Voltage Divider

Discussion in 'Homework Help' started by Rickreezi, Aug 9, 2013.

  1. Rickreezi

    Thread Starter New Member

    Aug 9, 2013
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    1. A voltage divider circuit is to be constructed to obtain an output voltage 3 V (ptp) from an a.c. supply of 12 V (ptp). If the value of one of the available resistor pair is 6 kΩ, then the value of the other resistor with the output voltage across it is;


    1. 1 kΩ
    2. 2 kΩ
    3. 3 kΩ
    4. 4 kΩ
    5. None of the above

    so i know the formular is 12 x(R2/R1+R2) but what do i do if i only have the value for 1 Resistor.

    MY answer was 2. but i did it with an unorthodox way. is there a formula i can use to find the value of the hidden resistor. and is that unknown resistor R2 or R1?
     
  2. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Algebra?

    .....
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Perhaps you would let us into the secret of your unorthodox way?

    Meanwhile what do you know about potential dividers?

    Are the resistors in series or parallel for instance?
     
  4. Rickreezi

    Thread Starter New Member

    Aug 9, 2013
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    Well they would have to be in series i guess, and the way i did it was:

    V1/V2 = R1/R2

    we already know that there is a total of 12v running through the circuit and 3 volts was running across one of the resistors, we can know that the second V of the other Resistor will be 9 since 12v-3v=9v

    3/9= 0.33333333333

    and then i multiply the answer of that to the known resistor which is 6kΩ

    0.3333333333 x 6= 2kΩ

    so the second resistor is 2kΩ

    but my problem is if 3v is v2 in the equation since its the output voltage which was stated in the question, that means that it should have been 9/3=3 instead of 3/9= 0.3333333333.

    so i only get the correct answer if i do v2/v1. :( thats my weird confusing way of getting the missing resistor value. is ther any other way? sorry im a beginner at this trying to do coursework and revise for an exam.
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    You are right to try to clarify this.

    Yes the resistors are in series across the applied 12 volts.

    That is the total resistance across the 12 volts is 6k + 2k = 8 k

    The same fraction of total voltage appears across the resistor as the fraction of total resistance it represents.

    So there is (6/8 * 12) = 9 volts across the 6 k and

    (2/8 * 12) = 3 volts across the 2k

    Yes the formula is


    {V_{R1}} = \frac{{R1}}{{R1 + R2}}*{V_{total}}

    This formula can be extended to any number of resistors in series as a potential divider with many steps.

    It is a very very important and fundamental configuration.

    NB it is always good to draw a (rough!) diagram.
     
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  6. Rickreezi

    Thread Starter New Member

    Aug 9, 2013
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    Thank you very much for your help and efforts im starting to understand more now. but would want to know what if you was not given the value of R2? how would you have known the total resistance across the circuit?
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Well this is homework help.
    Can you not rearrange the formula?

    You should also note that since R1 and R2 are in series, the formula can be derived by noting that the same current flows through both.
     
  8. Rickreezi

    Thread Starter New Member

    Aug 9, 2013
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    yeh thats what i needed basically, help rearranging the formula.thanks for your help appreciated.
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Is it that difficult?


    \begin{array}{l}<br />
 {V_{R2}}(R2 + R1) = {V_T}R2 \\ <br />
 {V_{R2}}R1 + {V_{R2}}R2 = {V_T}R2 \\ <br />
 R2({V_{R2}} - {V_T}) =  - {V_{R2}}R1 \\ <br />
 R2 = \frac{{R1{V_{R2}}}}{{({V_T} - {V_{R2}})}} = \frac{{6*3}}{9} = 2 \\ <br />
 \end{array}

    And by the way thank you for the friend request, but there is no benefit to that option on this forum, it is just part of the vBulletin software that runs it.
     
    Last edited: Aug 9, 2013
  10. WBahn

    Moderator

    Mar 31, 2012
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    First, your equation 12 x(R2/R1+R2) isn't what you think it is because you overlooked the fact that division takes precedence over addition. What it actually says is 12 x((R2/R1)+R2). What you should have written is 12 x(R2/(R1+R2)). It's a common mistake -- and one that we pretty much all make from time to time -- because we tend to see what we want and not what is actually there. I mention it because you do want to develop the habit of being careful about it, both to prevent miscommunication but also because there are lots of times when you need to enter equations in text, such as in a simulator or in program code, and there it makes a really big difference because the computer will strictly interpret what you actually write.

    Getting to the problem at hand, it will help if you understand exactly where that equation comes from. Consider the following circuit:

    [​IMG]

    First off, ignore the fact that it is a "voltage divider" and just look at it as a circuit with a voltage source and two series-connected resistors. By Ohm's Law if is I=V1/(R1+R2), where (R1+R2) is the equivalent resistance of the two resistors, since this is the resistance as seen by the voltage supply.

    Now, what is the voltage across R2? Again by Ohm's Law, it is Vout=I*R2 and we can then replace I with what we got above and the result is

    Vout = I*R2 = [V1/(R1+R2)]*R2 = V1*R2/(R1+R2)

    By knowing this, you can easily match up which resistor is which when given information like that given in the problem since R2 is the resistor that the output voltage is taken across and R1 is, therefore, the other one.

    Thus, in your problem, R1=6kΩ. You already know the answer to that one and have seen it worked out properly by studiot. Let's change it up a bit and say that you were given that the voltage output is taken across a 2kΩ resistor and you were asked to find the other resistance, given the same voltages. Now you know that R2=2kΩ and want to know R1. So you solve the equation for R1.

    They key in algebra is that you always do the same thing to both sides of an equation in order to keep them equal, except when you are doing something to one side that doesn't affect the value.

    <br />
V_{out}\; =\; V_{in}\frac{R_2}{(R_1+R_2)}<br />
\ <br />
V_{out} \cdot (R_1+R_2)\; =\; V_{in} \cdot R_2<br />
\ <br />
V_{out} \cdot R_1\; +\; V_{out} \cdot R_2\; =\; V_{in} \cdot R_2<br />
\ <br />
V_{out} \cdot R_1 \; +\; V_{out} \cdot R_2\; -\; V_{out} \cdot R_2\; =\; V_{in} \cdot R_2\; -\; V_{out} \cdot R_2<br />
\ <br />
V_{out} \cdot R_1 \; =\; V_{in} \cdot R_2\; -\; V_{out} \cdot R_2<br />
\ <br />
V_{out} \cdot R_1 \; =\; (V_{in} \; -\; V_{out}) \cdot R_2<br />
\frac{V_{out} \cdot R_1}{V_{out}} \; =\; \frac{(V_{in} \; -\; V_{out}) \cdot R_2}{V_{out}}<br />
\ <br />
R_1 \; =\; R_2 \cdot \frac{V_{in} \; -\; V_{out}}{V_{out}}<br />
\ <br />
R_1 \; =\; R_2 \cdot \left( \frac{V_{in}}{V_{out}} \; -\; \frac{V_{out}}{V_{out}} \right)<br />
\ <br />
R_1 \; =\; R_2 \cdot \left( \frac{V_{in}}{V_{out}} \; -\; 1 \right)<br />

    Now just plug in the known values of V_in, V_out, and R_2:

    <br />
R_1 \; =\; 2k\Omega \cdot \left( \frac{12V}{3V} \; -\; 1 \right)<br />
\ <br />
R_1 \; =\; 2k\Omega \cdot \left( 4\; -\; 1 \right)<br />
\ <br />
R_1 \; =\; 2k\Omega \cdot 3<br />
\ <br />
R_1 \; =\; 6k\Omega<br />

    Finally, do yourself a BIG favor and learn to track your units throughout your work, as I did above once I plugged values into the equations. Most mistakes you will make will mess up the units and, if you are using than and checking them, you will spot your mistakes almost immediately.
     
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  11. crutschow

    Expert

    Mar 14, 2008
    13,006
    3,232
    Here's an alternate way to calculate the unknown resistor value using Ohm's law, giving a minimum of calculations.

    You know that you want (12V-3V) = 9V across the 6kΩ resistor for 3V output. Thus the resistor current is 9V / 6kΩ = 1.5mA.

    Since the same current flows through the other resistor in the divider, then it's value must be 3V / 1.5mA = 2kΩ, as you determined.
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    In practice, this is the way I generally would approach it. The point I was trying to make is where the equation comes from that he was using and, since the OP seems to have very little background in the way of algebra skills, took it very incrementally step by step.
     
  13. Rickreezi

    Thread Starter New Member

    Aug 9, 2013
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    0
    Thank you very much for this well detailed explanation. helped me alot. Only thing im having a bit of trouble understanding how you did was that whole rearranging the formula to make R2 the subject. But will eventually get it. thanks Alot, you are very helpful WBahn.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    4,792
    I'm not sure what you mean by making R2 the subject. If you could indicate which line(s) starts doing something that you are having trouble following, I would be happy to walk through it in more detail.
     
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