How Do You Calculate Short Circuit Current in Thevenin Equivalent Circuits?

[Saladsamurai]

Homework Statement

I need to find the Thevenin equivalent of the circuit shown:

I found the open circuit voltage (V_AB) by noting that V_AB = V₃ and used the voltage division

V_{AB} = V_s\left ( \frac{R_3}{R_1||R_2 + R_3}\right ) = 12\V

I am having a little trouble finding the short circuit current I_AB (which comes about by replacing R_L with a wire).

The Attempt at a Solution

We can see that the current that comes out of the combined resistance of R₁||R₂ is what goes into R₃ and the length of wire AB. Now it was my reasoning that since the wire AB offers no resistance, then there is no incentive for current to flow through R₃ and hence the current through AB would just be:

I_{AB}=\frac{V_s}{R_1||R_2}

but this does not give the correct results.

Any thoughts on this?

 

[gneill]

You don't need the short circuit current -- that would be for the Norton equivalent. What you want is the equivalent resistance at AB with the voltage source shorted. Easy-peasy.

 

[Saladsamurai]

So this is just the equivalent resistance of the circuit, but not including R_L right?

I.e.

R_{TH} = (R_1||R_2) + R_3

?

 

[gneill]

I don't know what R6 is. But assuming you meant R2,... No. If Vs is shorted then R3 ends up in parallel with R1 and R2.

 

[Saladsamurai]

I don't know what R6 is. But assuming you meant R2,... No. If Vs is shorted then R3 ends up in parallel with R1 and R2.

Yes R2 (edited for clarity).

I guess I don't know the method you are using. It never crossed my mind to short V_s. The method I was shown was 1) Find V_oc, 2) Find I_sc and then obtain R_TH = V_oc/I_sc.

I know that you are probably using a shorter method, but with a test in the morning, I'll keep going the slow and steady way. It turns out my method in post #1 works, I'm just an idiot and can't do basic division . V_AB/I_AB =R_TH and my result checks out (3 kΩ IIRC).

As a matter of curiosity: If you "short" V_s, where does the current come from? I don't think there is any and hence i have trouble seeing what is in parallel with what.

 

[gneill]

Yes R2 (edited for clarity).
As a matter of curiosity: If you "short" V_s, where does the current come from? I don't think there is any and hence i have trouble seeing what is in parallel with what.

If you short V_s, there's no current. Just a resistive network. The resitance "seen" at AB is the Thevenin resistance.

In general, you find the Thevenin equivalent by (1) determining the open circuit voltage and (2) determining the equivalent resistance of the network that remains after shorting out all voltage supplies and opening all current supplies.

You find the Norton equivalent in a similar way. (1) determine the short circuit current, and (2) determine the equivalent resistance when all current supplies have been opened and all voltage supplies shorted.

As you can see, the Norton resistance is exactly the same as the Thevenin resistance (which makes interconverting the two quite a simple matter). And it also explains why finding the short circuit current is another path to finding the Thevenin resistance.

 

[Saladsamurai]

Well I'll be dipped! Thanks again gneill I think I see it now! Though "parallel" is defined to mean a shared voltage ... so it doesn't make 100% sense since there is no voltage ... but I kind of get it.