Finding the Thevenin voltage for the circuit.

In summary, Johnsy1312 found the thevenin's resistance to be 2.19 ohms, and attempted to find the thevenin's voltage by isolating each voltage terminal one at a time and adding them together. However, he didn't answer correctly. He showed his calculations in more detail and arrived at his formula for Vth by first calculating the voltage across the 4.91 Ω resistor and then isolating the 18 V terminal. He noted that the polarity of the currents and voltages affects the polarity of the voltage created by a potential drop across a resistor.
  • #1
johnsy1312
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0
1. For the following circuit, i must find the thevenin's resistance and voltage.
Using R_th = 1/r1 +1/r2 may be effective in finding resistance and V_th = Vout*(R/Rt) may be effective in finding voltage
i have managed to find the thevenin's resistance to be 2.19ohms. I attempted to find the thevenin's voltage by isolating each voltage terminal one at a time to find the voltage and then adding these two voltages together, but i didn't answer correctly.
 

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  • #2
Hi johnsy1312, Welcome to Physics Forums.

In future please retain and use the formatting template provided when starting a new thread.

johnsy1312 said:
1. For the following circuit, i must find the thevenin's resistance and voltage.



Using R_th = 1/r1 +1/r2 may be effective in finding resistance and V_th = Vout*(R/Rt) may be effective in finding voltage



i have managed to find the thevenin's resistance to be 2.19ohms. I attempted to find the thevenin's voltage by isolating each voltage terminal one at a time to find the voltage and then adding these two voltages together, but i didn't answer correctly.

Can you show your calculations in more detail? Your formula for the Thevenin resistance is not quite right (should be ##1/R_{th} = 1/r1 + 1/r2##) but it looks like you made the correct calculation anyway. Be careful of rounding though... it looks to me like you rounded up the final digit for no reason.

How did you arrive at your formula for Vth? Show your calculations.
 
  • #3
[itex]\frac{}{}[/itex]Firstly i isolated the 12V terminal and calculated the voltage:
[itex]V= \frac{18*3}{3+8}=4.91V[/itex]
Then i isolated the 18V terminal,
[itex]V= \frac{12*8}{3+8}=3.27V[/itex]
V_th = 4.91 + 3.27 = 8.18V
 
  • #4
gneill said:
Hi johnsy1312, Welcome to Physics Forums.

In future please retain and use the formatting template provided when starting a new thread.



Can you show your calculations in more detail? Your formula for the Thevenin resistance is not quite right (should be ##1/R_{th} = 1/r1 + 1/r2##) but it looks like you made the correct calculation anyway. Be careful of rounding though... it looks to me like you rounded up the final digit for no reason.

How did you arrive at your formula for Vth? Show your calculations.


[itex]\frac{}{}[/itex]Firstly i isolated the 12V terminal and calculated the voltage:
[itex]V= \frac{18*3}{3+8}=4.91V[/itex]
Then i isolated the 18V terminal,
[itex]V= \frac{12*8}{3+8}=3.27V[/itex]
V_th = 4.91 + 3.27 = 8.18V
 
  • #5
johnsy1312 said:
[itex]\frac{}{}[/itex]Firstly i isolated the 12V terminal and calculated the voltage:
[itex]V= \frac{18*3}{3+8}=4.91V[/itex]
Then i isolated the 18V terminal,
[itex]V= \frac{12*8}{3+8}=3.27V[/itex]
V_th = 4.91 + 3.27 = 8.18V

First, when you say isolate a terminal, I presume you mean suppress a voltage source.

Second, check your voltage divider equations; the resistance that appears in the numerator should be the one that you want to find the potential across. For example, with the 12 V source suppressed and the 18 V source driving the circuit, that would be the 8 Ω resistor.

Third, pay close attention to the polarities of the currents and voltages. The polarity of the voltage created by a potential drop across a resistor depends upon the current direction, which in turn depends upon the polarity of the source voltage.

attachment.php?attachmentid=68013&stc=1&d=1395828469.gif
 

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FAQ: Finding the Thevenin voltage for the circuit.

1. What is Thevenin voltage and why is it important in circuit analysis?

The Thevenin voltage is the equivalent voltage of a complex circuit at a particular node. It is important in circuit analysis because it simplifies the circuit to a single voltage source, making it easier to calculate current and voltage at various points.

2. How do I find the Thevenin voltage for a circuit?

To find the Thevenin voltage, you need to remove all sources (voltage and current) from the circuit and calculate the voltage at the desired node using the following steps:

  1. Choose the node where you want to calculate the Thevenin voltage.
  2. Remove all sources from the circuit.
  3. Calculate the equivalent resistance at the chosen node.
  4. Reconnect the sources and calculate the voltage at the chosen node.

3. Can I use Thevenin's theorem to analyze any circuit?

Thevenin's theorem can be used to analyze linear circuits, or circuits that follow Ohm's law. This means that the components in the circuit have a linear relationship between voltage and current. Non-linear components, such as diodes, cannot be analyzed using Thevenin's theorem.

4. How accurate is Thevenin's theorem in predicting circuit behavior?

Thevenin's theorem is an approximation method and may not always accurately predict the behavior of a complex circuit. It assumes that the circuit is linear and that the components behave predictably. In real-world circuits, there may be factors such as temperature and impedance that can affect the accuracy of the theorem.

5. Can I use Thevenin's theorem to calculate the voltage at any point in the circuit?

Thevenin's theorem can only be used to calculate the voltage at the chosen node. It cannot be used to calculate the voltage at any other point in the circuit. To calculate voltage at other points, you would need to use other methods such as Kirchhoff's laws or nodal analysis.

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