Finding Thevenin Equivalent at Terminals a-b

In summary: V and V_2=0V. The final Thevenin equivalent is V_{TH}=70V and R_{TH}=2857Ω In summary, the Thevenin equivalent at terminals a-b of the circuit is V_{TH}=70V and R_{TH}=2857Ω. This is found by removing the independent voltage source and adding a test voltage of 1V, an unknown test current I_{OC}, and two KVL loops. Solving for the test current, we get I_{OC}=0.00035A. Using the equation R_{TH}=\frac{V_{OC}}{I_{OC}}, we can find R_{TH} to be approximately
  • #1
VinnyCee
489
0

Homework Statement



Find the Thevenin equivalent at terminals a-b of the circuit below.

http://img153.imageshack.us/img153/6116/chapter4problem40ef4.jpg [Broken]

Homework Equations



v = i R, KCL, KVL, [itex]R_{TH}\,=\,\frac{V_{OC}}{I_{OC}}[/itex].

The Attempt at a Solution



To get [itex]R_{TH}[/itex], I removed the independent voltage source and I added a test voltage of 1V, an unknown test current [itex]I_{OC}[/itex] and two KVL loops.

http://img156.imageshack.us/img156/1597/chapter4problem40part2ne1.jpg [Broken]

[tex]V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V[/tex] <----- Right?

For KVL Loop 1)

[tex]10000\,I_1\,+\,(1\,V)\,=\,0[/tex]

[tex]I_1\,=\,-\frac{1}{10000}\,A\,=\,-0.0001\,A[/tex]For KVL Loop 2)

[tex](-1\,V)\,+\,20000\,I_2\,+\,4\,V_0\,=\,0[/tex]

[tex]I_2\,=\,\frac{5}{20000}\,A\,=\,0.00025\,A[/tex]

KCL at node between two resistors)

[tex]I_1\,+\,I_{OC}\,=\,I_2\,\,\longrightarrow\,\,I_{OC}\,=\,I_2\,-\,I_1[/tex]

[tex]I_{OC}\,=\,(0.00025\,A)\,-\,(-0.0001\,A)\,=\,0.00035\,A[/tex]

Now I use the equation mentioned above for [itex]R_{TH}[/itex] to get the Thevenin equivalent resistance)

[tex]R_{TH}\,=\,\frac{(1\,V)}{0.00035\,A}\,\approx\,2857\Omega[/tex]

Now I need to get the Thevenin equivalent voltage at the terminals a-b. I redrew the circuit to include the 70V independent voltage source and added two node voltages.

http://img249.imageshack.us/img249/6549/chapter4problem40part3rj2.jpg [Broken]

[tex]V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0[/tex] <----- Right?

[tex]V_2\,=\,4\,V_0\,=\,4\,(70\,-\,V_1)\,=\,280\,-\,4\,V_1[/tex]

KVL Loop 1)

[tex](-70\,V)\,+\,V_0\,+\,V_{TH}\,=\,0[/tex]

[tex]V_1\,-\,V_{TH}\,=\,70[/tex]

KVL Loop 2)

[tex]-V_{TH}\,+\,V_1\,+\,V_2\,=\,0[/tex]

[tex]3\,V_1\,+\,V_{TH}\,=\,280[/tex]

Now I have two equations in two variables and I can solve. I get [itex]V_1\,=\,87.5\,V[/itex] and [itex]V_{TH}\,=\,17.5\,V[/itex]. But wouldn't the Thevenin equivalent voltage include the contributions from both sources? (i.e. - [itex]V_{TH}[/itex] is actually [itex]V_1[/itex], [itex]V_{TH}\,=\,87.5\,V[/itex] and not [itex]V_{TH}\,=\,17.5\,V[/itex])
 
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  • #2
Does the [itex]R_{TH}[/itex] look correct? Shouldn't [itex]V_1[/itex] and [itex]V_{TH}[/itex] be equal?
 
  • #3


VinnyCee said:

Homework Statement



Find the Thevenin equivalent at terminals a-b of the circuit below.

http://img153.imageshack.us/img153/6116/chapter4problem40ef4.jpg [Broken]


Homework Equations



v = i R, KCL, KVL, [itex]R_{TH}\,=\,\frac{V_{OC}}{I_{OC}}[/itex].


The Attempt at a Solution



To get [itex]R_{TH}[/itex], I removed the independent voltage source and I added a test voltage of 1V, an unknown test current [itex]I_{OC}[/itex] and two KVL loops.

http://img156.imageshack.us/img156/1597/chapter4problem40part2ne1.jpg [Broken]

[tex]V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V[/tex] <----- Right?

For KVL Loop 1)

[tex]10000\,I_1\,+\,(1\,V)\,=\,0[/tex]

[tex]I_1\,=\,-\frac{1}{10000}\,A\,=\,-0.0001\,A[/tex]


For KVL Loop 2)

[tex](-1\,V)\,+\,20000\,I_2\,+\,4\,V_0\,=\,0[/tex]

[tex]I_2\,=\,\frac{5}{20000}\,A\,=\,0.00025\,A[/tex]

KCL at node between two resistors)

[tex]I_1\,+\,I_{OC}\,=\,I_2\,\,\longrightarrow\,\,I_{OC}\,=\,I_2\,-\,I_1[/tex]

[tex]I_{OC}\,=\,(0.00025\,A)\,-\,(-0.0001\,A)\,=\,0.00035\,A[/tex]

Now I use the equation mentioned above for [itex]R_{TH}[/itex] to get the Thevenin equivalent resistance)

[tex]R_{TH}\,=\,\frac{(1\,V)}{0.00035\,A}\,\approx\,2857\Omega[/tex]

Now I need to get the Thevenin equivalent voltage at the terminals a-b. I redrew the circuit to include the 70V independent voltage source and added two node voltages.

http://img249.imageshack.us/img249/6549/chapter4problem40part3rj2.jpg [Broken]

[tex]V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0[/tex] <----- Right?

[tex]V_2\,=\,4\,V_0\,=\,4\,(70\,-\,V_1)\,=\,280\,-\,4\,V_1[/tex]

KVL Loop 1)

[tex](-70\,V)\,+\,V_0\,+\,V_{TH}\,=\,0[/tex]

[tex]V_1\,-\,V_{TH}\,=\,70[/tex]

KVL Loop 2)

[tex]-V_{TH}\,+\,V_1\,+\,V_2\,=\,0[/tex]

[tex]3\,V_1\,+\,V_{TH}\,=\,280[/tex]

Now I have two equations in two variables and I can solve. I get [itex]V_1\,=\,87.5\,V[/itex] and [itex]V_{TH}\,=\,17.5\,V[/itex]. But wouldn't the Thevenin equivalent voltage include the contributions from both sources? (i.e. - [itex]V_{TH}[/itex] is actually [itex]V_1[/itex], [itex]V_{TH}\,=\,87.5\,V[/itex] and not [itex]V_{TH}\,=\,17.5\,V[/itex])

In the equation for loop 1 you have made a wrong substitution. You should have
[tex]V_1\,-\,V_{TH}\,=\,0[/tex]
instead of
[tex]V_1\,-\,V_{TH}\,=\,70[/tex]
Of course this leads to [itex]V_{TH}=V_1[/itex]
 
Last edited by a moderator:

1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a series resistor. This equivalent circuit is used to analyze the behavior of a complex circuit and make calculations easier.

2. How do you find the Thevenin equivalent at terminals a-b?

To find the Thevenin equivalent at terminals a-b, you need to follow these steps:

  1. Remove all the loads and open the terminals a-b.
  2. Calculate the open-circuit voltage (Voc) by using voltage division or any other appropriate method.
  3. Calculate the short-circuit current (Isc) by replacing all the voltage sources with a short circuit and using current division or any other appropriate method.
  4. The Thevenin equivalent voltage (Vth) is equal to the open-circuit voltage (Voc), and the Thevenin equivalent resistance (Rth) is equal to Voc/Isc.

3. Why is it important to find the Thevenin equivalent at terminals a-b?

Finding the Thevenin equivalent at terminals a-b is important because it allows us to simplify a complex circuit and analyze its behavior using only a voltage source and a series resistor. This makes calculations and circuit analysis much easier and more efficient.

4. Can the Thevenin equivalent circuit be used to replace the original circuit?

Yes, the Thevenin equivalent circuit can be used to replace the original circuit for the purpose of analysis. However, it cannot be used for actual circuit implementation as it is an approximation of the original circuit and may not accurately represent its behavior in all cases.

5. Are there any limitations to finding the Thevenin equivalent at terminals a-b?

Yes, there are some limitations to finding the Thevenin equivalent at terminals a-b. It is only applicable to linear circuits, which means that the components in the circuit must obey Ohm's law. It also assumes that the circuit is in a steady state, and does not take into account any transient effects. Additionally, it is not applicable to circuits with dependent sources or those with multiple sources that cannot be simplified into a single voltage source and series resistor.

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