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Potential dividers and the formula Vout = Vin * R2/(R1+R2)

  1. Feb 20, 2015 #1
    potential_divider_02.gif
    I believe that the formula Vout = Vin * R2/(R1+R2) is used to calculate how much voltage whatever is connected to Vout will be receiving.

    So if:
    • Vin = 12V
    • R1 = 10 ohms
    • R2 = 5 ohms
    Then Vout should be getting 4 volts

    But if we then connect an appliance with a resistance of 2 ohms to Vout
    • The circuit's overall resistance becomes 11.43 ohms
    • R1 receives 1.05A and so reduces the voltage by 1.05A * 10 ohms = 10.5 volts
    • Vout and R2 only get 1.5 volts
    So what's the point of the Vout formula? I imagine that anything you connect to Vout will have a resistance, resulting in Vout not getting whatever you calculated using the formula.
     
  2. jcsd
  3. Feb 20, 2015 #2

    CWatters

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    Sometimes Vout is indeed connected to something that has a very high resistance (for example an Operational Amplifier, the gate of a FET or a capacitor).

    If that's not the case then you have to take the load resistance into account...

    Lets call the 2 Ohm resistor R3. Then you can replace R2 in the equation with R2||R3 ...

    Vout = Vin * R2||R3 / (R1 + R2//R3)

    R2||R3 = 1.43 Ohms

    So
    Vout = 12 * 1.43 / (10 + 1.43)
    = 1.5V

    The equation also assumes that the source resistance of Vin is small. If not then you have to add it to R1.
     
    Last edited: Feb 20, 2015
  4. Feb 20, 2015 #3

    CWatters

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    PS: You might like to think about the range of values for which the resistance at Vout can be ignored. For example work out the effect on Vout if it was 20 Ohms, 200 Ohms or 2KOhms.

    PPS: What is the resistance of the meter used to measure the voltage at Vout? Could that effect the measured voltage? What if R1 and R1 were 1M Ohm?
     
  5. Feb 20, 2015 #4
    I've played around with different combinations of numbers, and the formula only seems to get close if R3 (the component at Vout) has a resistance that is much higher than that of R1. It doesn't seem to matter if R2 is high or low as long as R3>>>R1

    So would it be correct to say that R1 is always a low resistance resistor, R2 is usually variable, and Vout (R3) always has a much higher resistance than R1, and in this situation you would use this formula? And that this formula just supplies a 'good enough' estimation?
     
  6. Feb 20, 2015 #5
    R2 represents anything you put between the Vout terminals. The formula allows you to calculate the effect of R2 with whatever Voltage is applied and whatever value is selected for R1.
     
  7. Feb 21, 2015 #6
    ph1.12.10.gif
    The issue is that if R2 represents anything you put in the Vout terminals, the above diagram wouldn't make any sense
     
  8. Feb 21, 2015 #7

    CWatters

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    What matters is the relationship between R3 and R2 (rather than R3 and R1). R3 won't matter if R3 >> R2. That applies any time you have two resistors in parallel.

    If R3 isn't >> R2 then replace R2 in the equation with R2||R3.

    Works for the LDR as well.
     
  9. Feb 21, 2015 #8

    anorlunda

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    Q
    I don't get the point of your original post. Are your trying to understand circuit analysis formulas, or are you trying to design a regulated voltage supply?
     
  10. Feb 21, 2015 #9

    sophiecentaur

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    Yes. We can say that the formula is correct (as long as the load resistance is included in the value for the 'bottom' resistor). Anything else is just a matter of detail and what you actually want the circuit to do.
     
  11. Mar 8, 2015 #10
     

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  12. Mar 8, 2015 #11
    How is this? Simple divider variable load.jpg
     
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