Calculating Rth error?

[GBA13]

Homework Statement

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?

Homework Equations

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

 

[gneill]

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

 

[mfb]

What is Rth?

8*2/8+2

That would be $\frac{8\cdot2}{8}+2$, I guess you mean 8*2/(8+2). Where does that come from?

I think that both 2ohm resistors are in series

Right.

and then those are in parallel with the 6 ohm

Only if you remove all power sources, but then the network becomes pointless.

 

[gneill]

Right.Only if you remove all power sources, but then the network becomes pointless.

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