Use Norton’s Theorem to find the current in RL

[agata78]

Homework Statement

Use Norton’s Theorem to find the current in RL for the figure attached to this post?

Homework Equations

Norton's Resistance Equation (The first part of Norton's Theorem)
R_N = RL + ((R1 x R3) / (R2 + R3))

The Attempt at a Solution

R_N = 50 + (20+j5) x (40-j10) / (5+j10) + (40-j10)

R_N = 50 + (800 - 200J + 200J -J²50) / (5+j10 + 40-j10)

R_N = 50 + (800 - j²50 / 45)

R_N = 50 + (800 - j²50 / 45)

R_N = (50 x 45 + 800 - j²50) / 45

R_N = (2250 + 800 - j²50) / 45

R_N = 3050 - j²50 / 45

R_N = 5 (610 - 10j²) / 5 x 9

R_N = 610 - 10j² / 9

Am I correct so far? If I was given simple figures for the R1, R2 & R3 then this exercise would be very straight forward. But because of the complex figures for each Resistor, I need guidance so I know I am answering the question correctly. Many thanks!

 

[phinds]

We'd be able to help better if there actually were a figure attached.

 

[gneill]

Homework Statement

Use Norton’s Theorem to find the current in RL for the figure attached to this post?

Homework Equations

Norton's Resistance Equation (The first part of Norton's Theorem)
R_N = RL + ((R1 x R3) / (R2 + R3))

Can you elaborate on the above? How did you arrive at that expression for the Norton Resistance?

What are you intending to include in the Norton equivalent circuit? It looks as though you're bundling the load resistance RL into the model. Isn't that the resistor for which you want to find the current? Once a component is absorbed into the equivalent circuit it "disappears" from view, and you can no longer say or determine anything about it...

Since RL is your load (attached at indicated points a and b), you don't want to include it in the Norton equivalent. You want to keep it separate so that later you can determine the current though it.

 

[agata78]

Norton's Resistance Equation (The first part of Norton's Theorem)
R_N = R2 + ((R1 x R3) / (R1 + R3))

3. The Attempt at a Solution

R_N = 5+J10 + (20+j5) x (40-j10) / (20+j5) + (40-j10)

R_N = 5+J10 + (800 - 200J + 200J -(j²50)/ (5+j10 + 40-j10)

R_N = 5+J10 + (800 - j²50) / 45)

Am i on the correct path? I am a bit confussed, it seems to be not difficult one but gives me a headache!

Yes, I was incorrect previously. Once the RL is removed, it can no longer be worked on in further calculations.

 

[gneill]

Norton's Resistance Equation (The first part of Norton's Theorem)
R_N = R2 + ((R1 x R3) / (R1 + R3))

Yes, that looks better.

3. The Attempt at a Solution

R_N = 5+J10 + (20+j5) x (40-j10) / (20+j5) + (40-j10)

R_N = 5+J10 + (800 - 200J + 200J -(j²50)/ (5+j10 + 40-j10)

R_N = 5+J10 + (800 - j²50) / 45)

Am i on the correct path? I am a bit confussed, it seems to be not difficult one but gives me a headache!

You've got the right starting formula, so that's a plus but your calculations seem to be getting mucked up. Yes, doing complex number arithmetic by hand can be tedious. But there's no thing for it but practice. All I can say is, work carefully and perhaps check each step using an online complex calculator as you practice

Yes, I was incorrect previously. Once the RL is removed, it can no longer be worked on in further calculations.

Right. Later you'll tack it back onto the Norton source and determine the current through it.

 

[agata78]

RN = 5+J10 + (((20+j5) x (40-j10)) / ( (20+j5) + (40-j10)))
Then this would be correct yes??!

 

[gneill]

RN = 5+J10 + (((20+j5) x (40-j10)) / ( (20+j5) + (40-j10)))
Then this would be correct yes??!

Yes.

 

[agata78]

STEP 1: Calculate Norton Resistance (R_N)

R_N = 5+J10 + ((20+j5) x (40-j10)) / ((20+j5) + (40-j10))

R_N = 5+J10 + ((800 - 200J + 200J -(j²50)) / (20+j5 + 40 - j10)

R_N = 5+J10 + ((800 - j²50)) / (60 - j5)

R_N = ((5 + j10) (60 - j5) + 800 - j²50)) / (60 - j5)

R_N = (300 - 25j + j600 - j²50 + 800 - j²50) / (60 - j5)

R_N = (1100 + 575j - 100j²) / (60 - 5j)

STEP 2: Calculate Total Current (I_T)

I_T = Battery Voltage / Norton's Resistance

I_T = 50 / ((1100 + 575j - 100j²) / (60 - 5j))

I_T = 50 x ((60 - 5j) / (1100 + 575j - 100j²))

I_T = (3000 - 250j) / (1100 + 575j - 100j²)

STEP 3: Calculate Norton Current (I_N)

I_N = I_T x R₃ / R₂ + R₃

Thanks Neil. Yeah, once again I can see where I went wrong. Hopefully, I am heading in the right direction now. Agata

 

[agata78]

Can someone please help me out? Was my last post correct? Much appreciated!

 

[gneill]

Your Rn is on the right track, but you didn't complete the simplification. The final result should be a single complex value of the form a + bj. That is, the result should be clear of complex values in the denominator.

The same can be said for the other steps, they seem to have been left incomplete.

Remember that in complex numbers, j² = -1.

 

[agata78]

Thank you.
Why j² = -1

I actually missed that. It makes it much easier to calculate.

 

[agata78]

ok, i tried to go with your advice and calculate the rest.

Rn = (240+115j) / ( 12-j )
It= (120-10j ) / ( 48+23j )
In= ( 960 -340j ) / (432 - 207 j )

the final figures looks better but still it doesn't give me ( a+bj ) answer.

Im worried that i made a mistake in calculations. Could you have a look i can write all the calculations if necessery.
thank you

 

[gneill]

Your Rn is okay so far, but you still need to clear the complex value from the denominator. It's a basic operation for complex numbers that you'll need to know how to do. It involves multiplying the numerator and denominator by the complex conjugate of the denominator (effectively multiplying the whole thing by one). The complex conjugate of a complex value is obtained by changing the sign of the imaginary part. So for a complex value a + bj, the complex conjugate is a - bj. Note that multiplying out (a + bj)(a - bj) results in $a^2 + b^2$, a purely real value.

There is something amiss with your calculation of It. You assumed that It = supply voltage/Norton resistance, but that's not true. The Norton impedance (I call it an impedance rather than a resistance because it's a complex value not a pure real resistance) is obtained "looking into" the open terminals a-b. The power supply is "looking into" the other end of the network with a-b shorted. So you'd need to calculate that impedance that the voltage source "sees" first. After that you can obtain In by the method you've shown.

 

[agata78]

Ok.
Rn= ((240 + 115j) / (12-j)) x ( (12+j) / ( 12+j)) =( 2765 + 1620j ) / 145
Is my final answer for Rn is correct?

 

[gneill]

Ok.
Rn= ((240 + 115j) / (12-j)) x ( (12+j) / ( 12+j)) =( 2765 + 1620j ) / 145
Is my final answer for Rn is correct?

Yes, that's better. There are still some common factors to the numbers, so a tad more reduction is possible. You could write it as separate terms:

$\frac{553}{29} + \frac{324}{29}j$ [Ohms]

 

[agata78]

Yes i can see what you ve done, you diveded everything by 5. Yes it looks better.
Thank you. Just need to sit down now and try to find the It.

 

[agata78]

The value (I) for the current used in Norton's Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance R_N.

V_AB = V₁ x R₂ / R₁ + R₂


I = V_AB / R_N


Am I heading in the correct direction?

 

[agata78]

Use Norton’s Theorem to find the current in RL for the figure attached to this post?

Norton's Resistance Equation (The first part of Norton's Theorem)
RN = RL + ((R1 x R3) / (R2 + R3))
RN = 5+J10 + ((20+j5) x (40-j10)) / ((20+j5) + (40-j10))
=( 2765 + 1620j ) / 145
Is my final answer for Rn is correct? Anybody?
Next part:
The value (I) for the current used in Norton's Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance RN.

VAB = V1 x R2 / R1 + R2


I = VAB / RN
Am i right? Would like move further but I am stuck with this one. Please help me!

 

[gneill]

The Norton resistance does not include the load resistor. You remove the load and "look into" the circuit from where the load was connected.

 

[agata78]

I decide to move further myself.

Vab= (116- 248j) / 29

i= 144500- 99560j

I need to use Thevenin theorem as well but it seems like all figures are the same as for Norton theorem.

Would be nice to know if its correct?

 

[gneill]

I decide to move further myself.

Vab= (116- 248j) / 29

i= 144500- 99560j

I need to use Thevenin theorem as well but it seems like all figures are the same as for Norton theorem.

Would be nice to know if its correct?

Doesn't look right. Show a bit more of your work; What expression did you start with for Vab?

 

[agata78]

The value i for the current used in Norton's Theorem is found by determining the open circuit voltage at the terminals AB and dividing it by the Norton resistance r.
VAB =( V1 x R2) /( R1 + R2)
Vab= (50 x ( 40 -j10 )) / ( 20 + j5) + ( 40 - j10)
Vab= (200- j500) / (60-j5)
Vab= (( 200- j500 ) ( 60 +j5 )) / (60 -j5 ) ( 60 +j5)
Vab= (12000- 1000j - 30000j- j Xj x 2500 ) / ( 3600 + 300j -300j + 25 )
Vab= (12000-31000j + 2500 ) / 3625
Vab= (14500-31000j ) / 3625 everything divide per 125
Vab= (116 - 248j) / 29

 

[gneill]

Your expression for Vab is not correct. There's a voltage divider, but it involves resistors R1 and R3.

R2 is open-circuited at terminal a, so it passes no current and so produces no potential drop from the junction of R1 and R3.

 

[agata78]

I made mistake with my previous post:
Vab= V1 x R3 / R1 + R3

How can i deivide voltage?

 

[gneill]

I made mistake with my previous post:
Vab= V1 x R3 / R1 + R3

Much better.

How can i deivide voltage?

? Divide voltage ?

 

[agata78]

Vab= V1 x R3 / R1 + R3

Is this correct?

 

[agata78]

And then
Vab= (50 x ( 40 -j10 )) / ( 20 + j5) + ( 40 - j10)
Vab= (200- j500) / (60-j5)
Vab= (( 200- j500 ) ( 60 +j5 )) / (60 -j5 ) ( 60 +j5)
Vab= (12000- 1000j - 30000j- j Xj x 2500 ) / ( 3600 + 300j -300j + 25 )
Vab= (12000-31000j + 2500 ) / 3625
Vab= (14500-31000j ) / 3625 everything divide per 125
Vab= (116 - 248j) / 29

 

[gneill]

And then
Vab= (50 x ( 40 -j10 )) / ( 20 + j5) + ( 40 - j10)
Vab= (200- j500) / (60-j5) <------- 50 x 40 is not 200
Vab= (( 200- j500 ) ( 60 +j5 )) / (60 -j5 ) ( 60 +j5)
Vab= (12000- 1000j - 30000j- j Xj x 2500 ) / ( 3600 + 300j -300j + 25 ) Sign is incorrect
Vab= (12000-31000j + 2500 ) / 3625
Vab= (14500-31000j ) / 3625 everything divide per 125
Vab= (116 - 248j) / 29

Check the sign on that term.

EDIT: Just noticed the "200" instead of "2000" above!

 

[agata78]

Yes i can see where i made mistake, it should be +1000j.
After changing it i have:
Vab=(14500-29000j+2500 ) / 3600+25
Vab= (14500-29000j ) / 3625
Vab= (116-232j) / 29
Vab= 4-8j= 4(1-2j)
and then
i= Vab/ r

i = ((4-8j )) / ( 553+232j/29)
i= (4-8j ) x (29/ (553+324j ))
i= ((4-8j ) x 29 ) (553 +324j )
i= (116-232j) / (553+324j)Do you agree with my later calculations?

 

[gneill]

Sorry, a just noticed a previous error in the calculation. My apologies. Please go back to my post #28 and see the edit (in green).

You really should look into finding an online complex number calculator that you can use to confirm your operations.

 

[agata78]

Yes i can see it.
Vab= (50 x ( 40 -j10 )) / ( 20 + j5) + ( 40 - j10)
Vab= (2000- j500) / (60-j5)
Vab= (( 2000- j500 ) ( 60 +j5 )) / (60 -j5 ) ( 60 +j5)
Vab= (120000+ 10000j - 30000j- j Xj x 2500 ) / ( 3600 + 300j -300j + 25 )
Vab= (120000-20000j + 2500 ) / 3625
Vab= (122500-20000j ) / 3625 everything divide per 125
Vab= (980 - 160j) / 29
and then
i= Vab/ r

i = ((980 -160j )/ 29) / (( 553+324j)/29)
i= ((980-160j ) /29) x (29/ (553+324j ))
i= ((980-160j ) / (553 +324j )

 

[gneill]

Yes i can see it.
Vab= (50 x ( 40 -j10 )) / ( 20 + j5) + ( 40 - j10)
Vab= (2000- j500) / (60-j5)
Vab= (( 2000- j500 ) ( 60 +j5 )) / (60 -j5 ) ( 60 +j5)
Vab= (120000+ 10000j - 30000j- j Xj x 2500 ) / ( 3600 + 300j -300j + 25 )
Vab= (120000-20000j + 2500 ) / 3625
Vab= (122500-20000j ) / 3625 everything divide per 125
Vab= (980 - 160j) / 29

Huzzah! That is correct

and then
i= Vab/ r

i = ((980 -160j )/ 29) / (( 553+324j)/29)
i= ((980-160j ) /29) x (29/ (553+324j ))
i= ((980-160j ) / (553 +324j )

Good so far! Now normalize it (remove the imaginary in the denominator).

 

[agata78]

The best i can do is:
(490100-406000j) / 410785 divide per 5
(98020- 81200j) / 82157
20( 4901- 4060j) / 82157

It doeasnt look good, I thought the final number would look much different.

 

[gneill]

The best i can do is:
(490100-406000j) / 410785 divide per 5
(98020- 81200j) / 82157
20( 4901- 4060j) / 82157

It doeasnt look good, I thought the final number would look much different.

You're right, it doesn't look good Hint: a common denominator is 2833.

I've got to run now. Back in 7 or 8 hours.

 

[agata78]

20( 4901- 4060j) / 82157

(490100-406000j) / 29 x 2833 divide everything by 29

(16900- 14000j) / 2833

Thank you for your help. I still need to work on this one!

I need to use Thevenin theorem as well but it seems like all figures are the same as for Norton theorem.

 

[agata78]

Still don't know how to finish this one.

 

[gneill]

Go back to your expression:

(490100-406000j) / 410785

Now, with the denominator factor hint I gave, 410785/2833 = 145. So divide all the numerator numbers by 145.

In truth, your expression above is a perfectly good answer, too. In fact, in practical terms you could just use a calculator and present the result in decimal form:

1.193 - 0.988j

Add the appropriate units and you're done.

Sometimes you will get a problem that requests you to express the result in "reduced form", where you need to find the lowest common denominator as we've been doing.

 

[agata78]

Yes i know what you mean. Thank you for helping me with this.
My task is :

Use Thevenin's and Norton's theorems to find the current in Rl.

I checked out some good examples and it seems like Thevenin and Norton come out the same. Is it true? Am i right?

How to complete the task?

 

[gneill]

Yes i know what you mean. Thank you for helping me with this.
My task is :

Use Thevenin's and Norton's theorems to find the current in Rl.

I checked out some good examples and it seems like Thevenin and Norton come out the same. Is it true? Am i right?

How to complete the task?

Well, the current in the load should certainly come out the same. Thevenin and Norton simply replace a given network with sources with a single source (either voltage or current) and a single resistor. The equivalent circuit behaves just as the original sources and network as far as the load is concerned.

You've worked through finding the Norton model, and you can easily find the Thevenin model from that. The rest is just a matter of sticking your load onto one of the models and calculating the current.

 

[agata78]

It is getting a bit complicated now.

Voc= e

Isc= V/ R1= 50/ 20+j5= (1000 -j250) / 325

Rth= Voc/ Isc= ((980- 160j ) / 29 ) / ((1000- 250j ) /325 )

what do you think?

 

[gneill]

It is getting a bit complicated now.

Voc= e

Isc= V/ R1= 50/ 20+j5= (1000 -j250) / 325

Rth= Voc/ Isc=

Sorry, I don't follow what you're doing here.

The Thevenin impedance is identical in value to the Norton impedance. So if you've found one, you've found the other.

The Thevenin voltage is equal to the Norton current multiplied by the Norton impedance. That is,

$R_{th} = R_N$

$V_{th} = I_NR_N$

and of course this also means:

$R_N = R_{th}$

$I_N = V_{th}/R_{th}$

The Thevenin model is a voltage in series with an impedance. The Norton model is a current source in parallel with an impedance.

(Note that I use the term "impedance" here rather than resistance because we're working with an AC circuit with impedances)

 

[agata78]

At beginning i thought the same, but i just didnt understand your previous clue.

Vth= V/ R1

Thank you so much for your help.

 

[gneill]

At beginning i thought the same, but i just didnt understand your previous clue.
does it mean the question has been answered!

Thank you so much for your help.

It means that once you've slogged through the hard work of finding one of the equivalents, finding the other is very easy to do.

To find the current in the load RL, all you have to do is place RL on one of the equivalents and determine the current. Probably easiest to use the Thevenin model for that so you don't have to deal with current division with impedances.

 

[agata78]

I load= Vth / (Rth+ RL )
Am i right?

 

[gneill]

I load= Vth / (Rth+ RL )
Am i right?

Yup.

 

[agata78]

Vth= In x Rn

Vth= ((3380 -2800j ) / 2833 ) x( ( 553+324 j) / 29 )
Vth=(( 2776340-453280j) /82157 )

And then :

I load= Vth/ Rth+RL

Iload= (( 2776340-453280j) /82157 ) / ((553 +324j )/ 29) + 50

Before i go further am i correct?

 

[gneill]

That's fine. By the way, your Thevenin voltage can be reduced to $\frac{980}{29} - \frac{160}{29}j$ V.

 

[agata78]

I load = (1911100 - 638000j) / 4116985

Iload= 0.464-0.155j

Would you agree with me?

 

[gneill]

Yes I would agree. :thumbs:

 

[agata78]

Then it would be my current which i was looking for to answer my task.

Thank you so much!