Finding Thevenin Equivalent at Terminals a-b

[VinnyCee]

Homework Statement

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

http://img153.imageshack.us/img153/6116/chapter4problem40ef4.jpg

Homework Equations

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

The Attempt at a Solution

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

http://img156.imageshack.us/img156/1597/chapter4problem40part2ne1.jpg

V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V <----- Right?

For KVL Loop 1)

10000\,I_1\,+\,(1\,V)\,=\,0

I_1\,=\,-\frac{1}{10000}\,A\,=\,-0.0001\,AFor KVL Loop 2)

(-1\,V)\,+\,20000\,I_2\,+\,4\,V_0\,=\,0

I_2\,=\,\frac{5}{20000}\,A\,=\,0.00025\,A

KCL at node between two resistors)

I_1\,+\,I_{OC}\,=\,I_2\,\,\longrightarrow\,\,I_{OC}\,=\,I_2\,-\,I_1

I_{OC}\,=\,(0.00025\,A)\,-\,(-0.0001\,A)\,=\,0.00035\,A

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

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

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

V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0 <----- Right?

V_2\,=\,4\,V_0\,=\,4\,(70\,-\,V_1)\,=\,280\,-\,4\,V_1

KVL Loop 1)

(-70\,V)\,+\,V_0\,+\,V_{TH}\,=\,0

V_1\,-\,V_{TH}\,=\,70

KVL Loop 2)

-V_{TH}\,+\,V_1\,+\,V_2\,=\,0

3\,V_1\,+\,V_{TH}\,=\,280

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

 

[VinnyCee]

Does the R_{TH} look correct? Shouldn't V_1 and V_{TH} be equal?

 

[CEL]

Homework Statement

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

http://img153.imageshack.us/img153/6116/chapter4problem40ef4.jpg

Homework Equations

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

The Attempt at a Solution

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

http://img156.imageshack.us/img156/1597/chapter4problem40part2ne1.jpg

V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V <----- Right?

For KVL Loop 1)

10000\,I_1\,+\,(1\,V)\,=\,0

I_1\,=\,-\frac{1}{10000}\,A\,=\,-0.0001\,A


For KVL Loop 2)

(-1\,V)\,+\,20000\,I_2\,+\,4\,V_0\,=\,0

I_2\,=\,\frac{5}{20000}\,A\,=\,0.00025\,A

KCL at node between two resistors)

I_1\,+\,I_{OC}\,=\,I_2\,\,\longrightarrow\,\,I_{OC}\,=\,I_2\,-\,I_1

I_{OC}\,=\,(0.00025\,A)\,-\,(-0.0001\,A)\,=\,0.00035\,A

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

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

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

V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0 <----- Right?

V_2\,=\,4\,V_0\,=\,4\,(70\,-\,V_1)\,=\,280\,-\,4\,V_1

KVL Loop 1)

(-70\,V)\,+\,V_0\,+\,V_{TH}\,=\,0

V_1\,-\,V_{TH}\,=\,70

KVL Loop 2)

-V_{TH}\,+\,V_1\,+\,V_2\,=\,0

3\,V_1\,+\,V_{TH}\,=\,280

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

In the equation for loop 1 you have made a wrong substitution. You should have
V_1\,-\,V_{TH}\,=\,0
instead of
V_1\,-\,V_{TH}\,=\,70
Of course this leads to V_{TH}=V_1