Using Thevenins therom to solve a simple circuits

  • Context: Engineering 
  • Thread starter Thread starter Apple&Orange
  • Start date Start date
  • Tags Tags
    Circuits
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
Apple&Orange
Messages
28
Reaction score
2
Using Thevenins therom to solve a "simple" circuits

Homework Statement



The problems I am stuck on is problem two, figure 1 and figure 2. It's attached to this post.

Homework Equations





The Attempt at a Solution



My initial attempt at solving for Rthev for figure 1 was that Rthev=[itex]\frac{1}{4}[/itex]+[itex]\frac{1}{14}[/itex] = [itex]\frac{9}{28}[/itex] = [itex]\frac{28}{9}[/itex].

However, on the answers it says that the 12K is shorted out since the source is connected at the same points, so it became Rthev=[itex]\frac{1}{4}[/itex]+[itex]\frac{1}{2}[/itex]=[itex]\frac{3}{4}[/itex]=[itex]\frac{4}{3}[/itex]. At first this made sense, but when I tried doing the second question, the 10K wasn't shorted out when solving for the Rthev.

Could someone please clarify why the 12K was shorted, but the 10K wasn't?
 

Attachments

Physics news on Phys.org


Apple&Orange said:
Could someone please clarify why the 12K was shorted, but the 10K wasn't?

In determining the Thevenin equivalent circuit,

(i) voltage sources are replaced by their internal series resistance, which for an ideal voltage source is zero ohms (i.e., a short circuit).

(ii) current sources are replaced by their internal series resistance, which for an ideal current source is infinite ohms (i.e., an open circuit).

Fig 2 involves a constant current source.
 


Got it, thanks a lot!