How to Calculate Thevenin Equivalent with a Dependent Source?

[James889]

Hello,

I have the following circuit
[PLAIN]http://img143.imageshack.us/img143/2213/theveningdependantsrc.png

I need to calculate the Thevenin equivalent circuit. I don't really know
where to start. Can i zero the independent source and calculate the resistance?
Or do you start by connecting a test source across a and b.

//James

 

[xcvxcvvc]

I don't think you can. I would find the Thevenin voltage and Thevenin current. Their ratio equals the Thevenin resistance.

 

[CEL]

Hello,

I have the following circuit
[PLAIN]http://img143.imageshack.us/img143/2213/theveningdependantsrc.png

I need to calculate the Thevenin equivalent circuit. I don't really know
where to start. Can i zero the independent source and calculate the resistance?
Or do you start by connecting a test source across a and b.

//James

To calculate the resistance you must zero the independent source AND connect a test source between a and b.
Alternatively, you can calculate the open circuit voltage Voc and the short circuit current Isc.
The equivalent resistance is the quotient of Voc and Isc

 

[James889]

To calculate the resistance you must zero the independent source AND connect a test source between a and b.
Alternatively, you can calculate the open circuit voltage Voc and the short circuit current Isc.
The equivalent resistance is the quotient of Voc and Isc

Hi, thanks for your reply.

I decided to try the latter..

So the current divides at the connection between the 10ohm and the 5 ohm resistor.

$$\frac{10}{10+10}\cdot 10 = 5A$$

So 5A is the current flowing across both the 10ohm and the two 5 ohm resistors.

Is $$V_{oc}$$ equal to sum of the voltages across them?

 

[CEL]

Hi, thanks for your reply.

I decided to try the latter..

So the current divides at the connection between the 10ohm and the 5 ohm resistor.

$$\frac{10}{10+10}\cdot 10 = 5A$$

So 5A is the current flowing across both the 10ohm and the two 5 ohm resistors.

Is $$V_{oc}$$ equal to sum of the voltages across them?

No, Voc is the sum of the voltage across the second 5 ohm resistor and the voltage of the controlled source.

 

[James889]

So the short-circuit current must be 5A, as no current flows trough the 10 ohm resistor.
The voltage across the 5 ohm resistor is 25V, and the voltage given by the dependent voltage source must therefore be 5*K V = 25V. So $$V_{oc} = 50V$$

$$R_{th}~~ = \frac{50}{5} = 10\ohm$$

Yes?

 

[vela]

Your short-circuit current is wrong because your assumption that no current flows through the 10-ohm resistor is wrong. I'm not sure why you think no current flows through the 10-ohm resistor.

 

[James889]

Your short-circuit current is wrong because your assumption that no current flows through the 10-ohm resistor is wrong. I'm not sure why you think no current flows through the 10-ohm resistor.

Because the current takes the easiest possible way, around the periphery

 

[vela]

Using that argument, you'd conclude that when calculating the open-circuit voltage, all the current goes through the 5-ohm resistor and none through the 10-ohm resistor, which is also wrong. It's not an all-or-nothing situation. If it's easier to go around the periphery, then more current will go though that path than through the 10-ohm resistor, but both paths will still have some current flowing through it. There'd be no current in the 10-ohm resistor only if you placed a short in parallel with it, but that's not what you're doing here.

 

[James889]

ok, thanks for clearing that up.

With a short circuit the total resistance is

$$R_{tot}~=~\frac{1}{(1/10)+(1/10)+(1/5)} = 2.5\ohm$$

$$I_{sc} = \frac{5}{5+2.5} \cdot 10 = 6.67A$$

 

[vela]

No, that's not right. The dependent source complicates things, and I think you made a mistake analyzing the remaining circuit as well.

What I would do is first replace the current source and 10-ohm resistor by its Thevenin equivalent, and then write two loop equations for the circuit and apply Kirchoff's current law. You'll have three equations and three unknowns, which are straightforward to solve.

 

[James889]

Hi,

I made a new drawing just to make it easier to visualize.

[PLAIN]http://img413.imageshack.us/img413/3554/theveningdependantsrctr.png $$\begin{array}~~ 15I_1 + 5(I_2-I_1)-100 = 0 \\ -5(I_1-I_2) + 5(I_2-I_1) + 5I_2 = 0\end{array}$$

Then the current $$I_x$$ can be expressed as i2-i1

For V1: $$\frac{100}{15}-\frac{2v1}{5} = 0$$

 

[vela]

Hi,

I made a new drawing just to make it easier to visualize.

[PLAIN]http://img413.imageshack.us/img413/3554/theveningdependantsrctr.png[/quote]
Always a good idea.

$$\begin{array}~~ 15I_1 + 5(I_2-I_1)-100 = 0 \\ -5(I_1-I_2) + 5(I_2-I_1) + 5I_2 = 0\end{array}$$

Some of the signs are wrong. In the first equation, the plus sign should be a minus sign, [strike]and in the second equation, the 5I₂ term should have a minus sign[/strike].

Then the current $$I_x$$ can be expressed as i2-i1

For V1: $$\frac{100}{15}-\frac{2v1}{5} = 0$$

I'm not sure what you're doing here, but do you even need V₁? You're just trying to find I_sc.

 

[The Electrician]

in the second equation, the 5I₂ term should have a minus sign.

Are you sure about this, vela? It looks ok to me.

 

[vela]

Yeah, you're right. The second equation is okay as is.

 

[James889]

Yeah, you're right. The second equation is okay as is.

Is it really?

Im having some second thoughts about the controlled source. It isn't common to both meshes.
Shouldn't it be just $$5i_2$$ ?

 

[vela]

Yes, your second equation is correct as you wrote it. The controlled source is defined in the problem to depend on the total current through the 5-ohm resistor common to both loops, so the dependent source will vary with both i₁ and i₂.

 

[James889]

Ok,

So i have
1.$$20I_1 = 100 +5I_2$$
2.$$15I_2-10I_1 = 0$$

$$\begin{array}{rcl}20I_1 = 100 +5I_2 \\ I_1 = 5 + 0.25I_2$$

Substituting into 2, gives:

$$\begin{array}{rcl}I_2 = 4\\ I_1 = 6\end{array}$$

So now. $$V_{oc}$$ is the voltage across the 5ohm resistor plus the voltage generated by the dependent source?

 

[vela]

Yes, when the load isn't present because it's the open-circuit voltage. You calculated it earlier to be 50 V, which was correct. Now R is the ratio of the open-circuit voltage and the short-circuit current.

 

[James889]

[PLAIN]http://img513.imageshack.us/img513/88/thevenin.png

 

[vela]

You got it.

 

[James889]

I know i am amazing.
Thanks Vela for all the help!