I am using the AoE lab book to teach myself electronics. I am on the voltage divider section, and I am confused at their simplified explanation. I'd like to explain it a different way, but I am not sure how to understand the open circuit voltage.

For the following circuit:


simulate this circuit – Schematic created using CircuitLab

Calculating Vout is simple:

$$V_{out} = V_{in}\frac{R2}{R1 + R2}$$ $$V_{out} = 30\frac{10000}{20000} = 15V$$

Now we introduce a load into the circuit:


simulate this circuit

My approach to solving this myself was to turn the original circuit into a black box and imagine we are looking at it from the perspective of the output.

From reading, the short circuit current here will be

$$I_{sc} = \frac{V}{R} = \frac{15}{10000} = .0015A$$

The first of my questions is - why does only the first resistor matter for calculating the short circuit voltage? Shouldn't both resistors be...resisting in the circuit? Intuitively, is this because we are bypassing R2 via a short circuit and the voltage will just (for lack of a better term) ignore the resistor and follow the path of least resistance (ground)?

After this I need to derive the open circuit voltage. I am having trouble figuring out how to view the circuit in a way that makes this make sense. If I disconnect the load, the open circuit voltage would be the original Vout, would it not?

Basically, I am trying to derive the math that makes treating R1 and R2 in parallel make sense. I understand the thevenin resistance will be this value, and the thevenin voltage is equal to vout (15V) in this case. The explanations in the lab book, while helpful, sort of handwaved through this part (they began to talk about switching voltages and grounds to show which resistor matters) and I'd appreciate any help in understanding this section. I feel like it is critical to understanding circuit impedance (the next section) and I would like to fully understand it before moving on.

Thank you!


You're close. I'm not sure how the AoE book approaches things, but I'd say you're introducing the load resistor too early. The point of using a Thévenin equivalent circuit is to obtain a model for the (sub)circuit in question, and then put a load on there and see how the circuit behaves.

In your case we have the following simple circuit (note that, for the time being, I will ignore the exact numerical values of the resistances and just focus on the general solution).


simulate this circuit – Schematic created using CircuitLab

I'm calling the output node "out" instead of "Vout" because we will be worrying about both its voltage and its current.

As you correctly point out, to determine the open-circuit voltage \$V_{oc}\$, we leave the output node open-circuit (infinite impedance to ground) and calculate the voltage at the node "out", which in this case is a simple voltage divider:

$$ V_{oc} = V_s \cdot {R2 \over R1 + R2} $$

In calculating the short-circuit current \$I_{sc}\$, we short this output node to ground (zero impedance to ground). Since the output node is at ground potential, there is zero volts across R2 and therefore no current flows through it.

$$ I_{sc} = {V_s \over R1} $$

Note that your approach above had an error, since you were using \$V_{oc}\$ instead of \$V_s\$ in this calculation.

As a side note, we can alternatively view the above two circuit setups in terms of parallel resistors. (Remember that, if you have two resistors in parallel, then the resultant resistance is always less than or equal to the smaller of the two resistances.)

  • In the \$V_{oc}\$ calculation, the low-side resistance is R2 in parallel with an infinitely high resistor, which is equal to R2.
  • In the \$I_{sc}\$ calculation, the low-side resistance is R2 in parallel with zero ohms, which is equal to zero.

Finally, to get the Thévenin resistance \$R_{th}\$, we can just divide the open-circuit voltage by the short-circuit current:

$$ R_{th} = {V_{oc} \over I_{sc}} = {V_s \cdot {R2 \over R1 + R2} \over {V_s \over R1}} = {R1 \cdot R2 \over R1 + R2} = {R1 \parallel R2} $$

  • \$\begingroup\$ Ah, ok that makes a little more sense. So the short circuit voltage wouldn't be related to the voltage divider (out here) since R2 is being bypassed. This makes sense since technically, shorting the out would mean we can't "see" 15V and the only voltage we know is the input. This makes a lot of sense. Thanks for the explanation. \$\endgroup\$ – CL40 Nov 18 '19 at 6:58

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