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Homework Help: Determining current from charge function

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    the amount of charge flwoing past a point in a current in a circuit is found to vary with time as follows:

    Q(t) = 100[e^(+t/100) + 0.01t^2]

    what is the current flowing at t = 10 seconds?

    i am supposed to get 47.2 amperes

    2. Relevant equations

    C = Q/V ---> Q = CV where C is capacitance, V is electric potential, Q is charge

    V = IR ----> I = V/R = dW/dt where R is resistance, W is work, t is time

    3. The attempt at a solution

    first solve the function in terms of t = 10 seconds

    Q(10) = 100[e^(+10/100) + 0.01(10^2)] = 100[ 1.105 + 1] = 100(2.105) = 210.5 coulombs

    since Q = CV and V = IR ---> Q = CIR now if i let Q = 210.5 coulombs, i don't have values for R or C, i'm solving for I.

    am i using the wrong approach entirely or the wrong equation?

    help appreciated
    Last edited: Jun 11, 2008
  2. jcsd
  3. Jun 11, 2008 #2


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    The current flowing is numerically equal to the The rate of change of charge per unit time
  4. Jun 11, 2008 #3
    do you mean I = dQ/t? i thought I = dW/dt, anyways if i used I = dQ/t = 210.5/10 = 21.05 amperes, i'm supposed to get 47.2 amperes

    by the way, did i solve for charge Q correctly?
  5. Jun 11, 2008 #4


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    I = dQ/dt... meaning you have to differentiate Q(t).
    I(10) = dQ/dt | t=10, not I = Q(10)/10... they are completely different expressions.
  6. Jun 11, 2008 #5


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    but I think you'll get the 21.05A. Not sure how to squeeze the 47.2A answer though.
    But that is how you would do the problem,using [itex]I=\frac{dQ}{dt}[/itex]
  7. Jun 11, 2008 #6
    you are right, dW/dt is power, i must've gotten it confused with dQ/dt, thanks for clearing that up.

    did i solve for Q correctly though?
  8. Jun 11, 2008 #7


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    Looks correct to me.
  9. Jun 11, 2008 #8
    does anybody know why i am getting 21.05 ampere, when i should be getting 47.2 ampere?

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