What is the Thevenin Equivalent for a Circuit with a Voltage-Controlled Source?

In summary, Kirchoff's law states that the voltage across an open circuit is the sum of the voltages across the individual components. Thevenin's equivalent circuit uses Kirchoff's law to find the voltage across a node between two resistors. However, when only one resistor is present, Thevenin's equivalent circuit becomes complicated. Reducing the circuit to a single loop has an effect on the controlled voltage source, which is not clear from the text.
  • #1
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Homework Statement



I am being asked to draw the Thevenin equivalent for
1pduh1.jpg


and to calculate the voltage that would appear across RL if it were connected across the 2 output terminals.

Using Kirchoff's law I found that the voltage across RL is -12.5 V. However I am really suppose to do it using Thevenin but I can't figure out how to find the equivalent.

I found the voltage at the node between the two resistors to be 25 V because I1+I2=0, so (50-V1)/5+ (-V1)/5=0, which means V1 = 25V, and I1=5A. Then I found the voltage across the output terminals to be 50V-5Ax5ohm-5Ax10=-25V.
This would mean that Veq=25V...which I believe is wrong, and I also found Req to be 5ohm...which is probably also wrong...

Can someone help please? :bugeye:

Thanks :)
 
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  • #2
An easy way to do this is to notice that when nothing is connected between the ou tput terminals, the only closed circuit consists of the 50V battery and the two resistors. The voltage drop across the first is logically 25V, so the top terminal must be 15V.

As for the equivalent resistance, when all batteries are shorted out, the two resistors are arranged in parallel.
 
  • #3
You mean 25V for the top terminal right? And, the two 5 resistors are in parallel when the battery is shorted?

What is the effect of reducing the circuit to one loop on the controlled voltage source, which is affected by I1? I can't make sense of this...
 
  • #4
Isn't there another voltage drop when you pass the current controlled voltage source, so the voltage at the top terminal would be 25V-10I1? I calculated I1 to be 5A, so that means the voltage would be -25V at the top terminal...I think I am doing something wrong :S...and the equivalent resistance I calculate to be 2.5 ohm, I think that is right, but I am confused about how to go about the other stuff.
 

Related to What is the Thevenin Equivalent for a Circuit with a Voltage-Controlled Source?

What is the Thevenin Equivalent Problem?

The Thevenin Equivalent Problem is a method used to simplify complex electrical circuits into a single equivalent circuit consisting of a voltage source, a resistance, and a load. This simplification allows for easier analysis and understanding of the circuit's behavior.

How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, you need to follow these steps:

  1. Remove the load from the circuit and note the open-circuit voltage (Voc).
  2. Calculate the equivalent resistance (Req) of the circuit by shorting all voltage sources and opening all current sources.
  3. Draw the Thevenin equivalent circuit with the voltage source (Voc) in series with the equivalent resistance (Req).

What is the purpose of the Thevenin equivalent circuit?

The Thevenin equivalent circuit is used to simplify complex circuits into a single equivalent circuit that has the same behavior as the original circuit. This simplification allows for easier analysis and calculation of the circuit's output with different loads.

Can the Thevenin equivalent circuit be used for any circuit?

The Thevenin equivalent circuit can only be used for linear circuits, which means that the relationship between voltage and current in the circuit is proportional. Non-linear circuits, such as those with diodes or transistors, cannot be simplified using the Thevenin equivalent method.

What are the limitations of the Thevenin equivalent circuit?

The Thevenin equivalent circuit is limited in its application to linear circuits. It also assumes that the circuit is in a steady-state, which means there are no time-varying signals. Additionally, the Thevenin equivalent circuit is only valid for a specific load, and its accuracy decreases for different loads.

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