Circuit Simplification concept

  • Thread starter JieXian
  • Start date
  • #1
6
0
Hi, I have a question relating to the image below.

http://img830.imageshack.us/img830/2173/elec.png/ [Broken]


From step 2 to 3, I get how 120V becomes 20mA but how is the 6k ohms affected? Why is it reduced to 3k ohms?

And from step 1 to step 2, why isn't it like below? Case 2b instead of 2a?

http://img835.imageshack.us/img835/9250/elech.png [Broken]


I arrived at the answer using a different technique but I would like to understand this one too.

Thank you very much.
 
Last edited by a moderator:

Answers and Replies

  • #2
gneill
Mentor
20,936
2,877
In the first picture, between steps 2 and 3 the 120V source and its series 4k+2k Ohms are converted to their Norton equivalent: a 20mA source in parallel with 6k Ohms. Since all the components are in parallel, this 6k resistor can be combined with the existing 6k resistor at the left end of the circuit.

In the second drawing, the portion of the circuit for which the Thevenin equivalent is being found is the current source and everything to its right. The equivalent circuit (a voltage source in series with a 4k resistor) must be "tacked back on" at the same connection point.

Your figure 2b would put the Thevenin resistance of 4k in parallel with the voltage source. That is no what a Thevenin equivalent looks like; it must be in series.
 
  • #3
6
0
I've never done a "partial" Thevenin or Norton (most of them are reduced to a source, an equivalent resistor and an open circuit. But now that I know that it's those 2 theorems I think I'll get it.

Thank you.
 

Related Threads on Circuit Simplification concept

  • Last Post
Replies
1
Views
5K
Replies
9
Views
5K
  • Last Post
Replies
9
Views
821
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
645
  • Last Post
Replies
2
Views
1K
Top