Using Thevenins therom to solve a simple circuits

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SUMMARY

The discussion focuses on applying Thevenin's theorem to solve simple circuit problems, specifically addressing the confusion around shorting resistors in different scenarios. The user initially calculated the Thevenin resistance (Rthev) for figure 1 as Rthev=9/28 but later learned that the 12K resistor was shorted due to the voltage source connections, leading to a corrected Rthev of 4/3. The user sought clarification on why the 10K resistor was not shorted in the second problem, which involves a constant current source.

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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?
 

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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!
 

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