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