Thanks for the help thus far guys. I need more help on these thevenin problems, they are a real doozy. Trying to get ready for an exam next week.

Here is the problem

Now I know that you put a 1A current trough the circuit to find the Rth, but Prof keeps telling me I'm wrong. Is there something that I'm missing here.
Also what method would one go trying to find the Vth? I've tried the Mesh Method but I got 3 unknowns with 2 equations (Vz, I1, I2)

 

Using the 1A current injection method you should leave the 85V source in the circuit for analysis, rather than shorting it out as you have done.

Also, how did you come to the conclusion that the 1A injected current splits into two 0.5A amp branch currents at the node?

Vth is relatively simple to find. Notice that if the terminals a&b are left open then no current flows in the 1Ω. Hence Vz=0 and the dependent current source is also zero. You are then left with a voltage divider comprising the 6Ω & 32Ω resistors across the 85V source. The drop across the 32Ω will then be Vth.

 

Using the 1A current injection method you should leave the 85V source in the circuit for analysis, rather than shorting it out as you have done.

Also, how did you come to the conclusion that the 1A injected current splits into two 0.5A amp branch currents at the node?

Vth is relatively simple to find. Notice that if the terminals a&b are left open then no current flows in the 1Ω. Hence Vz=0 and the dependent current source is also zero. You are then left with a voltage divider comprising the 6Ω & 32Ω resistors across the 85V source. The drop across the 32Ω will then be Vth.

I assumed since I'd follow KCL that the one 1A current will split at that point.
Also I was told in my notes to short an Ind Volt Source and open an Ind Current Source when I inject the 1A in the circuit.

 

I assumed since I'd follow KCL that the one 1A current will split at that point.

You are right that it will split. You are wrong in assuming that it will split equally.

Just by inspection you can tell that it will not split into .5 and .5. Think for a moment. You have one branch that is 0 Ohm resistance (short) and another branch that is 32 Ohm resistance. The current will take route of least resistance which means that most of the current will go into the short and some little amount will go into 32 Ohm branch. At this point I think if you want to know the current that exit the node, you will have to do mesh-current analysis because you are injecting current into the circuit.

 

When you inject 1 amp into the a-b terminals, the dependent current source becomes a constant .7 amp source. You now have two current sources; use superposition to determine the voltage at a-b due to the two current sources. That voltage is numerically equal to Rth

T_N_K told you how to determine Vth, so you're all done after you get Rth.

 

Using the 1A current injection method you should leave the 85V source in the circuit for analysis, rather than shorting it out as you have done.

If he does this, then he will have to calculate the increase in voltage at terminals a-b due to the injected 1 amp. That increment will be numerically equal to Rth.

 

Also I was told in my notes to short an Ind Volt Source and open an Ind Current Source when I inject the 1A in the circuit.

Yep - that's OK. Electrician is correct. If you leave the source in as 85V you then have to subtract Vth from the value of Vab formed when injecting the 1A. The difference is then Rth. It's more complicated & unnecessary leaving the independent sources in.

 

Yep - that's OK. Electrician is correct. If you leave the source in as 85V you then have to subtract Vth from the value of Vab formed when injecting the 1A. The difference is then Rth. It's more complicated & unnecessary leaving the independent sources in.

Much thanks guys on the clarifications, I'll look at this later tomorrow, I'm pretty beat for studying for this exam, did around 24 or so problems today (Including these.)

Ok I looked at it, I got the Vth to be 71.57. (Can't believe I missed that, 0 volt running through 1ohm)

Now I'm at the super position of 1A and .7A.

For the 1A wouldn't I add the 1 resistance to the 32 then take the current divider I = 1 (33/39)?
For the .7A since there is no Voltage between a,b there would be no current as well, so I'm left with the 6ohm and 32ohm resistors. Wouldn't I just run a .7/2 A current through them both? then add them both?

 

When discussing resistors in parallel, it's helpful to use the vertical bar symbol above the backslash key. For example, 6Ω in parallel with 32Ω would be represented as 6||32.

So, for the first part when you inject 1 amp into the a terminal, that 1 amp will pass into a resistor Req which is the parallel combination 6||32 = 96/19. 1 amp in 96/19 ohms gives 96/19 volts. Now, the 1 amp also passes through the 1Ω resistor, causing a 1 volt drop there. So you will have 96/19 volts plus 1 volt, appearing at the a terminal; a total of 115/19 volts, due to the injection of 1 amp.

When calculating the voltage at terminal a due to the .7 amp source, remember that the 1 amp source is now disconnected and there is nothing connected across the a-b terminals; there will be no effect from the 1Ω resistor. The voltage due to the 7/10 amp will be what you get when you inject 7/10 amp into a single resistor Req = 6||32. Do you see why this is?

You are making a false assumption when you say "Wouldn't I just run a .7/2 A current through them both?"; you made this same false assumption earlier. When you have a current I applied to two resistors of different value, the current doesn't divide into I/2 in each resistor. It is crucial that you understand this, preferably before your exam!

There is a current divider formula, similar to the voltage divider formula:

http://en.wikipedia.org/wiki/Current_divider

but you don't need to know how the current divides to solve this problem. All you need to do is calculate the equivalent resistance, Req, of two resistors in parallel, and calculate the voltage drop when a current passes through that Req.

 

When discussing resistors in parallel, it's helpful to use the vertical bar symbol above the backslash key. For example, 6Ω in parallel with 32Ω would be represented as 6||32.

So, for the first part when you inject 1 amp into the a terminal, that 1 amp will pass into a resistor Req which is the parallel combination 6||32 = 96/19. 1 amp in 96/19 ohms gives 96/19 volts. Now, the 1 amp also passes through the 1Ω resistor, causing a 1 volt drop there. So you will have 96/19 volts plus 1 volt, appearing at the a terminal; a total of 115/19 volts, due to the injection of 1 amp.

When calculating the voltage at terminal a due to the .7 amp source, remember that the 1 amp source is now disconnected and there is nothing connected across the a-b terminals; there will be no effect from the 1Ω resistor. The voltage due to the 7/10 amp will be what you get when you inject 7/10 amp into a single resistor Req = 6||32. Do you see why this is?

You are making a false assumption when you say "Wouldn't I just run a .7/2 A current through them both?"; you made this same false assumption earlier. When you have a current I applied to two resistors of different value, the current doesn't divide into I/2 in each resistor. It is crucial that you understand this, preferably before your exam!

There is a current divider formula, similar to the voltage divider formula:

http://en.wikipedia.org/wiki/Current_divider

but you don't need to know how the current divides to solve this problem. All you need to do is calculate the equivalent resistance, Req, of two resistors in parallel, and calculate the voltage drop when a current passes through that Req.

I really need to stop the assuming thing don't I?
Force of habit.
After seeing the previous problem I asked, you pointed out the same thing with that one, except in voltage case.

I had a this problem earlier I did earlier today with a dependent CS and did the "assuming it splits evenly with an injected 1A at certain nodes" and it came out right, thing was though is that the 1A was directly next to and || to the dependent CS. I'm just thinking it was the circumstances of the circuit that made it allowable, or I just got lucky.

Either way now I understand now why the way it is, and I'll stop the assumptions.

I did a few more problems like this one in my textbook and they all came out very nicely.

Much thanks all!