CIRCUIT ANALYSIS: FInd the Thevenin equivalent of the circuit. 2 res - 1 IVS - 1 VCVS

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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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Answers and Replies

  • #2
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Does the [itex]R_{TH}[/itex] look correct? Shouldn't [itex]V_1[/itex] and [itex]V_{TH}[/itex] be equal?
 
  • #3
CEL
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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]
 
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