculate Z(load) and find the current in the load.

This can be expressed in the same format as the beginning voltages, ie;

i = sqrt2 * ?sin(100pi.t + ?)

And youre done! :)

 

[magician]

Also,

Vth = j6 * I + V2

 

[The_daddy_2012]

Still confused as to what this current actually represents. If it is the current of the thevenin equivalent circuit, then I would multiply it by j2.4 (Zth) which gives me 99.6 + j99.6Vth.

I woke up this morning thinking I'd cracked it, I'm now more confused than ever!

 

[The_daddy_2012]

Right, so now, on that logic. I have got.

j6 x (41.5 - j41.5) + 415

Which gives

j249 + 249 + 415 = 664 + j249?

If this is correct wink twice.

 

[magician]

Ok.

I haven't checked your maths, but using that, you will now have a Voltage, Vth connected to j2.4, connected to Z_load, with current, i through the load.

Calculate Z_load (pretty simple)

Hence current in load is:

Vth = ...

 

[The_daddy_2012]

The current in the load will be

Vth/(Zth+ZL) which from this I'm going to go for.

664 + j249/ j2.4 + 40.95 + j28.68

Therefore

IL= 7.36 - j3.95
In polar 8.35angle28.2

Therefore

IL= root2 x 8.35 sin (100 pi t + 28.2)

Is that the right way?

 

[magician]

Try:

Vth = i * 2.4j + i * Z_load

 

[magician]

And PF is lagging...

50< - cos^-1 . 0.7

 

[The_daddy_2012]

Ok, I'm going to start again and work my way through making sure I get each stage right. Then I'll post my calculations and results.

Thank you magician, you've been a massive help! :)

 

[The_daddy_2012]

Just a quick one, the -v1+ j4 x I + j6 x I + v2 that's from nodal analysis isant it?

 

[The_daddy_2012]

Ok, what I have from the stage where the voltages are converted to polar/rectangular

V1= j415
V2= 415

Using

-v1 + (j4 x i) + (j6 x i) + v2

i = 41.5 + j41.5

Therefore:

Vth = j6 x i + v2

Now I have:

Vth = 166 + j249

Zth = j2.4

ZL = 35 + j 35.7

Current flowing through load is

iL = Vth / (Zth + ZL)

iL = 5.71 + j0.89 or 5.77angle8.86 or root2 x 5.77 sin (100 pi t + 8.86)

I thought I had it right, but then did the superposition theorem in the next question and the answers are completely different.

 

[gneill]

Okay, so your Thevenin approach looks good, and your answer looks fine.

For the superposition version, can you first describe in broad strokes what your approach is?

 

[dave pallamino]

Hi Guys,

I've pretty much got my head around everything up to finding the current for the Thevenin model... -v1 + (j4 x i) + (j6 x i) + v2 = 0 ... -415 + (j4 * I) + (j6 * I) + 415 = 0

but I'm really struggling with how to approach the complex math to achieve i = 41.5 + j41.5

I'm not looking for a complete breakdown but please could someone show me how to approach this?

Thank you

 

[KatieMariie]

Hello,

I too am having a terrible time with this question, I've had a look through this post but the only thing i can gather is that the folders I've been sent really are glorified paper weights!

If someone could talk me though and teach me the theory behind this that would be amazing, as i really am at a loss and am too stubborn to move on until this question is completed.

Thank you!