Thevenins Equivalent Circuit

Thevenins Equivalent circuit

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ErnieM

Joined Apr 24, 2011
8,377
Thanks @ErnieM That has been a huge help!

So today I thought I had the solution at 37.5V but now after doing the method you have just explained to me I have different. Please let me talk you through how I have just done this and please advised me if I am wrong.

Again thank you so much for your time

So here we go:

I have split the circuit into 2 as you said and have indeed got a voltage of 60v and a resistance of 60 ohms!

Onto the 2nd half of the circuit I now have 60v, a 60 ohm Resistor in series with a 100 ohm resistor (160 ohms)
parrallel to a 100 ohm resistor.

160*100 = 61.54 ohms (which is the correct resistance that I got in the 1st place.
160+100

Now

would I be correct in saying that I now do V1(60V)*R2(100 ohms) = 6000 =23.07V
R1(160 ohms) + R2 (100 ohms) 260
I did this again and got your numbers. Apparently the meds my doc has me on are stringer then I though.

Have an inner ear infection that makes me quite dizzy when I get up down, moderately dizzy for shaking my head. It's better today.
 

Thread Starter

JIMMYR10

Joined Jan 25, 2015
11
That's good news!

So thanks to your help and everyone else I now have have:

61.54 ohms
23.08 V

And I now have a good understanding of the method to achieve this.

If I apply the same circuit via the Norton theorem at which point do I work out the current?

I know that my resistance will be the same the.

In previous exercises the circuits have been more basic therefore all I have had to do is add the resistors in series and divide by the voltage for the equivalent current.

Any help would be much appreciated @WBahn @ErnieM
 

WBahn

Joined Mar 31, 2012
29,976
Now

would I be correct in saying that I now do V1(60V)*R2(100 ohms) = 6000 =23.07V
R1(160 ohms) + R2 (100 ohms) 260
No, because you have forgotten to take into account the
That's good news!

So thanks to your help and everyone else I now have have:

61.54 ohms
23.08 V

And I now have a good understanding of the method to achieve this.

If I apply the same circuit via the Norton theorem at which point do I work out the current?

I know that my resistance will be the same the.

In previous exercises the circuits have been more basic therefore all I have had to do is add the resistors in series and divide by the voltage for the equivalent current.

Any help would be much appreciated @WBahn @ErnieM
You just place a 0Ω resistor (i.e., a short) across the terminals where you load would go and then determine the current in that short. That's the Norton current.
 

Thread Starter

JIMMYR10

Joined Jan 25, 2015
11
Sorry @WBahn it does not appear to show the whole of your last message

In regards to the Norton is the method the exact same up until you work out the Norton Current at the end?

Thanks
 

WBahn

Joined Mar 31, 2012
29,976
Sorry @WBahn it does not appear to show the whole of your last message

In regards to the Norton is the method the exact same up until you work out the Norton Current at the end?

Thanks
Here's the entire last post of mine: "You just place a 0Ω resistor (i.e., a short) across the terminals where you load would go and then determine the current in that short. That's the Norton current."

You can find the equivalent resistance the same way -- as you noted, they are the same in either case.

If you have the Thevenin circuit you can find the Norton current by dividing the voltage by the resistance. Notice that this is the same result that is achieved by shorting the output and asking how much current flows through the short.

In general you can ignore finding the resistance and just find the open circuit voltage (which IS the Thevenin voltage) and then find the short circuit current (which IS the Norton current). The ratio of the two IS the equivalent resistance.

Lots of ways to skin this particular cat.
 
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