Norton circuit

1. May 21, 2010

James889

Hello,

I have the following circuit:

[PLAIN]http://img15.imageshack.us/img15/6981/nortonm.png [Broken]

I need to determine the Norton equivalent for this circuit.
I wonder whether the easiest way to solve this is by zeroing the independent source, connecting a test voltage source and calculate the equivalent resistance.

The dependent source confuses me somewhat, being positioned in parallel with the resistance.

Any ideas?

Last edited by a moderator: May 4, 2017
2. May 21, 2010

Phrak

It confuses me too. What does the diamond represent? Is it a current source, or something else?

Last edited: May 21, 2010
3. May 21, 2010

James889

Oh, forgot to write it out.

It's a current source.

4. May 21, 2010

The Electrician

Convert the current source and its parallel 2 ohm resistor into a voltage source in series with a 2 ohm resistor. That should remove the confusion.

5. May 21, 2010

CEL

Where is V in your circuit?

6. May 21, 2010

James889

V is the voltage across the 3ohm resistor.

7. May 21, 2010

vela

Staff Emeritus
The potential difference from bottom to top or top to bottom?

Also, the units of K don't make sense. An ampere is equal to a volt per ohm, not a volt-ohm.

8. May 21, 2010

The Electrician

I think K is the transconductance: amps/volt.

9. May 22, 2010

James889

The drawing a made seem to have caused a bit of confusion, since i forgot to put all the details in

Anyway, here is how it is supposed to look:
[PLAIN]http://img709.imageshack.us/img709/9333/nortony.png [Broken]

Last edited by a moderator: May 4, 2017
10. May 22, 2010

vela

Staff Emeritus
I suggest you use a 1-A test current source instead of a 1-V test voltage source.

11. May 22, 2010

James889

Ok, so zero the independent source, connect a test current across the terminals and find the voltage across the 3 ohm resistor?

12. May 22, 2010

vela

Staff Emeritus
Yeah, that's how you'd start. Then once you know Vx, you can calculate the current through the 2-ohm resistor, and finally calculate the total voltage drop across the test source.

13. Dec 8, 2010

James889

So,

If you imagine the voltage source as a short, and a 1A current source applied to the two terminals.

You'd get

$$\frac{6}{6+3} \cdot 1 = 0.667A$$

$$0.667 \cdot 3 = 2$$

So $$V_x = 20 V$$

2 + 18(from the Voltage source) ?

14. Dec 8, 2010

vela

Staff Emeritus
You removed the 18-V source and replaced it with a short. It's not in the circuit anymore, so Vx is simply 2 V. (And you wouldn't just arbitrarily add 18 V in anyway, even if the source was still there.)

15. Dec 9, 2010

James889

Why not?, 18V appear across the 3 ohm resistor ?

16. Dec 9, 2010

vela

Staff Emeritus
1. You removed the 18-V source from the circuit.
2. Even if it were still in the circult, the entire 18 V wouldn't appear across just that one element.