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Math Help - find power and energy delivered

  1. #1
    Super Member bigwave's Avatar
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    find power and energy delivered

    the charge entering the positive terminal of an element is

    q=10\sin{4\ \pi\ t} \ mC

    while the voltage across the element (plus to minus) is

    v=2\cos4\ \pi\ t \ V

    find the power delivered to the element at t = 0.3s

    so I did thus...

    given that \ v=24.01

    then since  i = \frac{dq}{dt} = \frac{d}{dt}10\sin{4\pi t}<br />
\rightarrow<br />
40 \pi \cos(4 \pi t)

    so at t=0.3s then i = -101.664

    p=vi\rightarrow (24.01)(-101.664) =

    its obvioiusly not correct

    the correct answer is 164.5mW

    so somewhere??? also I didn't know what "plus to minus" meant... AC??
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  2. #2
    A Plied Mathematician
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    The statement "given that v = 24.01" appears to be incorrect. You need to be more careful about the parentheses encasing the arguments to functions. To be crystal clear, you'd write it this way:

    q(t)=10\sin(4\pi t)\;[\text{mC}],

    v(t)=2\cos(4\pi t)\;[\text{V}].

    Proceeding as you did, you'd get this:

    i(t)=40\pi\cos(4\pi t)\;[\text{mA}]. Therefore, the expression for the power is

    P(t)=i(t)\,v(t)=80\pi\cos^{2}(4\pi t)\;[\text{mW}]. Plugging in t=0.3 gives the correct result.

    Does that help?
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  3. #3
    Super Member bigwave's Avatar
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    Cool best to multiply the expresions of i(t) and v(t) first

    P(t)=i(t)\,v(t)=80\pi\cos^{2}(4\pi t)\;[\text{mW}]. Plugging in t=0.3 gives the correct result.

    80\pi\cos^{2}(4\pi 0.3)=164.496 or 164.5 [mW] with wolframalpha

    so it is best to multiply the expresions of i(t) and v(t) first before we apply the t=0.3s

    agree about the () just wish this text book would do so it is confusing.. also like the [] around the units.
    yes thnks for help...more EE ?? to come
    Last edited by bigwave; December 28th 2010 at 12:10 PM. Reason: added units
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  4. #4
    A Plied Mathematician
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    You're welcome. Looking forward to more EE.
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