# Voltage divider with short circuit part 2

simulate this circuit – Schematic created using CircuitLab

This question is a slightly modified follow-up to my previous question Voltage divider with a short, what is Vout?

In this voltage divider I was originally wondering what $V_{out}$ was when a short was created across points A & B.

With the short, the voltage at A is given by:

$$V_{A} = \frac{R4}{R1 + R4} \times V_{in} = \frac{200\Omega}{200\Omega+200\Omega} \times 5V = 2.5V$$

The CircuitLab simulator shows the voltage at $V_{out}$ as 2.5V, and I actually built this on a breadboard to test it out and indeed the voltage at $V_{out}$ is 2.5V.

What I don't understand, is why the voltage at $V_{out}$ is 2.5V, since it seems like all the current from A to B should be flowing over the short, i.e. the path of least resistance. My intuitive expectation is that $V_{out}$ should read 0, or at least have a floating value, but this isn't the case.

Perhaps this is rightly a Physics question, I don't know. But why is there a voltage on $V_{out}$?

The voltage at A, B, and Vout are all 2.5 volts. Since the voltage at A and B is the same, there is no current through R2 and R3. The voltage at your output is the voltage at A minus the voltage drop across R2, which is R2 times the current through R2, which is zero. 2.5 minus zero is 2.5

• Makes sense, but if there's no current through R2 how does a multimeter read 2.5V on Vout? Is it not necessary to have current flow to read voltage on a meter? – par May 13 '14 at 1:19
• @par, an ideal voltmeter is an open circuit so, no, a current is no necessary to read voltage with an ideal voltmeter. Real voltmeters are effectively an open circuit so insignificant current is required to read voltage. – Alfred Centauri May 13 '14 at 1:28
• Thank you. I have to go read up on how voltmeters work, right now it seems like magic. – par May 13 '14 at 1:40
• @par In practice you do need some current to measure voltage but modern volt meters are very high input impedance at least 1000000 ohms so the actual measured voltage will be somewhere between 2.4995 and 2.5. If we assume the meter is totally accurate. In the ideal case where we assume the meter takes zero current the answer is 2.5V in reality the meter should be taking less than $2.5 \mu A$ possibly a lot less. – Warren Hill May 13 '14 at 12:48

But why is there a voltage on Vout?

By Ohm's law, the current through R2, from left to right, is

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

Also, from Ohm's law

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

Now, assuming there is no other circuit connected to the $V_{out}$ node, the current through R2 must be the current through R3, i.e., R2 and R3 are in series. Thus

$$\frac{V_A - V_{out}}{R_2} = \frac{V_{out} - V_B}{R_3}$$

But, $V_A = V_B = 2.5V$ so the only possible solution is

$$V_{out} = 2.5V$$

simulate this circuit – Schematic created using CircuitLab

This is your circuit now. Vout will be the same as points A and B, unless a load is connected at Vout.

For a circuit analysis approach, the resistance looking into Vout forms a voltage divider with...an open circuit, or an infinitely high valued resistor.