Determining Thevenin equivalent resistance

[TheCanadian]

I am currently learning about Thevenin equivalent circuits and was posed with the question attached to this post. I was wondering why I3 is equivalent to the current formed by a short circuit between nodes a and b. Shouldn't the current between a and b be I1 + I2 + I3 since in a short circuit, all the current would go through that path? Why exactly would it be just I3?

Also, with regards to the same circuit shown in the image, isn't the Thevenin resistance computed by removing all independent voltage sources and then simply finding the resistance? That is:

$$ R_{Th} = [ (10+5)||5 + 10 ]||5 + 10 \Omega = 13.7 \Omega $$

(Where || indicates the resistance is being computed for the two parallel resistors.)

Is there anything wrong in my approach or computation? Any help is greatly appreciated!

 

[axmls]

When we write loop currents, the currents are exactly that: the current in that loop. I1 and I2 are not associated with the loop containing a and b, so they don't matter. It's the loop with I3 that's been short circuited.

As for your equivalent resistance, go through the circuit and combine one pair at a time, redrawing the circuit each time. Notice that you can't work from right to left--you have to work from left to right. Do you see that none of the resistors are in parallel or series on the right part? Remember: to be in parallel, they must share the same nodes at both terminals.

 

[Hesch]

Shouldn't the current between a and b be I1 + I2 + I3 since in a short circuit, all the current would go through that path?

You have sketched 3 current loops in such a way that only I3 passes from a to b. Thus I_sc = I3.

You could have sketched the currents so that all of them pass from a to b, but then your equations would be different, giving other results as for I1, I2, I3.

 

[TheCanadian]

When we write loop currents, the currents are exactly that: the current in that loop. I1 and I2 are not associated with the loop containing a and b, so they don't matter. It's the loop with I3 that's been short circuited.

As for your equivalent resistance, go through the circuit and combine one pair at a time, redrawing the circuit each time. Notice that you can't work from right to left--you have to work from left to right. Do you see that none of the resistors are in parallel or series on the right part? Remember: to be in parallel, they must share the same nodes at both terminals.

Thank you for explaining that about the Thevenin resistance. That part makes sense now.

Hmmm, I am still not quite understanding the short circuit current. Why exactly would the current in loop 3 match the current in the short circuit between a and b? I now see my initial approach was wrong, and that I1 and I2 are not associated with the short circuit current, but how exactly is I3 related to it? Wouldn't there now be a wire without any resistor connecting nodes a and b? In this case of a short circuit, wouldn't this path receive all the current, making I1 = I2 = I3 = 0?

 

[TheCanadian]

You have sketched 3 current loops in such a way that only I3 passes from a to b. Thus I_sc = I3.

You could have sketched the currents so that all of them pass from a to b, but then your equations would be different, giving other results as for I1, I2, I3.

Isn't I3 only through the 5 ohm resistor near nodes a and b, but not necessarily through nodes a and b?

 

[axmls]

Isn't I3 only through the 5 ohm resistor near nodes a and b, but not necessarily through nodes a and b?

Normally, yes. I3 is the current that flows through that loop. However, notice that the current now has two choices: go through the 5 Ohm resistor, or take the path with no resistance. You can then essentially replace that branch with the 5 Ohm resistor with the wire alone.

 

[Hesch]

Isn't I3 only through the 5 ohm resistor near nodes a and b, but not necessarily through nodes a and b?

When you short circuit a and b, you short circuit the 5Ω resistor: 5Ω || 0Ω = 0Ω.

So in your attached calculation, you calculate V_ab ( with ab open ). V_ab,open = I3*5Ω.

Next, you calculate I_sc ( with ab short circuited, substituting the 5Ω by 0Ω ).

Rth = V_ab,open / I_sc = 3.659Ω

Your attached calculation is absolutely correct.

 

[TheCanadian]

When you short circuit a and b, you short circuit the 5Ω resistor: 5Ω || 0Ω = 0Ω.

So in your attached calculcation, you calculate V_ab ( with ab open ). V_ab,open = I3*5Ω.

Next, you calculate I_sc ( with ab short circuited, substituting the 5Ω by 0Ω ).

Rth = V_ab,open / I_sc = 3.659Ω

Your attached calculation is absolutely correct.

Thank you! This is exactly what I was missing. It's a short circuit of the 5Ω resistor as opposed to the entire circuit! It's so simple yet I was just not seeing it. Thank you!