# find power and energy delivered

• Dec 27th 2010, 09:55 PM
bigwave
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

so somewhere??? also I didn't know what "plus to minus" meant... AC??
• Dec 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:

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

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

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?
• Dec 28th 2010, 12:09 PM
bigwave
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 (Cool)
• Dec 28th 2010, 12:14 PM
Ackbeet
You're welcome. Looking forward to more EE.