Thevenin Resistance - Is this a mistake in the book?

[NewtonianAlch]

Homework Statement

http://img818.imageshack.us/img818/4065/thvs.jpg

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?

 

[gneill]

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

 

[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?

 

[gneill]

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?

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?

 

[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!

 

[Femme_physics]

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

 

[gneill]

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

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.

 

[Femme_physics]

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

 

[gneill]

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

You are wrong

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

 

[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 want to 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!

 

[NascentOxygen]

Fair enough, but this seems very unrealistic to me. In reality what I said applies.

http://img140.imageshack.us/img140/9826/t2622.gif

Correct me if I'm wrong.

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.

 

[NascentOxygen]

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

 

[gneill]

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 want to 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!

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

 

[Femme_physics]

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

 

[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...

 

[Femme_physics]

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

Yea but purely from a student solving exercises perspective?

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...

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.

 

[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.

 

[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..