If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Voltage divider

A voltage divider is a simple series resistor circuit. It's output voltage is a fixed fraction of its input voltage. The divide-down ratio is determined by two resistors. Written by Willy McAllister.
A very common and useful series resistor circuit goes by the nickname voltage divider. We will work out how this circuit operates, and you will see where the nickname comes from.
A voltage divider looks like this:
Our goal is to come up with an expression relating output ${v}_{out}$ to input ${v}_{in}$. A good place to start is finding the current through $\text{R1}$ and $\text{R2}$.
Assumption: Assume $0$ current is flowing out of the divider. (Before we are done we will check to see what happens if this zero-current assumption doesn't hold).
With this assumption, $\text{R1}$ and $\text{R2}$ have the same current, and we can consider them to be in series.
${i}_{\text{1}}={i}_{\text{2}}\phantom{\rule{2em}{0ex}}$ and for now let's just call this $i$.
To find the current, we apply Ohm's law and what we know about resistors in series, (reminder: resistors in series add),
$v=i\phantom{\rule{0.167em}{0ex}}\text{R}\phantom{\rule{2em}{0ex}}$ Ohm's Law
${v}_{in}=i\phantom{\rule{0.167em}{0ex}}\left(\text{R1}+\text{R2}\right)$
Rearranging to solve for $i$,
$i={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{1}{\text{R1}+\text{R2}}$
We've solved for current $i$ in terms of ${v}_{in}$ and both resistors.
Next, we write an expression for ${v}_{out}$ using Ohm's Law,
${v}_{out}=i\phantom{\rule{0.167em}{0ex}}\text{R2}$
We can substitute for $i$ in the previous equation to get,
${v}_{out}=\left({v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{1}{\text{R1}+\text{R2}}\right)\text{R2}$
and we have derived the voltage divider equation:
The output voltage equals the input voltage scaled by a ratio of resistors: the bottom resistor divided by the sum of the resistors.
The ratio of resistors is always less than $1$ for any values of $\text{R1}$ and $\text{R2}$. This means ${v}_{out}$ is always less than ${v}_{in}$. Input voltage ${v}_{in}$ is scaled down to ${v}_{out}$ by a fixed ratio determined by the resistor values. This is where the circuit gets its nickname: voltage divider.

### Example - use the voltage divider equation to find ${v}_{out}$‍

We want to find ${v}_{out}$ using the voltage divider relationship.
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{\text{R2}}{\text{R1}+\text{R2}}$
We insert the actual input voltage and resistor values into the equation, remembering the equation tells us the bottom resistor, $\text{R2}$, goes in the numerator.
${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}{1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }+3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}$
${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}{4\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}$
${v}_{out}=12\phantom{\rule{0.167em}{0ex}}\text{V}\cdot \frac{3}{4}=9\phantom{\rule{0.167em}{0ex}}\text{V}$
Let's do an optional step to check the current.
$i=\frac{{v}_{in}}{\text{R}1+\text{R}2}=\frac{12\phantom{\rule{0.167em}{0ex}}\text{V}}{1\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }+3\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}=\frac{12\phantom{\rule{0.167em}{0ex}}\text{V}}{4\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }}=3\phantom{\rule{0.167em}{0ex}}\text{mA}$
Now we know the current, so we can compute how much power is being dissipated by our voltage divider,
$p=i\phantom{\rule{0.167em}{0ex}}v=3\phantom{\rule{0.167em}{0ex}}\text{mA}\cdot 12\phantom{\rule{0.167em}{0ex}}\text{V}=36\phantom{\rule{0.167em}{0ex}}\text{mW}$
Summary: Our voltage divider takes an input voltage (in this case $12\phantom{\rule{0.167em}{0ex}}\text{V}$, but it could be any value) and scales it down to create an output voltage that's $3/4$ of the input voltage. The $3/4$ ratio is determined by our choice of the two resistors. As long as ${v}_{in}$ is turned on, a current of $3\phantom{\rule{0.167em}{0ex}}\text{mA}$ flows down through the voltage divider, so it dissipates $12\phantom{\rule{0.167em}{0ex}}\text{V}×3\phantom{\rule{0.167em}{0ex}}\text{mA}=36\phantom{\rule{0.167em}{0ex}}\text{mW}$.

### Voltage divider practice problems

All of these problems use this circuit diagram,

#### Problem 1

Let ${v}_{in}=6\phantom{\rule{0.167em}{0ex}}\text{V}$, $\text{R}1=50\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$, and $\text{R}2=10\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$
Find ${v}_{out}$, the voltage across $\text{R}2$.
${v}_{out}=\phantom{\rule{1em}{0ex}}$
$\text{V}$

#### Problem 2

Let $\text{R}1=90\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$, $\text{R}2=10\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$,
let the output voltage ${v}_{out}=1.5\phantom{\rule{0.167em}{0ex}}\text{V}$,
Find ${v}_{in}$.
${v}_{in}=\phantom{\rule{1em}{0ex}}$
$\text{V}$

#### Problem 3

Let ${v}_{in}=5\phantom{\rule{0.167em}{0ex}}\text{V}$, ${v}_{out}=2\phantom{\rule{0.167em}{0ex}}\text{V}$, and $\text{R}1=30\phantom{\rule{0.167em}{0ex}}\text{k}\mathrm{\Omega }$
Find $\text{R}2$.
$\text{R}2=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }$

#### Problem 4 - Challenge

Let ${v}_{in}=1\phantom{\rule{0.167em}{0ex}}\text{V}$, ${v}_{out}=\frac{{v}_{in}}{2}$
Design a voltage divider that dissipates $10\phantom{\rule{0.167em}{0ex}}\mu \text{W}$.
$\text{R}1=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }\phantom{\rule{2em}{0ex}}$
$\text{R}2=\phantom{\rule{1em}{0ex}}$
$\mathrm{\Omega }$

A voltage divider doesn't do anything useful unless its output is connected to something. You should be aware of what happens when a divider is connected to a load. Remember we made an assumption at the beginning? We assumed the current flowing out of the output was $0$. That let us treat $\text{R1}$ and $\text{R2}$ as if they were in series, and we developed the voltage divider equation. Let's check what happens if the assumption is not true.

### Operating the voltage divider near its mid-range

To start this discussion, we let $\text{R1}=\text{R2}$. With matched resistors, the expected ${v}_{out}$ of the voltage divider is the mid-point of the divider's range, $0.5\phantom{\rule{0.167em}{0ex}}{v}_{in}$. To cause some current to flow out of the divider, we connect a resistor ${\text{R}}_{\text{L}}$. Does the divider still work? Does our voltage divider story collapse?
Resistor ${\text{R}}_{\text{L}}$ acts as a load on the output of the voltage divider, meaning that it causes a current ${i}_{\text{L}}$ to flow. The presence of ${\text{R}}_{\text{L}}$ means $\text{R1}$ and $\text{R2}$ are no longer strictly in series. Let's make ${\text{R}}_{\text{L}}$ fairly big, to make ${i}_{\text{L}}$ fairly small relative to ${i}_{2}$. Let ${\text{R}}_{\text{L}}$ be ten times bigger than $\text{R2}$,
${\text{R}}_{\text{L}}=10\phantom{\rule{0.167em}{0ex}}\text{R2}$
$\text{R2}$ and ${\text{R}}_{\text{L}}$ are in parallel with each other. Combine the two parallel resistors using the parallel resistor formula to get $\text{R2}\phantom{\rule{0.167em}{0ex}}||\phantom{\rule{0.167em}{0ex}}{\text{R}}_{\text{L}}$,
$\text{R2}\phantom{\rule{0.167em}{0ex}}||\phantom{\rule{0.167em}{0ex}}{\text{R}}_{\text{L}}=\frac{\text{R2}\cdot {\text{R}}_{\text{L}}}{\text{R2}+{\text{R}}_{\text{L}}}=\frac{\text{R2}\cdot 10\phantom{\rule{0.167em}{0ex}}\text{R2}}{\text{R2}+10\phantom{\rule{0.167em}{0ex}}\text{R2}}=\frac{10}{11}\phantom{\rule{0.167em}{0ex}}\text{R2}=0.91\phantom{\rule{0.167em}{0ex}}\text{R2}$
This is our loaded voltage divider circuit, redrawn to show the equivalent resistance of $\text{R}2$ in parallel with ${\text{R}}_{\text{L}}$,
The $10×$ load resistor has the effect of reducing the resistance at the bottom of the voltage divider by roughly $9\mathrm{%}$. What is the impact of this additional load on the divider's output voltage? Without the load, the expected output is $0.5\phantom{\rule{0.167em}{0ex}}{v}_{in}$. Now we figure out the output voltage in the presence of a load resistor.
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$
We designed our divider with $\text{R}1=\text{R}2$, so they cancel out,
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91}{1+0.91}$
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{0.91}{1.91}=0.48\phantom{\rule{0.167em}{0ex}}{v}_{in}$
The output voltage drops to $48\mathrm{%}$ of the input voltage. How big an error is this?
$\frac{0.48}{0.50}=0.96=96\mathrm{%}$
The actual output of the voltage divider is low by $4\mathrm{%}$ compared to the expected voltage. (Note the voltage error of $4\mathrm{%}$ is significantly less than the $9\mathrm{%}$ resistance change.) Does a few $\mathrm{%}$ error matter? That's for you alone to decide. It depends on how accurate the voltage divider needs to be for your application.
The nugget to tuck away from this analysis: If the effective load resistance is $10×$ greater than the bottom resistor in the voltage divider, you get roughly "one hand" of $\mathrm{%}$ error $\left(4-5\mathrm{%}\right)$ in the output voltage. This holds when the output voltage is near the center of its range (in the neighborhood of ${v}_{\text{in}}/2$).

### Operating the voltage divider near its extremes

If you design the voltage divider to operate near its extremes, with the output voltage close to $0$ or ${v}_{\text{in}}$, the percentage error in output voltage will be different. We repeat the analysis with the output voltage set to $90\mathrm{%}$ and $10\mathrm{%}$ of the divider range. We keep the load resistor ten times the bottom resistor, so the parallel combination of $\text{R}2$ and ${\text{R}}_{\text{L}}$ is still $0.91\phantom{\rule{0.167em}{0ex}}\text{R}2$.

#### Case 1: ${v}_{out}=90\mathrm{%}$‍  of ${v}_{in}$‍

Let ${v}_{out}=90\mathrm{%}$ of ${v}_{in}$. The expected output is $0.90\phantom{\rule{0.167em}{0ex}}{v}_{in}$.
First we design a voltage divider that gives us the desired output. Figure out $\text{R}2$ in terms of $\text{R}1$ for a $90\mathrm{%}$ voltage divider,
$\frac{{v}_{out}}{{v}_{in}}=0.90=\frac{\text{R}2}{\text{R}1+\text{R}2}$
$0.90\phantom{\rule{0.167em}{0ex}}\left(\text{R}1+\text{R}2\right)=\text{R}2$
$0.90\phantom{\rule{0.167em}{0ex}}\text{R}1=\text{R}2-0.90\phantom{\rule{0.167em}{0ex}}\text{R}2$
$0.90\phantom{\rule{0.167em}{0ex}}\text{R}1=0.10\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}2=\frac{0.90\phantom{\rule{0.167em}{0ex}}\text{R}1}{0.10}=9\phantom{\rule{0.167em}{0ex}}\text{R}1$
$\text{R}2$ is $9$ times bigger than $\text{R}1$.
Now we load the circuit with ${\text{R}}_{\text{L}}$ and see how the output voltage changes. The expression we derived above for the loaded voltage divider is,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$
We replace $\text{R}2$ with $9\phantom{\rule{0.167em}{0ex}}\text{R}1$,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\left(9\phantom{\rule{0.167em}{0ex}}\text{R}1\right)}{\text{R}1+\text{0}.91\phantom{\rule{0.167em}{0ex}}\left(9\phantom{\rule{0.167em}{0ex}}\text{R}1\right)}$
All the $\text{R}1$'s cancel out, leaving,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\left(9\right)}{1+0.91\phantom{\rule{0.167em}{0ex}}\left(9\right)}=\frac{8.19}{9.19}=0.89$
The actual output voltage is $89\mathrm{%}$ of ${v}_{in}$ instead of $90\mathrm{%}$.
The actual output voltage divided by the expected output is,
$\frac{0.89}{0.90}=0.99$
So the actual voltage is lower than the expected by only $1\mathrm{%}$.

#### Case 2: ${v}_{out}=10\mathrm{%}$‍  of ${v}_{in}$‍

Let ${v}_{out}=10\mathrm{%}$ of ${v}_{in}$. The expected output is $0.10\phantom{\rule{0.167em}{0ex}}{v}_{in}$.
Express $\text{R}1$ in terms of $\text{R}2$ for a $10\mathrm{%}$ voltage divider.
$\frac{{v}_{out}}{{v}_{in}}=0.10=\frac{\text{R}2}{\text{R}1+\text{R}2}$
$0.10\phantom{\rule{0.167em}{0ex}}\left(\text{R}1+\text{R}2\right)=\text{R}2$
$0.10\phantom{\rule{0.167em}{0ex}}\text{R}1=\text{R}2-0.10\phantom{\rule{0.167em}{0ex}}\text{R}2$
$0.10\phantom{\rule{0.167em}{0ex}}\text{R}1=0.90\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}1=\frac{0.90\phantom{\rule{0.167em}{0ex}}\text{R}2}{0.10}=9\phantom{\rule{0.167em}{0ex}}\text{R}2$
$\text{R}1$ is $9$ times bigger than $\text{R}2$.
Now we load the circuit with ${\text{R}}_{\text{L}}$ and evaluate the change in output voltage. The expression we derived above for the loaded voltage divider is,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{\text{R}1+0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$
We replace $\text{R}1$ with $9\phantom{\rule{0.167em}{0ex}}\text{R}2$,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91\phantom{\rule{0.167em}{0ex}}\text{R}2}{9\phantom{\rule{0.167em}{0ex}}\text{R}2+\text{0}.91\phantom{\rule{0.167em}{0ex}}\text{R}2}$
All the $\text{R}2$'s cancel out,
$\frac{{v}_{out}}{{v}_{in}}=\frac{0.91}{9+0.91}=\frac{0.91}{9.91}=0.092$
The actual output voltage is $9.2\mathrm{%}$ of ${v}_{in}$ instead of the expected $10\mathrm{%}$.
The actual output voltage divided by the expected output is,
$\frac{0.092}{0.10}=0.92$
So the actual voltage differs by $8\mathrm{%}$ from the expected. This is nearly twice the error compared to the mid-range divider.

### Lessons for a loaded voltage divider

With a $10×\text{R}2$ load resistor connected to a voltage divider:
• Near mid-range, the output voltage is reduced by $5\mathrm{%}$.
• Near the top of its range, the error goes down substantially, to around $1\mathrm{%}$.
• Near the bottom of its range, the error roughly doubles compared to mid-range. The output voltage is $8\mathrm{%}$ lower than expected.

### Controlling error in a loaded voltage divider

If your design requires the error to be significantly smaller, the load needs to be much larger than $10×\text{R2}$, like an additional $10×$ or more. You can get an additional $10×$ two ways. Increase the load resistance. Or, redesign the voltage divider to have smaller $\text{R1}$ and $\text{R2}$, (at the cost of more power dissipated by the voltage divider).

### Real-world resistor tolerance also impacts accuracy

Real-world resistors always have a $±$ tolerance on their value. If the accuracy of the voltage divider is critical to your application, use resistors with tight tolerances, and check for acceptable performance by analyzing the voltage divider at the anticipated tolerance extremes.

### What's in a nickname

We mentioned at the start that the nickname of this circuit is a voltage divider. In many situations, that is exactly what it does. However, we showed that under certain conditions when there is a load on the divider, the actual output voltage is slightly lower than the value predicted by the voltage divider equation. The lesson: Call a circuit by its nickname; just remember that it's only a nickname.

## Summary

Voltage divider:
${v}_{out}={v}_{in}\phantom{\rule{0.167em}{0ex}}\frac{\text{R2}}{\text{R1}+\text{R2}}$
where $\text{R2}$ is the resistor on the bottom of the divider.

## Want to join the conversation?

• in problem 2 and problem 4, suggest that 20k and 50k can be seen as correct answers, since the questions already use k and μ for avoiding too many 0s.
• Since this is the first time an answer is used in this course where "k" is part of the answer (20k), perhaps it is worth while putting in a mention of the possible notation in the text. Anyway, nice course, I am enjoying it. Thanks!
• regarding problem #3 in line 4 of the solution how do you go from -R2(R2x5/2)=30k to 3/2xR2=30k
how would I know to do that?
• What is the difference between a 'Voltage Divider' and a 'Voltage Regulator'?
• Good question. A Voltage Divider is the simple 2-resistor circuit we talk about here. Remember the caution that a voltage divider only behaves itself when the current leaving the center node is practically zero? That limits where you can use a voltage divider. If you need to pull a lot of current at the new lower voltage, you use a Voltage Regulator. This is an integrated circuit with 10-50 transistors specially designed to take a higher voltage and provide current at a very well-controlled lower voltage.

The original IC voltage regulator is the LM317. It's been around forever. https://en.wikipedia.org/wiki/LM317. If you scroll down a little you will see the typical circuit for using the LM317. Notice part of the circuit is a 2-resistor voltage divider that let's you adjust the output voltage. The divider determines the voltage. The regulator provides the robust current.
• If we put RL in the circuit, does that make any change to power dissipated by voltage divider ?
(1 vote)
• Hi Mogbulumide,

This is a hard question to answer without knowing your background. I am assuming you have some knowledge of electronics but have not yet studied AC circuits.

There are two answers to this question:

1) For Direct Current circuits - with DC an inductor acts as a short circuit. In reality you will "see" the resistance of the inductor's winding. In general you will not find inductors in pure DC circuits.

2) For Alternating Current circuits - yes the inductor can be part of the voltage divider. Although, we wouldn't call the circuit a voltage divider anymore. We would instead call the circuit a filter. The "voltage divider's" output would change depending on the frequency of the input signal.

This is where electronics gets interesting. You may be interested in this site:

http://sound.westhost.com/lr-passive.htm

Happy soldering,

APD
• When evaluating the expression
R1 || RL = R2 * 10R2 / R2 + 10R2 you get 10/11R2 = 0.91R2
Was this accomplished by factoring out R2 from the numerator and denominator?
R2(1 * 10) / R2(1 + 10) = 10/11 but wouldn't the R2/R2 just equal 1 or just cancel out? So how is it that it equals 0.91R2? We don't know the value of R2 so we just leave it in as a factor of the result?

When evaluating the expression
V-out = V-in(0.91R2/R1+0.91R2). You stated that R1 = R2 So they cancel out. leaving V-in(0.91/1+0.91) = 0.91/1.91=0.48V Assuming these computations were accomplished via factoring out the R2 this would assert the same computation as the first expression, but in this case R2 cancels out, whereas in the first expression it was left in. So how do I determine when I need to keep an a variable tied to a result, as in the first case of 0.91R2, and when to cancel it out, as in the second case 0.48v?

• The numerator of the resistor expression is R2 * 10R2 = 10 R2^2.
The denominator of the resistor expression is R2 + 10R2 = 11R2
When cancelling R2's you eliminate the one on the bottom but you still end up with one on top. 10R2/11.

In the voltage equation, all the R2's cancel out.
• Regarding Problem 4 - Challenge

Shouldn't the power be equal to P = I x Vout instead of Vin if it asks for the power dissipated by the divider? I was thinking that every element dissipates power and if it specifies the divider power it should be calculated using Vout. If not, why do we use Vin?
(1 vote)
• The voltage divider is made of two resistors, so the "power of the divider" has to account for both. You can compute the power a few different ways.

Assuming i_out = 0...

If you know the current i flowing through the two resistors, then p = i^2 (R1 + R2).

Or, you could compute power based on the voltages,
p_total = p_R1 + p_R2
p_R1 = (Vin - Vout) / R1
p_R2 = Vout^2 / R2

Or, you could compute power based on both i and v,
p_total = p_R1 + p_R2
p_R1 = i (Vin-Vout)
p_R2 = i Vout

Or, simplest of all,
p = i Vin
• So can this be used for a DC circuit? I am trying reduce the voltage of a battery from 9v to 3.7v. what is the best way to do this?
• Hello Swag,

I assume you are using a 9 VDC battery. If you use a voltage divider over half of the battery's energy will be lost in the resistors. It turns out that this simple circuit is very inefficient.

From an energy perspective you would be better off using a a “switching Regulator.” For an example see:

http://power.murata.com/data/power/dms-78xxsr.pdf

For an explanation of how the circuit operates see:

https://en.wikipedia.org/wiki/Buck_converter

Regards,

APD
• In problem 3, I don't get how you go from step 2 to step 3:

2V = 5V x ( R2 / 30kOhm + R2) -->

30kOhm + R2 = ( 5V / 2V ) x R2

It's some kind of algebra I don't understand, but I also don't know the name for it well enough to look it up.