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Use Norton’s Theorem to find the current in RL

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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)
RN = RL + ((R1 x R3) / (R2 + R3))

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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  • #2
phinds
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We'd be able to help better if there actually were a figure attached.
 
  • #3
gneill
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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)
RN = 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.
 
  • #4
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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.
 
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  • #5
gneill
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Norton's Resistance Equation (The first part of Norton's Theorem)
RN = R2 + ((R1 x R3) / (R1 + R3))
Yes, that looks better.
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!
You've got the right starting formula, so that's a plus :smile: 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 :wink:
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.
 
  • #6
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RN = 5+J10 + (((20+j5) x (40-j10)) / ( (20+j5) + (40-j10)))
Then this would be correct yes??!!
 
  • #7
gneill
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RN = 5+J10 + (((20+j5) x (40-j10)) / ( (20+j5) + (40-j10)))
Then this would be correct yes??!!
Yes.
 
  • #8
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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
 
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  • #9
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Can someone please help me out? Was my last post correct? Much appreciated!
 
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  • #10
gneill
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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
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Thank you.
Why j2 = -1

I actually missed that. It makes it much easier to calculate.
 
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  • #12
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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
gneill
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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
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Ok.
Rn= ((240 + 115j) / (12-j)) x ( (12+j) / ( 12+j)) =( 2765 + 1620j ) / 145
Is my final answer for Rn is correct?
 
  • #15
gneill
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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]
 
  • #16
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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
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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
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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 im stuck with this one. Please help me!
 
  • #19
gneill
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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
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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?
 
  • #21
gneill
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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?
 
  • #22
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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
 
  • #23
gneill
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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.
 
  • #24
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I made mistake with my previous post:
Vab= V1 x R3 / R1 + R3

How can i deivide voltage?
 
  • #25
gneill
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I made mistake with my previous post:
Vab= V1 x R3 / R1 + R3
Much better.

How can i deivide voltage?
??? Divide voltage ???
 

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