# Homework Help: Using Thevenins therom to solve a simple circuits

1. Sep 7, 2011

### Apple&Orange

Using Thevenins therom to solve a "simple" circuits

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

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=$\frac{1}{4}$+$\frac{1}{2}$=$\frac{3}{4}$=$\frac{4}{3}$. 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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2. Sep 7, 2011

### Staff: Mentor

Re: Using Thevenins therom to solve a "simple" circuits

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.

3. Sep 8, 2011

### Apple&Orange

Re: Using Thevenins therom to solve a "simple" circuits

Got it, thanks a lot!!