# Norton equivalent

Discussion in 'Homework Help' started by regexp, Dec 27, 2010.

1. ### regexp Thread Starter New Member

Nov 20, 2010
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Hi,

I have the following circuit

And i need to calculate the norton equivalent circuit.

After i shorted the voltage source and changed the current source into an open circuit, the thevenin resistance $R_{th} = 4ohm$

Applying the superposition principle the short circuit current is 2A plus the current generated by the voltage source. This is where i had to stop and think. Are there any current traveling through the 6ohm resistor?

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
1,149
Yes the current will be flow through 6ohm resistor.
And how can 2A is short circuit current?

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Look at this diagram

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4. ### regexp Thread Starter New Member

Nov 20, 2010
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Thanks, i didn't know you could add them like that.

Is there any other way besides source transformation?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, superposition also work very well

6. ### regexp Thread Starter New Member

Nov 20, 2010
24
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Hi,

Hm, the only way i get 1A from the voltage source is

if i consider the 6ohm and 3ohm resistor in parallel.

$\frac{6\cdot3}{6+3} = 2$

and then $\frac{2\cdot2}{2+2} = 1$

However arent they supposed to be in series when the voltage source appear in between?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
1,149
Hmm, To find Norton equivalent

First:
Calculate the output current, IAB, with a short circuit as the load.

Then find Rth
Replace all voltage sources with short circuits and all current sources with open circuits.

So first we need to find this short current by using superposition.
I first remove the current source

And then by inspection I write

Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

And now we replace voltage source with a short circuit

Is2 = 2A * (6Ω||2Ω) / ( 6Ω||2Ω + 2Ω ) = 2A * 2/4 = 1A

So our final solution is

Isc = Ics1 + Isc2 = 1.5A
and Rth= 4Ω

And of course we can get the same answer with source transformation method, or nodal/mesh analysis .

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Last edited: Dec 28, 2010
8. ### regexp Thread Starter New Member

Nov 20, 2010
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Hi,

Thank you very much. There is one part i don't really understand

Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

First, the total current is
6V/(3Ω||2Ω + 6Ω) = 6/7.2 = 0.83A

Then we use current division.

3/(3+2) * 0.83 = 0.498A

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,082
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You use current divider rule, but I use Ohms law.

First I find total current

Itot = [ 6V/(3Ω||2Ω + 6Ω) ]

next I find voltage drop across 3Ω||2Ω

V = Itot* 3Ω||2Ω =[ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω

And finally, if I know the voltage across 2Ω resistor I can easily find current that flow through 2Ω resistor.

I = Isc1 = V/2Ω = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A

I hope you understand now.

And maybe at the end I show how you can use nodal equation to find Isc
for this circuit:

2A = Va/6Ω + Vb/3Ω + Vb/2Ω (1)

Vb - Va = 6V ----> Vb = 6V + Va (2)

2 = Va/6 + 2Vb/6 + 3Vb/6

2 = (Va + 5Vb)/6

2= ( Va + 5*( Va+6 ) )/6= ( 6Va + 30) / 6

2 = (6Va +30) / 6

12 = 6Va + 30

-18 = 6Va

Va = -18/6 =-3V

Vb = 6V + (-3V) = 3V

And finaly Isc = Vb/2Ω = 3/2 = 1.5A

So as you can see you can use any method you want to solve this circuit.

Last edited: Dec 28, 2010

Nov 20, 2010
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