# 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? Commented Dec 17, 2022 at 20:41
• From ohm's law, if Z_2 is zero then V_out is also zero. Commented 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. Commented 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? Commented Dec 17, 2022 at 21:27
• I have the feeling you mean the case when Z2 is absent (infinity)... Commented Dec 17, 2022 at 21:28

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.