Calculating Rth error?

1. Dec 16, 2014

GBA13

1. The problem statement, all variables and given/known data
Hi Guys,

I am doing a number of questions on finding Rth and can across this example. When I looked at the answers I had a different one to them but I actually think they may be wrong.

Could you guys please take a look?

2. Relevant equations

3. The attempt at a solution
So they have Rth found (once voltages are short circuited) as Rth = 8*2/8+2 = 1.6 Ohms but that implies that one of the 2ohm resistors is it series with the 6 ohm resistor but I don't think it is. I think that both 2ohm resistors are in series and then those are in parallel with the 6 ohm so Rth is 2.4 ohms.
I may be wrong but what do you guys think?

Thanks

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2. Dec 16, 2014

Staff: Mentor

As you say, the text solution is incorrect. But your result is also incorrect, likely due to a calculation issue. Check your calculation.

3. Dec 16, 2014

Staff: Mentor

What is Rth?
That would be $\frac{8\cdot2}{8}+2$, I guess you mean 8*2/(8+2). Where does that come from?
Right.
Only if you remove all power sources, but then the network becomes pointless.

4. Dec 16, 2014

Staff: Mentor

GBA13 is looking for the Thevenin resistance of the network as seen at the open terminals ( $V_{th}$ ).