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Voltage drop and power

  1. Feb 3, 2010 #1
    Just curious: is the voltage drop across a load an expression of how much energy that device uses? If it is, what is the difference between voltage drop and power? Power is also used to determine the energy usage of a device, right?

    These aren't very important questions. I've just been thinking about them for some time. :P
  2. jcsd
  3. Feb 3, 2010 #2
    No sweat, I have already found my answer.
  4. Feb 3, 2010 #3
    Actually, I'm still sort of confused. I thought I found the answer in my old topics, but that didn't help. So I would still like someone to answer these questions if it's possible. My apologies.
  5. Feb 3, 2010 #4


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    A voltage is a difference in electric potential. Electric potential is potential energy per unit charge. Therefore, a given amount of charge moving across a given potential difference (or voltage) will lose a certain amount of potential energy.

    Power, of course, is the rate at which energy is produced or expended. Therefore, in order to determine the power dissipated by a load, you must know the rate at which charge passes across it (i.e. the current). This is why power = voltage x current. Think about it:

    [tex] \left(\frac{\textrm{energy}}{\textrm{charge}}\right)\left(\frac{\textrm{charge}}{\textrm{time}}\right) = \frac{\textrm{energy}}{\textrm{time}} [/tex]
  6. Feb 5, 2010 #5


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    Vdrop2 is proportional to Pdiss
  7. Feb 5, 2010 #6
    Not necessarily. The voltage drop across a reactive component can be large, like an inductor or a transformer primary with no secondary load, and the reactive VA (volt-amp) component dissipates no power.

    In general, if a sinusoidal voltage Vo sin ωt is applied across a load, and the required current is I0 cos ωt, then there is no power dissipation.

    Bob S
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