Can someone please help me solve these two problems? Also please explain by writing all the steps if possible so I can do similar problems myself.

Thank you so much.

James

 

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Sorry I didn't read that.

I am not very good with this. First time I am studying it. Been sitting here for 2 hours trying to figure out how to do this. I only have two examples infront of me and these questions are completely different to them.

For question 1 between A and B the 15 ohm resistor will go to take off I. Also -j40 Vs needs to be removed. I can't really finish the diagram to move onto the next stage.

Can anyone help me with that please?

 

If you are new to this then Q1 is quite a challenge. So I will give you a reasonable "lead".

Without solving anything, I have attached the steps required to reduce the network to determine ZT across terminals AB. This is not the only approach - just a suggested one.

If you've not used delta-star transformation you will have to look that up in a book.

 

Thanks a lot for that.

I tried to have a go not sure if I am right or wrong though.

We calculate Z A*xy and Z xy-B now dont we?

so for Z A*xy 20*j30/20+j30 than you convert them into polar for division and convert them back to cartesian. i got (13.83 + j9.22) for this.

and for Z xy-B you do j20.40/40+j20 and same steps with answer ( 8 + j16)

I doubt this is right but anyone please check this for me ?

Thank you

 

Rest of calculations to show that I am trying.

Vay= j30/ 20 + j30 * 15<0 = (10.36 + j6.91) V

Vby = 40 / 40+j20 * 15 < 0 = (6.01 + j12.01) V

Vab = Vay - Vby = 6.7 < -49.54 = Vth


Is any of this right? Can someone please confirm for me?

 

Look at the circuit (with the 15Ω removed) and assume there is a voltage Vx at the mid-point node formed by the junction of the j30Ω, the 20Ω and the j20Ω.

Using the voltage divider rule the voltage at A or VA will be

VA=Vx*(-j40)/(20-j40)

And the voltage at B or VB will be

VB=Vx*(40)/(40+j20)

What do you notice about the ratios (-j40)/(20-j40) and (40)/(40+j20) ???

What can you then conclude about the values of VA and VB and hence what is VAB?

As to the Zthevenin - how did you do the delta-star transformation?

 

Aint the ratios same for both of them. I mean when you divide them you get the same thing for VA and VB. Isn't VAB going to be zero??

I didn't do the delta-star transformation. I tried to do it with your diagrams but I guess that's wrong then. Can you help me with transformation please?

 

Aint the ratios same for both of them. I mean when you divide them you get the same thing for VA and VB. Isn't VAB going to be zero??

I didn't do the delta-star transformation. I tried to do it with your diagrams but I guess that's wrong then. Can you help me with transformation please?

Yes VAB will be zero.

Rather than just take my word for it you should prove that to yourself using an alternate method - such as mesh analysis. You don't need Delta-Star transform to do that.

The Star-Delta (& vice-versa) transform is a well-known circuit analysis operation

see

http://en.wikipedia.org/wiki/Y-Δ_transform

You should attempt to do that part by yourself and come back with some clearer working - even if it is only showing that particular part of the overall analysis. As stated previously, this isn't a homework service.

 

Thanks for your reply.

I am a bit confused here if VAB is zero isn't everything else is going to be zero? because you mutiply and divide in various steps to calculate things.

I have attached an image of what I tried to do.

 

Thanks for your reply.

I am a bit confused here if VAB is zero isn't everything else is going to be zero? because you mutiply and divide in various steps to calculate things.

I have attached an image of what I tried to do.

Consider the Wheatstone Bridge at balanced condition - you have two nodes at the same potential but there are non-zero voltages elsewhere. This is a similar situation.

You failed to show any working on your delta-star diagram. You continue to show a singular reluctance to reveal what working processes are going on when you attempt to solve a problem.

In any case I've given a starting point for you in the attachment.

 

my attempt so far not too sure im on the right track..