Find Thevenin Equivalent Resistance for Network

Click For Summary
To find the Thevenin Equivalent resistance, the voltage sources were short-circuited, and the current source created an open loop. The initial conclusion was that resistors R1 and R2 were in parallel, leading to a calculated resistance of 2.7K ohms, which was incorrect. The correct approach involves recognizing that there are five resistors in the circuit, with only two in parallel after removing the sources. The accurate calculation for the resistance across the terminals of Vo is approximately 2.74K ohms. Proper identification of all resistors and their configurations is essential for accurate results.
qwerty2x
Messages
15
Reaction score
0

Homework Statement




Find the Thevenin Equivalent resistance for the network below:
http://img27.imageshack.us/img27/8508/circuit.png


Homework Equations



R=(1/R1+1/R2)^-1


The Attempt at a Solution



Hi all,

to find the resistance i short circuited the volage sources and created an open loop across the current source. Then it looks like the current will not travel through the uppr loop beacuse its in parallel with a short circuit(where the current will rather go). I concluded that by breaking the circuit at Vo we have R1 and R2 in parallel. Which leaves us with the resistance being 2.7K ohm. but that's incorrect according to my book.

Help appreciated thnks
 
Last edited by a moderator:
Physics news on Phys.org
You removed the sources in the correct way, but I'll need to see your calculations if I am going to be able to see your mistake. Can you show your calculations? Also, what do you mean by R_1 and R_2? There are five resistors, not just two.
 
Last edited:
thanks for the reply,

resistance=((1/8.3k)+(1/4.1k))^-1 (just 2 resistors in parallel)
so resistance is 2.74kOhms
 
The 2 resistors in the top half of the circuit are not in parallel (after removing the current source and shorting the voltage sources).

Note, the resistance is to be calculated across the + and - terminals of Vo.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K