Engineering Finding ZL for a series-parallel RL circuit

[flirt]

Im starting to give up on this one, every time i try to figure it out i get an answer that is miles off.

Im looking to find ZL, can someone show me how this is done? I think my problem is with the dependent voltage.

http://img31.imageshack.us/i/unledak.png/

 

[Zryn]

What working have you got so far and what is the answer you are trying to reach?

 

[Zryn]

Perhaps you have solved it already?

If you do KVL to determine Iphi you then know the value of the dependent source and can then use a voltage divider to find the voltage across the j3R inductor (giving you the parallel voltage across Z) and thus the current through the 1R resistor (Ohms Law), and then KCL that node to find the current through Z and then Ohms Law with the parallel voltage to get a value for Z.

 

[Ghost101]

I have a similar problem, can you do

Iphi = (100<0')/(25+j10)
= 3.7139<-21.8014'AThen the dependant source voltage is 5x that value..

Then you use Voltage divider to see how much goes into the j3?

Does the 100<0' power supply directly affect ZL? Its a tough question :(

can anyone confirm / deny my method? what do you do after?

 

[Ghost101]

I'm having trouble visualizing how the independent voltage source interacts with the right side of the circuit because of that annoying dependent voltage source,
because usually you can short circuit each Voltage source and see it's effect but short circuiting the dependent source means there is no relationship between the left side and the right?
Im guessing I(first branch) + I(second branch) gives the total current in the third area?
any ideas?

 

[Zryn]

I have a similar problem, can you do

Iphi = (100<0')/(25+j10)
= 3.7139<-21.8014'A

Not quite. By Ohms Law you are trying to get the current through the resistor and inductor combination, but you need to take into account the voltage at the node on the other side, which is conveniently 5*Iphi, giving you still only one variable.

Then you use Voltage divider to see how much goes into the j3?

Yep.

Does the 100<0' power supply directly affect ZL? Its a tough question :(

Its an ambiguous question when you say directly. The power from the dependent and the independent power supplies both affect ZL with the current they provide. Unless you can show that a power source is supplying or absorbing 0 power, it will be affecting the circuit. What do you mean by directly though?

I'm having trouble visualizing how the independent voltage source interacts with the right side of the circuit because of that annoying dependent voltage source

Consider them both as sources of power. If you had 2 batteries in there instead of 1 battery and a Op-Amp dependent voltage source circuit you would still get power out of both.

Im guessing I(first branch) + I(second branch) gives the total current in the third area

Don't guess . Do KVL and KCL and play around with it.

Also, if you would like to talk about a circuit, its generally good to start your own thread, put up a picture of your circuit and say what you have figured out so far. There's bucket loads of people who will help you if they can!

 

[Ghost101]

Thanks dude! much appreciated!