# Load resistor in Thévenin-Norton equivalence for impedance

Lets say I have this circuit in the picture and I want to calculate Thévenin/Norton equivalent impedance from the terminal a-a' standpoint. Would the resistor $$\R_2\$$ be accounted for the Tévenin equivalent or should I "ignore" the load resistor?

$$Z_N = R_1//R_3 + j\omega L, \text{ignoring}$$ $$Z_N = (R_1//R_3 + j\omega L)//R_2, \text{including}$$

I know that if I want to calculate the thévenin voltage would be to know the drop across $$\R_2\$$, but I'm not clear if it would be the same with the Resistance.

• The meaning of request, "find the Thevenin equivalent of this circuit", is to be read more fully as, "find the Thevenin equivalent of this circuit as it would be seen by a load placed at these two terminals." The assumption here is that the load isn't already placed at those terminals. You only want to know what those terminals "look like" without the load attached to them. Once you add the load, then those terminals would present a different Thevenin equivalent to the next load added to them. Etc.
– jonk
Sep 2 at 1:29

I don't have a working crystal ball, so if someone else is presenting this circuit to you and you want to know what they are thinking, then I can't tell you much about that. But I can say how to proceed from $$\u_{_\text{B}}\$$ towards the terminals at a and a'.

Easiest first step is to assign the zero reference to a'. Providing a reference simplifies the math.

At this point, $$\R_1\$$ and $$\R_3\$$ form a divider, so you can easily Theveninze that much. This means you now have a different $$\u_{_\text{B}}^{\:'}\$$ and followed by an $$\R_{_\text{TH}}\$$ in series with an $$\s\,L\$$. So $$\v_{\text{a}}=u_{_\text{B}}^{\:'}=u_{_\text{B}}\frac{R_3}{R_1+R_3}\$$ and $$\Z_{_\text{TH}}=\frac{R_1\,R_3}{R_1+R_3}+j\,\omega\,L\$$.

That's before adding in $$\R_2\$$. If you want to know what $$\v_{\text{a}}\$$ looks like after adding in $$\R_2\$$, then you have another divider to work out.

This isn't rocket science.

It depends on whether you want to find the Thevenin equiv circuit of the circuit with R2 included or the Thevenin equiv circuit of the circuit without R2.

If R2 is a load then you would probably want to see how the Thevenin equiv circuit interacts with that load and so in that case you would remove R2 before calculating the Thevenin equiv circuit and then reconnect R2 afterwards.

If you want to see how the Thevenin equiv circuit interacts with an un shown load which would be connected to the output then you would calculate the Thevenin equiv circuit with R2 included.

• I disagree., (quite strongly actually). Thevenin and Norton circuit reductions are done by first removing the load. Otherwise the reduction/equivalent means something else. Sep 1 at 21:41
• @RussellH You have every right to disagree. Sep 1 at 22:08
• @James is correct. It all depends on whether R2 is to be regarded as the load or not. Sep 2 at 5:30

If R2 is "always" in the circuit between a & a' , yes you can compute the Norton equivalent as Rp = R1||R3||R2 and the Thévenin equivalent voltage, Vth.

Then add any additional load you need. Otherwise, there is no point in including R2 if it can be yanked out later.

I ignored L for the DC case as it has no influence. (after all every wire and resistor has "some" inductance without exception, in units of nH/mm :)

• But you shouldn't ignore it because the source isn't known to be a DC source. Sep 2 at 5:32
• Agreed. The fact that R2 is connected across the output terminals does not make it a "load resistor". Sep 2 at 10:37
• @CharlesB.Cameron OK but not relevant to this question on assumptions and procedures. When in doubt, always state the assumptions. Sep 2 at 12:38

You definitely should not include the load resistor.The Thevenin theorem tells us that a circuit made of linear components can be replaced with the Thevenin voltage in series with the Thevenin-Norton resistance.In order to find the Thevenin voltage and the Thevenin-Norton resistance we remove the load.