Rth = (8 || 12) + 36 = 74.2 Ohms

[roinujo1]

Homework Statement

Find the thevenin equivalent of the cicuit shown below with V=500 V and i=10 A

Homework Equations

V_oc=V_th
I_sc=V_N

The Attempt at a Solution

[/B]
So, to solve this, I used my instructors method of first finding the R_th. So, what I did was set the independent sources to 0(based on the fact that R_source doesn't change based on V_th), and tried to solve for the resistance. I used the y-Δ transformation on the three at the bottom and got my R_source. However, I am confused as to if this is right or if I am allowed to do that. I got Rsource to be 74.2 ohms, and that doesn't look correct, so I can't really move from there.

 

[gneill]

Hi roinujo1,

You're right, your result doesn't look correct. You'll have to show the details of your work if we're to spot what went wrong. I will mention, though, that if the sources are suppressed the 8 Ω resistor becomes parallel to the 12 Ω resistor, so the "Y" disappears. You might want to spend a bit more time redrawing and simplifying the circuit before bringing out the Y-Δ machinery .

 

[roinujo1]

Thank you soo much for the reply.So, I did some reworking and got an answer to the problem. For the resistance, I got R_th=7.5 ohms.
Before we move on, is this correct? And, is it in parallel because if we remove the independent sources the only "current that should be flowing is from the a to the b nodes? I really am having trouble using the open and short circuit stuff.

So, after that, i used mesh current analysis to find what the I_shortcircuit is. To do this, I simplified the 3 resistors at the bottom to 10 ohms:

Sorr for the shitty drawing. So basically, I found i1 and i2 and said i1 is=ix and ix=I_sc. So, I found the V_th by using I_sc/R_th.

 

[gneill]

Your result of 7.5 Ω for the Thevenin resistance looks good. But I'm having my doubts about your simplification in the latest diagram. In particular, the reduction of the three resistors to the 10 Ω value looks dubious, as the current source and 30 Ω resistor are connected to related nodes.

Here's a trick you can try. Convert the current source and its 30 Ω resistor to their Thevenin equivalent. Then note that the "new" voltage source sits atop the 500 V voltage source which gives it a fixed potential of 500 V above the reference node b. It's perfectly legal to duplicate the 500 V source and move the connection of the new Thevenin model away from that node, provided that the new 500 V source is connected to node b as well. Like this:

Convert current source to voltage source, then:

Now you can proceed to simplify this new layout.

 

[roinujo1]

Thanks for the response again! I really wish I could use this technique, I don't understand it enough and chicked out. What I ended up doing is mesh analysis, but this time using the original 3 resistors without combining them:

And I found i1=69.97 A,i2=56.67 A, and i3=26.67 A

i2=i_{short circuit}

I then used ohms law to get the V_th=R_th*i_{short circuit} which i got to =425.03 V approximately.

How about that?

 

[gneill]

That looks good.

 

[roinujo1]

Thank you so much for the help! Could you by chance help me understnad why the 8 and 12 are in parallel?Are they in parallel because if we remove the independent sources the only "current" that should be flowing is from the a to the b nodes?

 

[gneill]

Thank you so much for the help! Could you by chance help me understnad why the 8 and 12 are in parallel?Are they in parallel because if we remove the independent sources the only "current" that should be flowing is from the a to the b nodes?

The procedure for finding the Thevenin resistance involves suppressing the fixed sources. For voltage sources that means replacing them with a short circuit (basically a wire), and for current source, replacing them with an open circuit (removing them). In this problem, suppressing the voltage source places the 8 and 12 Ohm resistors in parallel.

The circuit with the sources suppressed: