Power dissipated by resistors

Discussion in 'General Electronics Chat' started by Phrazemaster, Dec 6, 2010.

  1. Phrazemaster

    Thread Starter New Member

    Dec 3, 2010
    15
    0
    Hi, this may be an obvious question, but my electronics text does not explain it.

    When calculating the power across a resistor, do you use the voltage the resistor is supposed to drop as the basis, or do you use the total voltage of the circuit?

    So I have a 15v power source, a resistor, and an LED. The LED is supposed to run at 3.5vf, 10ma, so I know I want a voltage drop of 11.5v across a resistor in series with it. So far so good?

    OK, so now do I use the 11.5v as the resistor's voltage drop, and also use the LED's 10ma, like so:

    P = IV = .001A x 11.5V = .0115W

    - or -

    P = IV = .001A x 15v = .015W

    I realize this won't affect which resistor power rating to buy in this case; .25W would be perfect. But in other situations this might change the power rating for the resistor needed.

    Is this making sense?

    Thank-you.

    Mike
     
  2. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    864
    40
    If you want to know the power dissipated by the resistor you use the voltage drop across the resistor itself. Using the voltage of the supply you would calculate the power supplied by the power supply itself.

    Also, 10 mA = 0.010 A.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Yep, it's 11.5v * 0.01A = 115mW.
    For reliability's sake, always double the power requirement; so 230mW.
    A 1/4 Watt (250mW) resistor would work just fine.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,144
    1,790
    By applying Ohm's Law there are several formulations that you can use all of which produce the identical result:

    1. P(Watts) = E(Volts) x I(Amps)
    2. P(Watts) = I^2(Amps^2) x R(Ohms)
    3. P(Watts) = E^2(Volts^2) / R(Ohms)
    It is always voltage drop across the part and current through the part. Nothing else matters.
     
  5. Phrazemaster

    Thread Starter New Member

    Dec 3, 2010
    15
    0
    Thank-you everyone; this makes sense!

    Just trying to take the dry info from the text and apply it to real life is challenging at times.
     
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