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Electrical Physics Question help

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data

    The Potential drop across a light bulb in a circuit is 5.6 volts. After 12C of charge have passed through the bulb how much electrical energy will the bulb have received?

    Any help would be much appreciated




    3. The attempt at a solution
    im not really sure of what formula to use..
    Q=IxT or Q=IxT but seeing that theres not time in the question im pretty confused..
     
    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2

    rl.bhat

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    Power P = I*V
    Energy = P*T = I*V*T = Q/T*V*T = Q*V
     
  4. Mar 26, 2009 #3

    CompuChip

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    Welcome to PF SussRuss.
    Maybe it helps if you consider that 1 V = 1 J / C.
     
  5. Mar 26, 2009 #4
    thanks for replies, thanks for welcoming :p

    so P=I*V
    P=5.6V*12C
    =67.2W
    correct??
     
  6. Mar 26, 2009 #5

    CompuChip

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    You were asked for the energy, not the power.
    Note that I is a current, which is measured in Ampere, not in Volts.
    rl.bhat also gave you a formula for energy.
     
  7. Mar 26, 2009 #6
    ooo so im confused now.
     
  8. Mar 26, 2009 #7

    CompuChip

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    What is confusing you?
    Do you have a problem with clearly placing the concepts of energy, current, voltage and charge relative to each other?
     
  9. Mar 26, 2009 #8
    i guess thats it
     
  10. Mar 26, 2009 #9

    CompuChip

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    Unfortunately I don't have the time to make a long explanation out of it, but here's a quick (and somewhat simplified) review.

    The particles involved in electricity are electrons and protons, which have a certain property called charge. Charge is measured in Coulombs (C). Electrons can carry some amount of energy, measured in Joules (J), the standard unit of energy. This amount of energy can vary from point to point, for example in an electrical circuit, or battery, etc. How much energy the electrons carry is measured by the voltage, which has units of J / C, i.e. energy per charge. An intuitive picture is that of electrons being little guys with backpacks. Since each electron has the same charge, the charge measures something like how many electrons there are around. They all carry some total energy in their backpacks, and the voltage tells you what is the average energy carried by each electron (up to a constant factor).
    What is important now is the voltage difference. When the electrons are left free to move (for example, if you close your electrical circuit) they will try to cancel the voltage difference by moving energy around. We can use this by putting a resistor (like a resistor component, or a light bulb, or a coffee machine) somewhere along the path. Because the electrons really want to get rid of the voltage difference, they will pass through the resistor, although they will lose a part of their energy (which is then used to make the light bulb glow or make your coffee water boil). As a consequence, there will be a voltage drop across the resistor: the average energy per electron is less on one side of the resistor than on the other.

    The current, I, is measured in amperes (A) which is 1 Coulomb / second. So the current does not tell you something about the energy the electrons are carrying around, but about the amount of them that passes in one second. Since electrons (usually) do not pile up somewhere, the current will be the same throughout the entire circuit (unless, for example, you make them split up somewhere by putting your resistors in parallel instead of in series). You see that the current can be defined as the total charge divided by the total time: I = Q / T. E.g. if 12 C of charge passes in 10 seconds, the current is 12/10 Coulomb / second = 1.2 Ampere.

    So we have electrons moving around our circuit, depositing energy as they go through our coffee machine, which we use to heat the water. How much energy they leave, depends both on how much they have, and how many there are, of course. Since we know how many electrons pass in a second, and how many energy each carries, we can calculate the total energy left in the resistor in one second. This is the power, P, measured in Watts (W) which is Joules / second. From my intuitive argument, you see that P = I * V.

    If you now know the total time T that the current runs, you can calculate the total energy: simply multiply the amount of energy per second by the time: E = P * T. For example, if your vacuum cleaner uses 1200 Joules in a second (i.e. it says 1200 W on the side) and you use it for 30 seconds, you are using an energy of 1200 * 30 Joules = 36 000 Joules.

    Playing with the formulas gives the sequence
    E = P*T (energy is energy per second times number of seconds)
    = I*V*T (P = I * V, because energy per second is charge per second times energy per charge)
    = Q/T*V*T (I = Q / T, because current is charge per second)
    = Q*V (because the T cancels out).

    So depending on what you are given, you can use any of these.

    In this case, you know the total charge passing and the voltage drop across the resistor, and you are asked for the energy.

    Note that if they said that the 12 C passes in 3 seconds, you could have first calculated the power as well (or you could calculate the energy first and then calculate the power from that).
     
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