Using Thevenins therom to solve a simple circuits

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In summary, the conversation discusses the use of Thevenin's theorem to solve simple circuits. The specific problem being discussed involves problem two and figures 1 and 2, attached to the post. The attempt at solving for Rthev in figure 1 resulted in confusion due to the discrepancy between the 12K and 10K resistors being shorted out. The explanation for this is that in determining the Thevenin equivalent circuit, voltage sources are replaced by their internal series resistance (zero ohms) and current sources are replaced by their internal series resistance (infinite ohms). Figure 2 involves a constant current source.
  • #1
Apple&Orange
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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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  • #2


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.
 
  • #3


Got it, thanks a lot!
 

FAQ: Using Thevenins therom to solve a simple circuits

What is Thevenin's theorem and how is it used to solve circuits?

Thevenin's theorem states that any linear circuit can be simplified into an equivalent circuit with just a single voltage source and a single resistance in series. This makes it easier to analyze and solve complex circuits.

How do I determine the Thevenin voltage and resistance in a circuit?

The Thevenin voltage is equal to the open circuit voltage across the two terminals of the circuit, while the Thevenin resistance is equal to the equivalent resistance seen from those terminals when all independent sources are turned off.

3. Can Thevenin's theorem be used for non-linear circuits?

No, Thevenin's theorem can only be applied to linear circuits. Non-linear elements such as diodes, transistors, and capacitors cannot be simplified using this theorem.

4. What are the advantages of using Thevenin's theorem to solve circuits?

Thevenin's theorem allows for the simplification of complex circuits, making it easier to analyze and solve. It also helps in determining the maximum power transfer and the effects of loading on the circuit.

5. Are there any limitations to using Thevenin's theorem?

Thevenin's theorem is only applicable to linear circuits and cannot be used for non-linear circuits. It also assumes that the circuit is in a steady state and does not take into account the effects of time-varying signals.

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