# How does a voltage divider work?

I have a simple question regarding the voltage divider.

Let's say we have the case as in the image below:

I know the formula for the division of voltage at $$\V_{out}\$$.

My question is what happens if $$\Z_2\$$, or the second impedence becomes 0? The formula says the output voltage and the input voltage should be the same, but intuitively I don't understand how they could be the same. Shouldn't the output voltage be a little less than the input, even if there is no second impedance, just because of the presence of that first one. Why is this not the case?

• Have you tried Ohm’s law and KVL? Dec 17, 2022 at 20:41
• From ohm's law, if Z_2 is zero then V_out is also zero. Dec 17, 2022 at 20:44
• Energy will take the path of the lowest impedance. So Vout will be 0. If you say Vout has a resistance of Rout, two parallel resistances exist. 1/Rtot = 1/R2+1/Rout. You can’t divide by 0, but say its like 0.00001… Rtot wil be 0 also. Dec 17, 2022 at 20:48
• @RemyHx no it won't. Energy will distribute itself around the various paths as a function of their various conductances. Ask yourself; will a 10 k resistor in parallel with a 10.1 k resistor hog all the current, power and energy? Dec 17, 2022 at 21:27
• I have the feeling you mean the case when Z2 is absent (infinity)... Dec 17, 2022 at 21:28

## 2 Answers

My question is what happens if Z2, or the second impedence becomes 0? The formula says the output voltage and the input voltage should be the same,

No, if Z2 becomes 0, the output will be 0.

The 'formula' for the output of an unloaded voltage divider as shown is Vout/Vin = Z2/(Z1+Z2) so if Z2 = 0 the output will be zero, provided only that Z1 $$\\ne\$$ 0.