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

## 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, $R_{TH}\,=\,\frac{V_{OC}}{I_{OC}}$.

## 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 [Broken]

$$V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V$$ <----- 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 [Broken]

$$V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0$$ <----- 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 $V_{TH}\,=\,17.5\,V$)

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

## 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, $R_{TH}\,=\,\frac{V_{OC}}{I_{OC}}$.

## 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 [Broken]

$$V_0\,=\,-1\,V,\,\,4\,V_0\,=\,-4\,V$$ <----- 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 [Broken]

$$V_0\,=\,70\,-\,V_1,\,\,V_2\,=\,4\,V_0$$ <----- 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
$$V_1\,-\,V_{TH}\,=\,0$$
$$V_1\,-\,V_{TH}\,=\,70$$
Of course this leads to $V_{TH}=V_1$