# Homework Help: Thevenin Resistance - Is this a mistake in the book?

1. Apr 11, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
http://img818.imageshack.us/img818/4065/thvs.jpg [Broken]

3. The attempt at a solution

Calculating the Thevenin Resistance there I get 16Ω - the book however says it's 3Ω.

The way I see 3Ω being calculated is if 4Ω were in parallel with 6Ω + 6Ω - however they all appear to be in series. I'm assuming the 1Ω resistor is replaced by an open-circuit?

Last edited by a moderator: May 5, 2017
2. Apr 11, 2012

### Staff: Mentor

Suppressing the sources, the voltage source is replaced by a short and the current source is removed. How's it look now?

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• ###### Fig1.gif
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3. Apr 11, 2012

### NewtonianAlch

I'm still not seeing how that gives 3 ohms unless I take 12 in parallel with 4, ignoring 1 ohm resistor.

Since the 1 ohm resistor is the load, I thought we remove the load when calculating Thevenin Resistance?

4. Apr 11, 2012

### Staff: Mentor

Yes, remove the load and "look into" the circuit at the terminals where it was connected. Clearly the 6+6 = 12 Ohms is in parallel with the 4 Ohm resistor, no?

5. Apr 11, 2012

### NewtonianAlch

Ohhhhh...

I didn't keep the ends of the terminal when I first drew it, I just ignored it hence why I just added it in series, but now I see...thanks!

6. Apr 11, 2012

### Femme_physics

Isn't this entire circuit shortcircuited, anyway? Where it says 3A.. that's the only loop where current will flow through.

7. Apr 11, 2012

### Staff: Mentor

No, a current supply does not mean short circuit; A current supply will produce any potential difference across itself that is required to supply the mandated current value. The 18V supply will also contribute current to the circuit.

8. Apr 11, 2012

### Femme_physics

Fair enough, but this seems very unrealistic to me. In reality what I said applies. Correct me if I'm wrong.

9. Apr 11, 2012

### Staff: Mentor

You are wrong

You can prove it to yourself by solving the circuit equations to see how much current the voltage supply will provide.

10. Apr 11, 2012

### Femme_physics

Actually whenever I see a shortcircuited path there's no point in applying the same old principles because a shortcircuit I figured changes everything-- you just look at the current's clear path. Why would a current wanna flow through resistance if it has a clear path?

Also they appear to have confused about the way current flow where they marked 3A. This exercise is all messed up!

11. Apr 11, 2012

### Staff: Mentor

http://img140.imageshack.us/img140/9826/t2622.gif [Broken]
Current will flow in both loops. I'm puzzled how you could conclude it wouldn't? Only were the potential across the current source equal to +18V could there be no current through that 6Ω and its voltage source.

Last edited by a moderator: May 5, 2017
12. Apr 11, 2012

### Staff: Mentor

That 3A represents an independent current source, it's not an arrow marked on a short-circuit in the schematichttp://img851.imageshack.us/img851/3541/iconexclaim.gif [Broken]

Last edited by a moderator: May 5, 2017
13. Apr 11, 2012

### Staff: Mentor

Perhaps you are mistaking the "3A" for a meter reading through a wire rather than as the fixed current source that it represents?

14. Apr 11, 2012

### Femme_physics

In other words this mini-circuit can be in fact a part of a much bigger circuit?

15. Apr 11, 2012

### NewtonianAlch

Even if it were a short-circuit path, in real life, some current is still going to flow through the other resistors.

I don't see how it's being viewed as a short-circuit anyway, it's just a source, like the voltage source, except it's current...

16. Apr 12, 2012

### Femme_physics

Yea but purely from a student solving exercises perspective?

It just defies the principles I studied which says that I can neglect the unnecessary resistors if I have a shortcircuit. I just look at the clear, free-of-resistor laden path.

I get that in reality a really minimal amount of current still flows via the resistors, but it's negligible I reckon.

17. Apr 12, 2012

### NewtonianAlch

That's true, for student exercises for the most part we consider short-circuits as ideal ones. But it's a current source that's supplying current, not a short-circuit.

18. Apr 12, 2012

### Femme_physics

Ahh....I see. Current source makes all the difference. I guess we didn't study the difference between current source and voltage source in my college. Thank you, though..