# Homework Help: Thevenin with dependant source

1. May 11, 2010

### James889

Hello,

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

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

Last edited by a moderator: May 4, 2017
2. May 11, 2010

### xcvxcvvc

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

3. May 11, 2010

### CEL

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

Last edited by a moderator: May 4, 2017
4. May 13, 2010

### James889

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?

5. May 13, 2010

### CEL

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

6. May 13, 2010

### 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?

7. May 13, 2010

### vela

Staff Emeritus
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.

8. May 13, 2010

### James889

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

9. May 13, 2010

### vela

Staff Emeritus
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.

10. May 13, 2010

### James889

ok, thanks for clearing that up.

With a short circuit the total resistance is

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

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

11. May 13, 2010

### vela

Staff Emeritus
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.

12. May 13, 2010

### James889

Hi,

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

[PLAIN]http://img413.imageshack.us/img413/3554/theveningdependantsrctr.png [Broken]

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

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

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

Last edited by a moderator: May 4, 2017
13. May 13, 2010

### vela

Staff Emeritus

Last edited by a moderator: May 4, 2017
14. May 13, 2010

### The Electrician

15. May 13, 2010

### vela

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

16. May 15, 2010

### James889

Is it really?

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

17. May 15, 2010

### vela

Staff Emeritus
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.

18. May 15, 2010

### 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?

19. May 15, 2010

### vela

Staff Emeritus
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.

20. May 15, 2010

### James889

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

Last edited by a moderator: May 4, 2017
21. May 15, 2010

### vela

Staff Emeritus
You got it.

22. May 15, 2010

### James889

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