5
\$\begingroup\$

I just encountered the voltage divider a few days back. It was working, yes. But i'm confused that in my circuit,
I needed to output 1/2 voltage so I used equal values for the two resistors. And from what I've read, I could change the ratios to change the voltage. And I tested that, sure enough, it works.

But this is what confuses me, I was taught that resistors don't lower voltage, they lower current. And if they do have some voltage drop, it surely can't drop it by that much. So why does it work?

Any help will really be appreciated.

\$\endgroup\$
7
\$\begingroup\$

Resistors are the subject of Ohm's Law

The resistance, current and voltage are all tied together by the formula:

\$I=\frac{V}{R}\$

The current flowing through the resistors is determined by the voltage across the resistors divided by the total resistance.

So, in a voltage divider, you have a known voltage across the resistors - say 5V.

enter image description here

If the total resistance is 20KΩ (two 10KΩ resistors in series), that is a total current of \$\frac{5}{20000}\$ which equals 250µA.

Now, if the resistors in the voltage divider are the same value then it stands to reason that the voltage applied across the whole divider is split in half across the two resistors, as, according to the same formula (turned around for voltage):

\$V=IR\$

Which is \$0.00025 \times 10000\$ - or 2.5V.

If the resistors were 15KΩ and 5KΩ then it would be \$0.00025 \times 15000\$ for one, and \$0.00025 \times 5000\$ for the other - that's 3.75V and 1.25V respectively.

\$\endgroup\$
2
\$\begingroup\$

The same current flows through both resistors, so by Ohm's law the voltages across the resistors are proportional to their resistance values.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.