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Thread: 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

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

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

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

    find the power delivered to the element at $\displaystyle t = 0.3s$

    so I did thus...

    given that $\displaystyle \ v=24.01$

    then since $\displaystyle i = \frac{dq}{dt} = \frac{d}{dt}10\sin{4\pi t}
    \rightarrow
    40 \pi \cos(4 \pi t)$

    so at $\displaystyle t=0.3s$ then $\displaystyle i = -101.664$

    $\displaystyle 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:

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

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

    Proceeding as you did, you'd get this:

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

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

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

    $\displaystyle 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; Dec 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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