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Capacitors and Resistors

  1. Jul 31, 2013 #1
    1. At time t = 0, a 23 ohms resistor is linked to a 4.4V battery. How much charge has gone through the resistor after 5 seconds?

    2. Q = CΔV
    ΔV = IR

    3. ΔV = IR
    I = ΔV/R
    = 4.4/24 A.

    I'm not sure what to do next.
     
  2. jcsd
  3. Jul 31, 2013 #2

    rude man

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    Q = ∫I dt
     
  4. Jul 31, 2013 #3

    Curious3141

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    How did the 23Ω become a 24 when you wrote it down in the formula?

    Anyway, what is the relationship between a constant current, charge and time?
     
  5. Jul 31, 2013 #4

    vanhees71

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    Note that the current is not constant in this case!
     
  6. Jul 31, 2013 #5

    Curious3141

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    EDIT: At first, I thought I'd made a ghastly mistake because I'd forgotten a capacitor or something. But this is just a simple circuit with a battery and a resistor. Why can't the current be assumed to be constant?
     
  7. Jul 31, 2013 #6
    There is only a resistor and a battery, how is the current not constant?
    There is no indication of capacitance in the question details.
     
  8. Jul 31, 2013 #7
    The current is constaant. Find the current and find that charge using the time given. Straight foward!
     
  9. Jul 31, 2013 #8

    vanhees71

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    Ok, then I didn't understand the question. I thought there is a resistor and a capacitor. So the Q=CU equation doesn't make any sense.
     
  10. Jul 31, 2013 #9
    1 A = ?? coulombs/sec
     
  11. Jul 31, 2013 #10
    I figured it out. Thanks guys!
     
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