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Math Help - How to solve this exponential Word problem

  1. #1
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    How to solve this exponential Word problem

    A capacitor is an RC circuit discharges according to the equation
    where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second


    V(x) = Vmax e^-2t/3
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  2. #2
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    Quote Originally Posted by daunder View Post
    A capacitor is an RC circuit discharges according to the equation
    where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second


    V(x) = Vmax e^-2t/3

    Well you are looking for a t such that

    V(t)=.75V_{max}

    So we want to solve the equation

    .75V_{max}=V_{max}e^{-\frac{2}{3}t}

    Notice that V_{max} is on both sides of the equation so it reduces out to gives

    .75=e^{-\frac{2}{3}t}

    all that is left is to solve for t
    Last edited by TheEmptySet; March 31st 2009 at 05:37 PM. Reason: I made a mistake :( thanks e
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  3. #3
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    Quote Originally Posted by daunder View Post
    A capacitor is an RC circuit discharges according to the equation
    where V is the voltage, t is the time in seconds and Vmax is the initial voltage, Determine how long it will take for the capacitor to discharge 25% of its initial charge, to the nearest hundredth of a second

    <br />
V(x) = V_{max} e^{\frac{(2t)}{3}}<br />
    I'd use v(t) as it's a function of time. I assume the equation at the bottom is given to you? You want the time when 25% is gone at which point there will be (1-0.25=0.75) left.

    If so you know at time t V(t) = 0.75V_{max} which will lead to V_{max} cancelling.
    <br />
e^{-\frac{2t}{3}} = \frac{V(t)}{V_{max}} = 0.75

    <br />
-\frac{2t}{3} = ln(0.75)

    and solve for t
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    I'd use v(t) as it's a function of time. I assume the equation at the bottom is given to you? You want the time when 25% is gone at which point there will be (1-0.25=0.75) left.

    If so you know at time t V(t) = 0.75V_{max} which will lead to V_{max} cancelling.
    <br />
e^{-\frac{2t}{3}} = \frac{V(t)}{V_{max}} = 0.75

    <br />
-\frac{2t}{3} = ln(0.75)

    and solve for t
    K i didn't write the question properly
    It said Determine how long it will take for the capacitor to discharge to 25% of its initial charge

    so, is this correct ?
    0.25 = e^-2t/3

    ln 0.25 = -2t/3 ln e
    t = -3ln0.25/2
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by daunder View Post
    K i didn't write the question properly
    It said Determine how long it will take for the capacitor to discharge to 25% of its initial charge

    so, is this correct ?
    0.25 = e^-2t/3

    ln 0.25 = -2t/3 ln e
    t = -3ln0.25/2
    Yep, that's correct
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  6. #6
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    Thanks for your help
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