Find the Thevenin’s equivalent circuit

Thread Starter

sonutulsiani

Joined Jan 27, 2010
27
Find the Thevenin’s equivalent circuit

I got V=30. I am confused if R = 4 ohm or 1 ohm?

I did the same circuit in multisim program and it said R=4 ohm

But if it's a short circuit won't the 3 ohm parallel to short circuit be 0 ohm?
 

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hgmjr

Joined Jan 28, 2005
9,027
Find the Thevenin’s equivalent circuit

I got V=30. I am confused if R = 4 ohm or 1 ohm?

I did the same circuit in multisim program and it said R=4 ohm

But if it's a short circuit won't the 3 ohm parallel to short circuit be 0 ohm?
The simulator has yielded the correct answer, in case you were wondering.

hgmjr
 

Heavydoody

Joined Jul 31, 2009
140
Did you use nodal analysis to find Vth? If so, when you short the right side terminals you will add one element to your nodal equation (1 ohm resistor) for finding the Isc. I solved it in my head without changing anything in the circuit (other than the imaginary short) and came up with 4 ohms as well.
 

Thread Starter

sonutulsiani

Joined Jan 27, 2010
27
But I am asking after combining 12 and 4 ohms which results in 3 ohm resistor.

So now 3 is parallel to short circuit, which results in 0 after combining both of them? and then R=0+1=1
 

The Electrician

Joined Oct 9, 2007
2,971
When you're finding Rth, you short the 32V source and the equivalent 3Ω resistor is not in parallel with the shorted 32V source. The shorted 32V source is in series with the 4Ω resistor, and that combination is in parallel with the 12Ω resistor. The shorted 32V source is not in parallel with the 12Ω resistor because the 4Ω resistor is in the way, so to speak.

When you find Isc, the short is across the a-b terminals, and that short is also not in parallel with the 3Ω equivalent.
 

Thread Starter

sonutulsiani

Joined Jan 27, 2010
27
Ok I short the 32 Vsource and then the 4 ohm and 12 ohm resistors are parallel right?
Which results in 3 ohm resistor. I am not finding the current, I am trying to find the equivalent Thevenin resistance.
 

Heavydoody

Joined Jul 31, 2009
140
I think some of the confusion is coming from a difference in method. I have been taught to solve these somewhat differently from the way AAC teaches. But it hasn't failed me yet (not that I have a ton of experience mind).

While it is clear you need to find an equivalent resistance, not current, you can derive that resistance from a current quite simply. Here is how I solved this.

As the circuit is drawn, find the single unknown node (where the 4 and twelve ohm resistors meet the 2 amp source). Your node here is actually a surface which extends around the 1 ohm resistor since no current flows out of that resistor. What results from this single equation is Vth.

Next, short out terminals a and b. Change nothing else in the circuit. Perform nodal analysis again, except now you need to include the 1 ohm resistor in your equation since current does flow through it. The resulting voltage minus your reference (0) divided by 1 ohm is your short circuit current (Isc).

Finally, divide Vth by Isc and you get Rth and you are done. Pretty simple and straight forward here, no source transformations, no combining resistors, no superposition, yada yada yada.
 

The Electrician

Joined Oct 9, 2007
2,971
Ok I short the 32 Vsource and then the 4 ohm and 12 ohm resistors are parallel right?
Yes, but they are not then in parallel with a short.

Replace the 32V source with a wire (a short) and then slide the 4Ω resistor around to the left until it's in the place where the 32V source was.

Now do you see that there is no short in parallel with the 4 and 12 ohm resistors?
 
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