Power drain in circuits

  1. Alright, so, is my thinking correct in that:

    1. Within a series circuit, a larger resistor will drain more power than a smaller resistor.
    2. Within a parallel circuit, a smaller resistor will drain more power than a larger resistor.
    3. Between two circuits, each with the same voltage source, the one with a smaller equivalent resistance will drain more power.

    I'm doing some MCAT prep and for some reason the power drain concept is difficult for me but I've think I've got the hang of it now.

    Any thoughts?
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,043
    Science Advisor
    Homework Helper

    Hi PStudent111! :smile:

    Yes.

    The easy way to check this is …
    For 1, the current (I) is the same, so you use P = IV = I2R, which is proportional to R.

    For 2 and 3, the voltage (V) is the same, so you use P = IV = V2/R, which is inversely proportional to R. :wink:
     
  4. Not necessarily. This is only true if something in the circuit maintains the same current when the resistance is changed.

    For example suppose you changed the resistor to one with an exceptionally high resistance, Lets say you replaced it with a 1m air gap. Virtually no current would flow through the 1m air gap so the power loss would approach zero. However no power would be delivered to the load either so it wouldn't be a very useful circuit.

    Instead of using the word "drain" it might be better to use the word "dissipate".
     
  5. Drakkith

    Staff: Mentor

    I think the OP means that there are 2 resistors in series. In such a case the current will be the same through both resistors and the one with a higher resistance will dissipate more power.
     
  6. Perhaps. He did say "a larger" rather than "the larger".
     
  7. sophiecentaur

    sophiecentaur 13,774
    Science Advisor
    Gold Member

    This is a good example of where some calculations are better than arm waving. If you draw the circuit out, specifying the conditions ( supply voltage, resistor values etc.) and work out the current and power for the supply and the various resistors in the circuit, for various values, you will see the pattern. I mean work the values out yourself and not use an emulator. The sums are easy enough.
    Of course, using symbols rather than numbers, the algebra will give you a better idea of what is happening - but it is not necessary if you really don't like Maths.
     
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