Thevenin Theorem Help (Existence of Dependent Source) X.X

What I did was to short-circuit the 20V, but then I was told I couldn't do that due to the existence of a dependent source.
Hence how should I proceed from here?

To find Thevenin's equivalent at terminals a & b you need to find

1. The open circuit voltage Vab = Vth.
2. The current flowing from a to b when a & b are shorted together.
3. Hence find Rth = Vab(open circuit)/Iab(short circuit)

To find the open circuit Vab [Vth], apply Kirchoff's voltage law in the left hand loop to find the current and then Vab = 6*i.

To find the short circuit current between a & b apply Kirchoff's rules for the two-loop network formed when a and b are shorted.

If you're not sure how to start with the first part

from KVL

20 = 6*i + 2*i + 6*i

Hence i = ?

Then Vab(open circuit) = Vth = 6*i

There's a slightly more "direct" way to get Vab(open circuit) if you can see the following works.

Remember Vab = 6*i, and thus i=Vab/6

So using KVL

20 = 6*(Vab/6) + 2*(Vab/6) +6*(Vab/6)

Gives Vab = Vth directly after some re-arranging.

You should try to find Iab (short circuit) and then Rth yourself.

t_n_k said:

from KVL

20 = 6*i + 2*i + 6*i

Hence i = ?

Click to expand...

Shouldn't it be:

20 = 6*i - 2*i + 6*i

and also:

20 = 6*(Vab/6) - 2*(Vab/6) +6*(Vab/6)

Sorry - my mistake.

Thanks "electrician" - well spotted as usual!

When the day comes that I don't make mistakes like that myself, we'll know that the Millenium has arrived!

Both good hints guys. It will be interesting to see what xiahbaby can do with the information.

hgmjr

Why consider the Vth equal to the voltage across the 6 ohm resistor when there is the existence of the 10 ohm resistor?

Given that the 10 ohm resistor is omitted because it is connected to an open circuit, then Voltage at a and b should also be omitted isn't? confusing.

thanks friends!

jake_up@100 said:

Why consider the Vth equal to the voltage across the 6 ohm resistor when there is the existence of the 10 ohm resistor?

Given that the 10 ohm resistor is omitted because it is connected to an open circuit, then Voltage at a and b should also be omitted isn't? confusing.

thanks friends!

Click to expand...

The voltage drop across a resistor is given by V=IR. For the 10 ohm resistor there is no current flowing across it (on one end is an open circuit), so there can be no voltage drop. That means that the voltage on one side of the resistor is the same as on the other side, hence the same as node c.

how to get Isc ????????? please help me

You are resurrecting a very old thread and hijacking it. This is considered poor behavior.

You essentially have the same thread going here.

http://forum.allaboutcircuits.com/showthread.php?t=78840

Let's stick with that one and let this one return to the dust bin.