# Use Norton’s Theorem to find the current in RL

1. Oct 5, 2013

### agata78

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution

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

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

RN = 50 + (800 - j250 / 45)

RN = 50 + (800 - j250 / 45)

RN = (50 x 45 + 800 - j250) / 45

RN = (2250 + 800 - j250) / 45

RN = 3050 - j250 / 45

RN = 5 (610 - 10j2) / 5 x 9

RN = 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!

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Last edited: Oct 5, 2013
2. Oct 5, 2013

### phinds

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

3. Oct 5, 2013

### Staff: Mentor

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.

4. Oct 5, 2013

### agata78

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

3. The attempt at a solution

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

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

RN = 5+J10 + (800 - j250) / 45)

Am i on the correct path? Im 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.

Last edited: Oct 5, 2013
5. Oct 5, 2013

### Staff: Mentor

Yes, that looks better.
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
Right. Later you'll tack it back onto the Norton source and determine the current through it.

6. Oct 6, 2013

### agata78

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

7. Oct 6, 2013

Yes.

8. Oct 6, 2013

### agata78

STEP 1: Calculate Norton Resistance (RN)

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

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

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

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

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

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

STEP 2: Calculate Total Current (IT)

IT = Battery Voltage / Norton's Resistance

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

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

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

STEP 3: Calculate Norton Current (IN)

IN = IT 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

Last edited: Oct 6, 2013
9. Oct 7, 2013

### agata78

Last edited: Oct 7, 2013
10. Oct 7, 2013

### Staff: Mentor

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.

11. Oct 7, 2013

### agata78

Thank you.
Why j2 = -1

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

Last edited: Oct 7, 2013
12. Oct 7, 2013

### 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 doesnt 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

13. Oct 7, 2013

### Staff: Mentor

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.

14. Oct 8, 2013

### agata78

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

15. Oct 8, 2013

### Staff: Mentor

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]

16. Oct 8, 2013

### 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.

17. Oct 8, 2013

### 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 RN.

VAB = V1 x R2 / R1 + R2

I = VAB / RN

Am I heading in the correct direction?

18. Oct 12, 2013

### 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

19. Oct 12, 2013

### Staff: Mentor

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

20. Oct 12, 2013

### 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?