I have attached a schematic of a circuit I'm having trouble with.

The question is:

Using Thevenin's Theorem, determine the current I in the 15Ω resistor and the voltage VAB of the AC circuit.
Hint: Use a delta-star transformation when determining the Thevenin impedance ZT.

I am fine with thevenin usually, however, this one is bothering me slighty.
I am having trouble with this because of the location of AB.
When short circuiting AB, am I supposed to take a impedance equivalent either side of AB and calcuate the voltage that way? I have been trying but been getting some results that don't really make sense.

thanks.

 

To calculate the Thevenin resistance you want to remove the load, open current sources, short voltage sources, and calculate the resistance from A to B. The easy way complete the task is to label the nodes and redraw the circuit.

I've attached a pdf file showing you how I would go about redrawing the circuit.

So after you get the thevenin resistance, calculate the open circuit voltage and redraw the thevenin equivalent circuit to solve the problem.

 

To calculate the Thevenin resistance you want to remove the load, open current sources, short voltage sources, and calculate the resistance from A to B. The easy way complete the task is to label the nodes and redraw the circuit.

I've attached a pdf file showing you how I would go about redrawing the circuit.

So after you get the thevenin resistance, calculate the open circuit voltage and redraw the thevenin equivalent circuit to solve the problem.

Thanks for your reply, I worked out the equivalent resistance and I got (61.22 + j7.45)Ω is that correct?
Now I am quite unsure how to calculate the thevenin voltage from this, because of the location of AB.
Please help me

 

I'm not sure about the exact number of the resistance, but if you followed the process you should be ok.

As far as calculating the thevenin voltage, you need to first remove the load. The remaining circuit is a voltage source in series with an inductor in series with two parallel branches. Calculate the voltage divider between the inductive load and the parallel branches. Once you know the voltage across the parallel branches you can easily Calculate V_ab.

 

I'm not sure about the exact number of the resistance, but if you followed the process you should be ok.

As far as calculating the thevenin voltage, you need to first remove the load. The remaining circuit is a voltage source in series with an inductor in series with two parallel branches. Calculate the voltage divider between the inductive load and the parallel branches. Once you know the voltage across the parallel branches you can easily Calculate V_ab.

I would have though the parallel branches have the same voltage, but when I use the voltage divider, the voltages come out different, is that meant to happen.
Do I then subtract the Vb from Va?

 

They do have the same voltage. Basically you have (40+j20) in parallel with (20-j40). Calculate the parallel resistance and then the voltage divider.

I'm going to call the voltage in the parallel branches V_1.

\(V_{1}=15v\frac{\frac{(20-j40)(40+j20)}{(20-j40)+(40+j20)}}{\frac{(20-j40)(40+j20)}{(20-j40)+(40+j20)}+j30}\)

So then you can find V_a and V_b.

\(V_{a}=V_{1}*\frac{-j40}{20-j40}\)

\(V_{b}=V_{1}*\frac{40}{40+j20}\)

And finally V_thev is the difference between the two.

 

Thank you, that now makes sense to me! Appreciated!

 

They do have the same voltage. Basically you have (40+j20) in parallel with (20-j40). Calculate the parallel resistance and then the voltage divider.

I'm going to call the voltage in the parallel branches V_1.

\(V_{1}=15v\frac{\frac{(20-j40)(40+j20)}{(20-j40)+(40+j20)}}{\frac{(20-j40)(40+j20)}{(20-j40)+(40+j20)}+j30}\)

So then you can find V_a and V_b.

\(V_{a}=V_{1}*\frac{-j40}{20-j40}\)

\(V_{b}=V_{1}*\frac{40}{40+j20}\)

And finally V_thev is the difference between the two.

I worked Va and Vb out to both be (2.87 - j11.465)V, does that make sense? That would mean Vab is equal to 0 volts?

 

\(V_{a}-V_{b}=V_{1}*[\frac{-j40}{20-j40}-\frac{40}{40+j20}]\)

\(=V_{1}*[\frac{-j40*(40+j20)-40*(20-j40)}{(40+j20)*(20-j40)}]\)

\(=0\)

So assuming we have made no mistakes to this point, there is no open circuit voltage and thus no current flowing.

Double check everything to make sure. Sometimes they throw in these problems to trick you.