Voltage divider with a short, what is Vout

short-circuitvoltage divider

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

Best Answer

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

I've redrawn figure 2 for clarity.

schematic

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.

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