How Do You Calculate Thevenin Equivalent Circuits and Current Flow?

In summary, the conversation discusses finding the Thevenin equivalent circuit for an unknown circuit connected to a variable resistor. It is found that the maximum power dissipation occurs when the value of the resistor is 25 ohms, and the Thevenin equivalent circuit is a 22.4V power supply and a 25 ohm resistor. When the resistor value is 5 ohms, the current is calculated to be 0.74 Amps. However, the conversation also points out an error in the calculations, as the result for the Thevenin equivalent circuit should be a 15.81V power supply connected to a 25 ohm resistor. The conversation continues with a discussion on the calculations and clarifying the correct result
  • #1
Solidsam
23
0

Homework Statement



A variable resistor (RL) is connected to an unknown circuit. It is found that the
maximum power that can be dissipated in the resistor (RL) is 10W, and that this
occurs when the value of the resistor (RL) is 25ohms.
Derive the Thevenin equivalent circuit for the unknown circuit, and hence find the
current which flows when RL = 5ohms.

Homework Equations



Maximum power dissaption occurs when RL=RS which was 25ohm

Which means P=V^2/R= V=sqrootof P*R = sqrootof 10W*(25+25)ohms=22.4V

Equivalent circuit is a 22.4 V power supply and a 25 ohm resistor

When RL is 5 I=V/(Rs+RL)=22.4/(25+5)=0.74Amps


are my calculations correct?
 
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  • #2
Solidsam said:
Which means P=V^2/R= V=sqrootof P*R = sqrootof 10W*(25+25)ohms=22.4V

P does equal V2/R, but the 10W of power is being dissipated by RL, not RL + RS.
 
  • #3
lewando said:
P does equal V2/R, but the 10W of power is being dissipated by RL, not RL + RS.

so do i use P=(V^2/R) or P=I^2*R


if i use P=(V^2/R) i get V = sqrootof(25*10)=15.81V

if i use P=I^2*R i get I=sqrootof(10/25)=0.632A V=R*I=15.81V


Then I=V/(RL+RS)= 15.81/(25+5)=0.527A

are my calculations correct?
 
  • #4
Your calculations are numerically correct but your result is wrong. You have gotten off the track with this line:
Then I=V/(RL+RS)= 15.81/(25+5)=0.527A

First off, what did you get for the Thevenin equivalent of the unknown circuit?
 
  • #5
a 15.81 voltage power supply connected to a 25ohm resistor?
 
  • #6
If you had a current going through RL=25ohms producing a voltage of 15.81V, what voltage would be produced across RS?
 
Last edited:

1. What is a Thevenin equivalent circuit?

A Thevenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a series resistance. It behaves the same as the original circuit when connected to an external load.

2. How is a Thevenin equivalent circuit calculated?

To calculate the Thevenin equivalent circuit, we need to follow these steps:

  1. Remove the load resistor from the original circuit.
  2. Determine the open-circuit voltage (Voc) across the load resistor terminals.
  3. Calculate the equivalent resistance (Req) between the load resistor terminals.
  4. The Thevenin voltage (Vth) is equal to Voc, and the Thevenin resistance (Rth) is equal to Req.

3. What are the applications of a Thevenin equivalent circuit?

Thevenin equivalent circuits are commonly used in circuit analysis and design. They allow us to simplify complex circuits and make calculations easier. They are also useful in troubleshooting and predicting the behavior of a circuit under different conditions.

4. Can a Thevenin equivalent circuit be used to replace any complex circuit?

No, a Thevenin equivalent circuit can only be used to replace linear, passive circuits. This means that the circuit must only contain resistors, capacitors, and inductors, and it cannot have any active elements such as transistors or operational amplifiers.

5. What is the difference between a Thevenin equivalent circuit and a Norton equivalent circuit?

A Norton equivalent circuit is another simplified representation of a complex circuit, but instead of a voltage source, it has a current source in series with a resistance. The main difference between the two is the type of source used. Thevenin circuits are useful for analyzing voltage-dependent circuits, while Norton circuits are better for current-dependent circuits.

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