Calculating Rth error?

  • Thread starter GBA13
  • Start date
  • #1
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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
 

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Answers and Replies

  • #2
gneill
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As you say, the text solution is incorrect. But your result is also incorrect, likely due to a calculation issue. Check your calculation.
 
  • #3
35,908
12,733
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.
 
  • #4
gneill
Mentor
20,945
2,886
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}## ).
 

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