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Potential Dividers

  1. Jan 4, 2008 #1

    I'm sitting my 'AS Level Physics B unit 1' in a week or so, and am having some trouble with potential dividers.

    Here is an example of a question. Could you please take me through how they work, my knwoledge of them is practically nothing, and the textbook I have doesn't really cover it well. Thanks for any help =]

  2. jcsd
  3. Jan 4, 2008 #2


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    Part A: The first component (X) is a photoresistor. That article will explain Part C as well.

    Part B: You can use Ohm's Law to determine the voltage. In a series DC circuit like the one depicted, the resistance is additive so the total current through the circuit is given by Ohm's Law: I = V/R, where I is the current, V is the voltage (1.5 V in your problem), and R is the total resistance (1.8K ohms plus 2.2K ohms in your problem).

    Once you have the total current in the circuit, and remembering that in a series circuit, the current is the same throughout (i.e. the current through each resistor is the same) you can calculate the voltage drop across each resistor by rearranging Ohm's Law into V = I*R.

  4. Jan 4, 2008 #3
    In the exam we are given the formula...

    R1 = (R1/R1+R2) x output voltage

    That is the formula we are given how can I apply that to this question? I don't know what would be R1 and R2? Can I use this formula for all potential divider questions?
  5. Jan 4, 2008 #4


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    That will work too, it's called the Voltage Divider Rule (but your equation is just re-arranged for R1 in this case). I just wanted you to understand the concepts behind what is going on in the circuit without just applying a "magical formula".

    You can read a little more about the Voltage Divider Rule here.

    EDIT: Just noticed that the equation you were given in class looks incorrect, since it doesn't reference the input voltage. Check the link for the re-arragned equation for R1.

    Last edited: Jan 4, 2008
  6. Jan 4, 2008 #5
    I wrote it out mate sorry about that.

    output voltage across R1 = (R1/R1+R2) x input voltage

    I don't know what to put into the formula, like what would go in for R1 and R2 in this case? I understand the basics on how they work and what they do, but if I know what goes where i'll be able to work it out in my head =]
  7. Jan 4, 2008 #6


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    Ok, that makes more sense now.

    The attached picture is basically the same as your diagram (Picture source is Wiki).

    Based on that picture you can just plug in the values given for your circuit.

    Just remember, what ever resistor's voltage drop you are looking for, use that resistor in the numerator.

    So, in your example you can call the resistors R1 and R2 with R1 being the "X" one and R2 being the other one (the 2.2K ohm one).

    Then using the Voltage Divider Rule your setup looks like this:

    [tex]V_{out,R_2} = \frac{R_2}{(R_1 + R_2)} \cdot V_{in}[/tex]

    You are given R1's value (1.8K Ohms) and R2's value (2.2K Ohms). Additionally, the source is also give as 1.5 Volts.

    Now it simple substitution & mathematics...


    Attached Files:

  8. Jan 4, 2008 #7
    Thank you so much I just need 1 thing cleared up. WHat do you mean by..

    "Just remember, what ever resistor's voltage drop you are looking for, use that resistor in the numerator."

    And so it doesnt matter which resistpr value you make R1 or R2?
  9. Jan 4, 2008 #8


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    I mean that if you are looking for the voltage drop across a specific resistor, then that resistor is used in the general equation for the voltage divider rule at the term in the numerator on the RHS of the equation (fractional part).

    If you'll notice in the equation I setup for you,

    [tex]V_{out,R_2} = \frac{R_2}{(R_1 + R_2)} \cdot V_{in}[/tex]

    the term on the LHS of the equation is [tex]V_{out,R_2}[/tex]. I used the subscript R2 to emphasis that the output voltage is read across R2 (i.e. the voltage drop across it). Notice that R2 is also in the numerator on the RHS of the equation (fractional part).

    Does that make sense now?

  10. Jan 4, 2008 #9
    Thanks so much, for all your help. I can go into my exam with some confidence. I'll just do a few past paper questions to get the feel but thanks! =]
  11. Jan 4, 2008 #10


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    You're quite welcome...good luck.

  12. Jan 7, 2008 #11
    i have a doubt in voltage divider bias. . . . . . when NPN transistor connected in Ce mode is used as an amplifier due to the resistor conneted in series with emitter , transistor is stabilised . . . . but while using the divider in single stage CE amplifier a emitter bypass capacitance is connectd in parallel with the resistor ,which conneted in series with emitter . . . which provides a low reactance path to the amplified ac signal . . .there fore no current flows thru the resistor . . . . but due to the flow of current in the resistor ,only the transistor is stabilised . . . . . .Actually what is happening there . . or am i going wrong somewhere
    Last edited: Jan 7, 2008
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