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 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?

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

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.

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? Thank you for your time.

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.

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, It depends on your application. 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

If we put RL in the circuit, does that make any change to power dissipated by voltage divider ?

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

In Problem 3, it's intuitive to follow the previous formatting convention and give an answer of 20k ohms. The answer box however refuses to accept anything other than "20000". Is it possible to add 20k as a possible answer?

Sorry. No k. I wish it was possible. The automatic answer-checker just looks for numeric answers (it was originally developed to check math problems). I'm glad you got the answer right!

Main content

Voltage divider

Google Classroom

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, start subscript, o, u, t, end subscript to input v, start subscript, i, n, end subscript. A good place to start is finding the current through start text, R, 1, end text and start text, R, 2, end text.

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, start text, R, 1, end text and start text, R, 2, end text have the same current, and we can consider them to be in series.

i, start subscript, start text, 1, end text, end subscript, equals, i, start subscript, start text, 2, end text, end subscript 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, equals, i, start text, R, end text Ohm's Law

v, start subscript, i, n, end subscript, equals, i, left parenthesis, start text, R, 1, end text, plus, start text, R, 2, end text, right parenthesis

Rearranging to solve for i,

i, equals, v, start subscript, i, n, end subscript, start fraction, 1, divided by, start text, R, 1, end text, plus, start text, R, 2, end text, end fraction

We've solved for current i in terms of v, start subscript, i, n, end subscript and both resistors.

Next, we write an expression for v, start subscript, o, u, t, end subscript using Ohm's Law,

v, start subscript, o, u, t, end subscript, equals, i, start text, R, 2, end text

We can substitute for i in the previous equation to get,

v, start subscript, o, u, t, end subscript, equals, left parenthesis, v, start subscript, i, n, end subscript, start fraction, 1, divided by, start text, R, 1, end text, plus, start text, R, 2, end text, end fraction, right parenthesis, start text, R, 2, end text

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 start text, R, 1, end text and start text, R, 2, end text. This means v, start subscript, o, u, t, end subscript is always less than v, start subscript, i, n, end subscript. Input voltage v, start subscript, i, n, end subscript is scaled down to v, start subscript, o, u, t, end subscript 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, start subscript, o, u, t, end subscript

We want to find v, start subscript, o, u, t, end subscript using the voltage divider relationship.

v, start subscript, o, u, t, end subscript, equals, v, start subscript, i, n, end subscript, start fraction, start text, R, 2, end text, divided by, start text, R, 1, end text, plus, start text, R, 2, end text, end fraction

We insert the actual input voltage and resistor values into the equation, remembering the equation tells us the bottom resistor, start text, R, 2, end text, goes in the numerator.

v, start subscript, o, u, t, end subscript, equals, 12, start text, V, end text, dot, start fraction, 3, start text, k, end text, \Omega, divided by, 1, start text, k, end text, \Omega, plus, 3, start text, k, end text, \Omega, end fraction

v, start subscript, o, u, t, end subscript, equals, 12, start text, V, end text, dot, start fraction, 3, start text, k, end text, \Omega, divided by, 4, start text, k, end text, \Omega, end fraction

v, start subscript, o, u, t, end subscript, equals, 12, start text, V, end text, dot, start fraction, 3, divided by, 4, end fraction, equals, 9, start text, V, end text

Let's do an optional step to check the current.

i, equals, start fraction, v, start subscript, i, n, end subscript, divided by, start text, R, end text, 1, plus, start text, R, end text, 2, end fraction, equals, start fraction, 12, start text, V, end text, divided by, 1, start text, k, end text, \Omega, plus, 3, start text, k, end text, \Omega, end fraction, equals, start fraction, 12, start text, V, end text, divided by, 4, start text, k, end text, \Omega, end fraction, equals, 3, start text, m, A, end text

Now we know the current, so we can compute how much power is being dissipated by our voltage divider,

p, equals, i, v, equals, 3, start text, m, A, end text, dot, 12, start text, V, end text, equals, 36, start text, m, W, end text

Summary: Our voltage divider takes an input voltage (in this case 12, start text, V, end text, but it could be any value) and scales it down to create an output voltage that's 3, slash, 4 of the input voltage. The 3, slash, 4 ratio is determined by our choice of the two resistors. As long as v, start subscript, i, n, end subscript is turned on, a current of 3, start text, m, A, end text flows down through the voltage divider, so it dissipates 12, start text, V, end text, times, 3, start text, m, A, end text, equals, 36, start text, m, W, end text.

Voltage divider practice problems

All of these problems use this circuit diagram,

Problem 1

Let v, start subscript, i, n, end subscript, equals, 6, start text, V, end text, start text, R, end text, 1, equals, 50, start text, k, end text, \Omega, and start text, R, end text, 2, equals, 10, start text, k, end text, \Omega

Find v, start subscript, o, u, t, end subscript, the voltage across start text, R, end text, 2.

v, start subscript, o, u, t, end subscript, equals

start text, V, end text

[Show Answer]

Problem 2

Let start text, R, end text, 1, equals, 90, start text, k, end text, \Omega, start text, R, end text, 2, equals, 10, start text, k, end text, \Omega,
let the output voltage v, start subscript, o, u, t, end subscript, equals, 1, point, 5, start text, V, end text,

Find v, start subscript, i, n, end subscript.

v, start subscript, i, n, end subscript, equals

start text, V, end text

[Show Answer]

Problem 3

Let v, start subscript, i, n, end subscript, equals, 5, start text, V, end text, v, start subscript, o, u, t, end subscript, equals, 2, start text, V, end text, and start text, R, end text, 1, equals, 30, start text, k, end text, \Omega

Find start text, R, end text, 2.

start text, R, end text, 2, equals

\Omega

[Show Answer]

Problem 4 - Challenge

Let v, start subscript, i, n, end subscript, equals, 1, start text, V, end text, v, start subscript, o, u, t, end subscript, equals, start fraction, v, start subscript, i, n, end subscript, divided by, 2, end fraction

Design a voltage divider that dissipates 10, mu, start text, W, end text.

start text, R, end text, 1, equals

\Omega

start text, R, end text, 2, equals

\Omega

[Show Answer]

Review the assumption (advanced)

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 start text, R, 1, end text and start text, R, 2, end text 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 start text, R, 1, end text, equals, start text, R, 2, end text. With matched resistors, the expected v, start subscript, o, u, t, end subscript of the voltage divider is the mid-point of the divider's range, 0, point, 5, v, start subscript, i, n, end subscript. To cause some current to flow out of the divider, we connect a resistor start text, R, end text, start subscript, start text, L, end text, end subscript. Does the divider still work? Does our voltage divider story collapse?

Resistor start text, R, end text, start subscript, start text, L, end text, end subscript acts as a load on the output of the voltage divider, meaning that it causes a current i, start subscript, start text, L, end text, end subscript to flow. The presence of start text, R, end text, start subscript, start text, L, end text, end subscript means start text, R, 1, end text and start text, R, 2, end text are no longer strictly in series. Let's make start text, R, end text, start subscript, start text, L, end text, end subscript fairly big, to make i, start subscript, start text, L, end text, end subscript fairly small relative to i, start subscript, 2, end subscript. Let start text, R, end text, start subscript, start text, L, end text, end subscript be ten times bigger than start text, R, 2, end text,

start text, R, end text, start subscript, start text, L, end text, end subscript, equals, 10, start text, R, 2, end text

start text, R, 2, end text and start text, R, end text, start subscript, start text, L, end text, end subscript are in parallel with each other. Combine the two parallel resistors using the parallel resistor formula to get start text, R, 2, end text, vertical bar, vertical bar, start text, R, end text, start subscript, start text, L, end text, end subscript,

start text, R, 2, end text, vertical bar, vertical bar, start text, R, end text, start subscript, start text, L, end text, end subscript, equals, start fraction, start text, R, 2, end text, dot, start text, R, end text, start subscript, start text, L, end text, end subscript, divided by, start text, R, 2, end text, plus, start text, R, end text, start subscript, start text, L, end text, end subscript, end fraction, equals, start fraction, start text, R, 2, end text, dot, 10, start text, R, 2, end text, divided by, start text, R, 2, end text, plus, 10, start text, R, 2, end text, end fraction, equals, start fraction, 10, divided by, 11, end fraction, start text, R, 2, end text, equals, 0, point, 91, start text, R, 2, end text

[Parallel resistor formula]

[What do the vertical bars mean?]

This is our loaded voltage divider circuit, redrawn to show the equivalent resistance of start text, R, end text, 2 in parallel with start text, R, end text, start subscript, start text, L, end text, end subscript,

The 10, times load resistor has the effect of reducing the resistance at the bottom of the voltage divider by roughly 9, percent. What is the impact of this additional load on the divider's output voltage? Without the load, the expected output is 0, point, 5, v, start subscript, i, n, end subscript. Now we figure out the output voltage in the presence of a load resistor.

v, start subscript, o, u, t, end subscript, equals, v, start subscript, i, n, end subscript, start fraction, 0, point, 91, start text, R, end text, 2, divided by, start text, R, end text, 1, plus, 0, point, 91, start text, R, end text, 2, end fraction

We designed our divider with start text, R, end text, 1, equals, start text, R, end text, 2, so they cancel out,

v, start subscript, o, u, t, end subscript, equals, v, start subscript, i, n, end subscript, start fraction, 0, point, 91, divided by, 1, plus, 0, point, 91, end fraction

v, start subscript, o, u, t, end subscript, equals, v, start subscript, i, n, end subscript, start fraction, 0, point, 91, divided by, 1, point, 91, end fraction, equals, 0, point, 48, v, start subscript, i, n, end subscript

The output voltage drops to 48, percent of the input voltage. How big an error is this?

start fraction, 0, point, 48, divided by, 0, point, 50, end fraction, equals, 0, point, 96, equals, 96, percent

The actual output of the voltage divider is low by 4, percent compared to the expected voltage. (Note the voltage error of 4, percent is significantly less than the 9, percent resistance change.) Does a few percent 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, times greater than the bottom resistor in the voltage divider, you get roughly "one hand" of percent error left parenthesis, 4, minus, 5, percent, right parenthesis in the output voltage. This holds when the output voltage is near the center of its range (in the neighborhood of v, start subscript, start text, i, n, end text, end subscript, slash, 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, start subscript, start text, i, n, end text, end subscript, the percentage error in output voltage will be different. We repeat the analysis with the output voltage set to 90, percent and 10, percent of the divider range. We keep the load resistor ten times the bottom resistor, so the parallel combination of start text, R, end text, 2 and start text, R, end text, start subscript, start text, L, end text, end subscript is still 0, point, 91, start text, R, end text, 2.

Case 1: v, start subscript, o, u, t, end subscript, equals, 90, percent of v, start subscript, i, n, end subscript

Let v, start subscript, o, u, t, end subscript, equals, 90, percent of v, start subscript, i, n, end subscript. The expected output is 0, point, 90, v, start subscript, i, n, end subscript.

First we design a voltage divider that gives us the desired output. Figure out start text, R, end text, 2 in terms of start text, R, end text, 1 for a 90, percent voltage divider,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, 0, point, 90, equals, start fraction, start text, R, end text, 2, divided by, start text, R, end text, 1, plus, start text, R, end text, 2, end fraction

0, point, 90, left parenthesis, start text, R, end text, 1, plus, start text, R, end text, 2, right parenthesis, equals, start text, R, end text, 2

0, point, 90, start text, R, end text, 1, equals, start text, R, end text, 2, minus, 0, point, 90, start text, R, end text, 2

0, point, 90, start text, R, end text, 1, equals, 0, point, 10, start text, R, end text, 2

start text, R, end text, 2, equals, start fraction, 0, point, 90, start text, R, end text, 1, divided by, 0, point, 10, end fraction, equals, 9, start text, R, end text, 1

start text, R, end text, 2 is 9 times bigger than start text, R, end text, 1.

Now we load the circuit with start text, R, end text, start subscript, start text, L, end text, end subscript and see how the output voltage changes. The expression we derived above for the loaded voltage divider is,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, start fraction, 0, point, 91, start text, R, end text, 2, divided by, start text, R, end text, 1, plus, 0, point, 91, start text, R, end text, 2, end fraction

We replace start text, R, end text, 2 with 9, start text, R, end text, 1,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, start fraction, 0, point, 91, left parenthesis, 9, start text, R, end text, 1, right parenthesis, divided by, start text, R, end text, 1, plus, start text, 0, end text, point, 91, left parenthesis, 9, start text, R, end text, 1, right parenthesis, end fraction

All the start text, R, end text, 1's cancel out, leaving,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, start fraction, 0, point, 91, left parenthesis, 9, right parenthesis, divided by, 1, plus, 0, point, 91, left parenthesis, 9, right parenthesis, end fraction, equals, start fraction, 8, point, 19, divided by, 9, point, 19, end fraction, equals, 0, point, 89

The actual output voltage is 89, percent of v, start subscript, i, n, end subscript instead of 90, percent.

The actual output voltage divided by the expected output is,

start fraction, 0, point, 89, divided by, 0, point, 90, end fraction, equals, 0, point, 99

So the actual voltage is lower than the expected by only 1, percent.

Case 2: v, start subscript, o, u, t, end subscript, equals, 10, percent of v, start subscript, i, n, end subscript

Let v, start subscript, o, u, t, end subscript, equals, 10, percent of v, start subscript, i, n, end subscript. The expected output is 0, point, 10, v, start subscript, i, n, end subscript.

Express start text, R, end text, 1 in terms of start text, R, end text, 2 for a 10, percent voltage divider.

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, 0, point, 10, equals, start fraction, start text, R, end text, 2, divided by, start text, R, end text, 1, plus, start text, R, end text, 2, end fraction

0, point, 10, left parenthesis, start text, R, end text, 1, plus, start text, R, end text, 2, right parenthesis, equals, start text, R, end text, 2

0, point, 10, start text, R, end text, 1, equals, start text, R, end text, 2, minus, 0, point, 10, start text, R, end text, 2

0, point, 10, start text, R, end text, 1, equals, 0, point, 90, start text, R, end text, 2

start text, R, end text, 1, equals, start fraction, 0, point, 90, start text, R, end text, 2, divided by, 0, point, 10, end fraction, equals, 9, start text, R, end text, 2

start text, R, end text, 1 is 9 times bigger than start text, R, end text, 2.

Now we load the circuit with start text, R, end text, start subscript, start text, L, end text, end subscript and evaluate the change in output voltage. The expression we derived above for the loaded voltage divider is,

We replace start text, R, end text, 1 with 9, start text, R, end text, 2,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, start fraction, 0, point, 91, start text, R, end text, 2, divided by, 9, start text, R, end text, 2, plus, start text, 0, end text, point, 91, start text, R, end text, 2, end fraction

All the start text, R, end text, 2's cancel out,

start fraction, v, start subscript, o, u, t, end subscript, divided by, v, start subscript, i, n, end subscript, end fraction, equals, start fraction, 0, point, 91, divided by, 9, plus, 0, point, 91, end fraction, equals, start fraction, 0, point, 91, divided by, 9, point, 91, end fraction, equals, 0, point, 092

The actual output voltage is 9, point, 2, percent of v, start subscript, i, n, end subscript instead of the expected 10, percent.

The actual output voltage divided by the expected output is,

start fraction, 0, point, 092, divided by, 0, point, 10, end fraction, equals, 0, point, 92

So the actual voltage differs by 8, percent from the expected. This is nearly twice the error compared to the mid-range divider.

Lessons for a loaded voltage divider

With a 10, times, start text, R, end text, 2 load resistor connected to a voltage divider:

Near mid-range, the output voltage is reduced by 5, percent.
Near the top of its range, the error goes down substantially, to around 1, percent.
Near the bottom of its range, the error roughly doubles compared to mid-range. The output voltage is 8, percent 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, times, start text, R, 2, end text, like an additional 10, times or more. You can get an additional 10, times two ways. Increase the load resistance. Or, redesign the voltage divider to have smaller start text, R, 1, end text and start text, R, 2, end text, (at the cost of more power dissipated by the voltage divider).

Real-world resistor tolerance also impacts accuracy

Real-world resistors always have a plus minus 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:

where start text, R, 2, end text is the resistor on the bottom of the divider.

[Copyright notice]

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Centa Liu
Posted 7 years ago. Direct link to Centa Liu's post “in problem 2 and problem ...”
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.
Button navigates to signup pageComment on Centa Liu's post “in problem 2 and problem ...”
(7 votes)
Answer
- Ken Roberts
  Posted 7 years ago. Direct link to Ken Roberts's post “Since this is the first t...”
  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!
  Button navigates to signup page
  (4 votes)
hilltopelectrical
Posted 7 years ago. Direct link to hilltopelectrical's post “regarding problem #3 in l...”
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?
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(4 votes)
Answer
- Dhanush Satheesh
  Posted 7 years ago. Direct link to Dhanush Satheesh's post “https://www.khanacademy.o...”
  https://www.khanacademy.org/science/physics/circuits-topic/circuits-resistance/v/circuits-part-1
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  (1 vote)
raresmose
Posted 4 years ago. Direct link to raresmose's post “Regarding Problem 4 - Cha...”
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?
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(2 votes)
Answer
- Willy McAllister
  Posted 4 years ago. Direct link to Willy McAllister's post “The voltage divider is ma...”
  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
  Comment on Willy McAllister's post “The voltage divider is ma...”
  (5 votes)
Hagar Usama
Posted 6 years ago. Direct link to Hagar Usama's post “What is the difference be...”
What is the difference between a 'Voltage Divider' and a 'Voltage Regulator'?
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(2 votes)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Good question. A Voltage ...”
  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.
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  (4 votes)
Lê Minh Đức
Posted 7 years ago. Direct link to Lê Minh Đức's post “If we put RL in the circu...”
If we put RL in the circuit, does that make any change to power dissipated by voltage divider ?
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(1 vote)
Answer
- APDahlen
  Posted 7 years ago. Direct link to APDahlen's post “Hi Mogbulumide, This is...”
  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
  Comment on APDahlen's post “Hi Mogbulumide, This is...”
  (6 votes)
daniel.r.ribera
Posted 6 years ago. Direct link to daniel.r.ribera's post “When evaluating the expre...”
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?

Thank you for your time.
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(2 votes)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “The numerator of the resi...”
  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.
  Comment on Willy McAllister's post “The numerator of the resi...”
  (4 votes)
swagB
Posted 7 years ago. Direct link to swagB's post “So can this be used for a...”
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?
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(2 votes)
Answer
- APDahlen
  Posted 7 years ago. Direct link to APDahlen's post “Hello Swag, It depends o...”
  Hello Swag,
  
  It depends on your application.
  
  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
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  (3 votes)
Filippos
Posted 3 years ago. Direct link to Filippos's post “Problem 4 Would it be wr...”
Problem 4

Would it be wrong to calculate the equivalent resistance (Rs=R1+ R2) of the divider using P=(Vin)^2 / Rs ?
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(1 vote)
Answer
Rodney J Toussaint
Posted 7 years ago. Direct link to Rodney J Toussaint's post “In problem 2, how are we ...”
In problem 2, how are we allowed to take the reciprocal of 1/10 to get 15V in?
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(2 votes)
Answer
Jack Clews
Posted 7 years ago. Direct link to Jack Clews's post “In Problem 3, it's intuit...”
In Problem 3, it's intuitive to follow the previous formatting convention and give an answer of 20k ohms. The answer box however refuses to accept anything other than "20000". Is it possible to add 20k as a possible answer?
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(1 vote)
Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Sorry. No k. I wish it wa...”
  Sorry. No k. I wish it was possible. The automatic answer-checker just looks for numeric answers (it was originally developed to check math problems). I'm glad you got the answer right!
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  (2 votes)

Electrical engineering

Course: Electrical engineering > Unit 2

Voltage divider

Example - use the voltage divider equation to find v, start subscript, o, u, t, end subscript

Voltage divider practice problems

Problem 1

Problem 2

Problem 3

Problem 4 - Challenge

Review the assumption (advanced)

Operating the voltage divider near its mid-range

Operating the voltage divider near its extremes

Case 1: v, start subscript, o, u, t, end subscript, equals, 90, percent of v, start subscript, i, n, end subscript

Case 2: v, start subscript, o, u, t, end subscript, equals, 10, percent of v, start subscript, i, n, end subscript

Lessons for a loaded voltage divider

Controlling error in a loaded voltage divider

Real-world resistor tolerance also impacts accuracy

What's in a nickname

Summary

Want to join the conversation?

Electrical engineering

Course: Electrical engineering > Unit 2

Example - use the voltage divider equation to find voutv_{out}vout​v, start subscript, o, u, t, end subscript

Voltage divider practice problems

Problem 1

Problem 2

Problem 3

Problem 4 - Challenge

Review the assumption (advanced)

Operating the voltage divider near its mid-range

Operating the voltage divider near its extremes

Case 1: vout=90%v_{out} = 90\%vout​=90%v, start subscript, o, u, t, end subscript, equals, 90, percent of vinv_{in}vin​v, start subscript, i, n, end subscript

Case 2: vout=10%v_{out} = 10\%vout​=10%v, start subscript, o, u, t, end subscript, equals, 10, percent of vinv_{in}vin​v, start subscript, i, n, end subscript

Lessons for a loaded voltage divider

Controlling error in a loaded voltage divider

Real-world resistor tolerance also impacts accuracy

What's in a nickname

Summary

Want to join the conversation?

Example - use the voltage divider equation to find v, start subscript, o, u, t, end subscript

Case 1: v, start subscript, o, u, t, end subscript, equals, 90, percent of v, start subscript, i, n, end subscript

Case 2: v, start subscript, o, u, t, end subscript, equals, 10, percent of v, start subscript, i, n, end subscript