# Voltage divider with a short, what is Vout?

Figure 1:

            A                 B
Vin---====-----====--+--====-----GND
R1       R2   |   R3
|
Vout


In this basic voltage divider, $V_{out}$ is calculated as:

$$V_{out} = \frac{R3}{R1 + R2 + R3} \times V_{in}$$

Figure 2:

            A                 B
+-----------------+
Vin---====--+--====--+--====--+--GND
R1       R2   |   R3
|
Vout


In Figure 2, a short is made between points A and B.

Will there be a voltage at $V_{out}$?

If so, how is it calculated?

EDIT: A follow-up to this question is posted here: Voltage divider with short circuit part 2

• Welcome to EE.SE. FYI. We have a CircuitLab schematic editor on EE.SE. There is a Schematic button on the toolbar, when you edit the question. So, no need to do the ASCII art. Commented May 12, 2014 at 22:43

A correct answer has been given in words. But you also asked "how is it calculated".

I've redrawn figure 2 for clarity.

simulate this circuit – Schematic created using CircuitLab

Now, it might not be obvious that this is the same circuit as your figure 2 but, in fact, it is.

Assuming the Vout node is not connected to any other circuit, we can calculate Vout as follows:

By inspection, R2 and R3 have the same voltage across and thus, the current 'down' through each resistor is, by Ohm's law:

$$I_{R2} = \frac{V_{out}}{R_2}$$

$$I_{R3} = \frac{V_{out}}{R_3}$$

Now, applying Kirchhoff's Current Law (KCL) at the output node, we have

$$I_{R2} + I_{R3} = 0$$

Combining the previous equations, we have

$$\frac{V_{out}}{R_2} + \frac{V_{out}}{R_3} = 0$$

The only solution to this equation is

$$V_{out} = 0$$

Now, as you gain more experience with solving circuits, this result will not require any calculation at all - the result will become part of your circuits intuition and will be obvious by inspection.

Once the short circuit is applied from A to B, then A is at the same potential as GND. Any current flowing through R1 will bypass R2 and R3, since there is a low-impedance path around them.

Since there is no current flowing through R2 an R3, the voltage at the Vout node will be zero.

• This isn't a homework problem, I'm just learning electronics and am playing with a resistor ladder connected to a keypad that when you press a button shorts across two pins. At any rate, I see how it could be a homework problem as it's pretty basic. Thank you for your answer I'll go work on it :)
– par
Commented May 12, 2014 at 22:37
• @par Excellent! That'll teach me to make assumptions :) I've edited the answer. Commented May 12, 2014 at 22:54
• Just to give you another way to look at this, when the switch is closed, R2 and R3 form a divider between 0 V and 0 V, and R1 becomes irrelevant to setting Vout. Any (unloaded) divider between equal potential points will have that same potential at the output, so you get the answer by inspection. Commented May 12, 2014 at 23:20
• @ThePhoton, +1. I almost wrote that as an answer. By superposition, $V_{out} = V_A \frac{R_3}{R_2 + R_3} + V_B \frac{R_2}{R_2 + R_3} = 0 \frac{R_3}{R_2 + R_3} + 0 \frac{R_2}{R_2 + R_3}$ Commented May 13, 2014 at 1:11