# calculate R-thevenin

Discussion in 'Homework Help' started by amichayfeldman, Jun 26, 2013.

1. ### amichayfeldman Thread Starter New Member

Jun 23, 2013
10
0
hello!

how i calculate the R thevenin ,if i got positive V open circuit, and i got negative I-short-circuit. is it correctly to calculate with the absolute value of I-short-circuit?? (beacause the minus sign of the current is just to show direction?)

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Show as a circuit diagram.

Jun 23, 2013
10
0
what i did

4. ### amichayfeldman Thread Starter New Member

Jun 23, 2013
10
0
I forgot to add wire under the 3Ix source..

Last edited: Jun 26, 2013
5. ### screen1988 Member

Mar 7, 2013
310
3
The image is not clear but your answer in Voc is wrong.
Voc is the voltage across the source 30Ix and the resistor 10Ω in series. Therefore,
Voc = -30Ix + 10*(-1.5Ix) = -45Ix.

To find Isc you can use KCV at the top-right corner of the circuit.
How can you know 1.5Ix * 20 = -30?
You forgot the voltage across current sources.

Last edited: Jun 26, 2013
6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097

Rth = 300V/10Ω = 30A

I also don't like you transformation the 15IX CCVS into 1.5Ix CCCS.

Jun 23, 2013
10
0
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
I use nodal analysis and I write the two nodal equation

30A = Vx/10Ω + 3Ix + Va/10Ω

but Ix = Va/10Ω

so we have

30A = Vx/10Ω + 3*Va/10Ω + Va/10Ω

and one additional equation for supernode

Vx - Va = 15*Va/10Ω

and we have a solution for Vth

The Vab = Vth = Voc = 600/13 = 46.1538V

• ###### cache.PNG
File size:
27.2 KB
Views:
16
Last edited: Jun 27, 2013
screen1988 likes this.
9. ### amichayfeldman Thread Starter New Member

Jun 23, 2013
10
0
ok. great job

but i don't understand why my way is incorrect?

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
You haven't given a clear development of what your method involves.
In addition it should be clear why the short circuit case is a trivial exercise.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Contrary to conventional wisdom, it is interesting that one can find the Thevenin voltage using superposition. It is often argued one cannot use superposition in circuits where dependent sources are included.

The process as I see it ...

For the 30A current source, the contribution to ix is 15A
For the 3*ix dependent current source, the contribution is -3*ix/2 or (-1.5*ix) Ampere
For the 15*ix dependent voltage source, the contribution is -15*ix[Volts]/20[Ω] or -0.75*ix Ampere.

So from superposition one may write

ix=15 -1.5*ix-0.75*ix

or

ix*(1+1.5+0.75)=15

and

ix=15/3.25=4.615385 Ampere.

giving

Vth=10*ix=46.15385 Volt.

Jony130 likes this.
12. ### screen1988 Member

Mar 7, 2013
310
3
Jony,
Can you tell me where is the Vb?
I am not quite understand your solution.
And in finding Voc can we use the circuit in paper that amichayfeldman used?
I can't find Ix from that circuit.

Last edited: Jun 27, 2013
13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Look again at my post 8. A correct the error.

I don't think so. I don't like his method. But maybe some more knowledgeable member in circuit theory can help you.

screen1988 likes this.
14. ### screen1988 Member

Mar 7, 2013
310
3
Thanks, it is easy to understand now.
I am curious to know how can we find Voc from the circuit.

It made me think that Voc is a function of Ix but that is wrong!!!

File size:
14.3 KB
Views:
50
15. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
You get wrong answer, because for me this "source transformation" done by amichayfeldman don't work as expected.
Conversion 3Ix current source into voltage one should work. But I don't know why this also don't work.
But I complicity don't like the way he transform the 15Ix CCVS into 1.5Ix CCCS.

screen1988 likes this.
16. ### screen1988 Member

Mar 7, 2013
310
3
Uhm, I see what is wrong here. A voltage source with parallel resistor can't be transformed to a current source with the resistor in series.
I didn't notice it.

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
I have some more thoughts about this 3Ix CCCS into CCVS 30Ix source transformation. And I think that I know the answer why in this case this transformation don't work. Because after transformation the Ix current that flow through 10R resistor will change his value.
See the example

Before source transformation IR1 ≈ 833.33mA and after transformed a current source with a parallel resistor into voltage source with a series resistor. IR1 change his value IR1' ≈ 166.7mA.
So our Ix current change his value so this source transformation is not aloud.
Because we have a current control sources in the example.

• ###### 1.PNG
File size:
10 KB
Views:
23
Last edited: Jun 27, 2013
18. ### screen1988 Member

Mar 7, 2013
310
3
Hi Jony, I think it will work.
Thevenin-Norton only make sure that voltage and current at terminals unchanged.
After converting 3Ix CCCS into CCVS 30Ix source I think we can continue to figure out Voc.

PS. In your image, I calculated the voltage at the node is 11.66666V. The current flows through R1 is not the same in the two circuits but the voltage and current looking into pair "current source and R1" and "voltage source and R1" are the same.

Last edited: Jun 27, 2013
19. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
I would like to see how?

In witch node you find 11.67V??

Yes, but in our case Ix is control current, so after transformation we change transformation current. So how can this transformation holds ?
This can work only if after transformation Ix don't change his value.

File size:
33.2 KB
Views:
25
20. ### screen1988 Member

Mar 7, 2013
310
3
Hum...I got stuck because after transforming from current source to voltage source the control Ix disappear.
The node that you wrote 8.33333
Maybe I need to go sleep now. I can't focus on anything.