find power and energy delivered

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December 27th 2010, 09:55 PM
bigwave

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??
December 28th 2010, 04:41 AM
Ackbeet

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?
December 28th 2010, 12:09 PM
bigwave

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 (Cool)
December 28th 2010, 12:14 PM
Ackbeet

You're welcome. Looking forward to more EE.