Trig Product and Sum Help

Mar 2016
14
0
England
Hi

I am struggling with the following question.

The instantaneous power, p, in an electric circuit is given by p = iv,where v is the voltage and i is the current.

Calculate the maximum value of power in the circuit if

\(\displaystyle v = 0.02\sin (100\pi t)\) volts

\(\displaystyle i = 0.6\sin (100\pi t + \frac{\pi}{4} )\) amps

Calculate the first time that the power reaches a maximum value

I have started with the two multiplied together which makes

\(\displaystyle p = V(max) * I(max) \sin (100\pi t) \sin (100\pi t + \frac{\pi}{4} ) \)

Using the trigonometric formula 2.sin A.sin B = Cos(A-B) - Cos(A+B)

\(\displaystyle p = \frac{V(max) I(max)}{2} \cos 100\pi t - \frac{V(max) I(max)}{2} \cos (2 * 100\pi t + \frac{\pi}{4}) \)

therefore \(\displaystyle p = \frac{0.02 * 0.6 }{2} \cos 100\pi t - \frac{0.02 * 0.6V}{2} \cos (2 * 100\pi t + \frac{\pi}{4}) \)

I dont know what the next move is, or even if i am i going along the right lines??

Thanks for your help
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
check your algebra

$A=100\pi t$

$B=100 \pi t + \dfrac \pi 4$

$A-B =-\dfrac \pi 4$

$\cos(-\pi/4)=\cos(\pi/4)$

so one of your trig terms ends up being a constant.

You can then easily find the maximum of the other trig term.
 
Mar 2016
14
0
England
Thanks for your help romsek,

i have re-calculated this and come up with

\(\displaystyle p = \frac{0.02 * 0.6}{2} [ \cos ( 2 * 100 \pi t + \frac{\pi}{4}) + \cos ( - \frac{\pi}{4}) \)

hence


\(\displaystyle p = \frac{0.02 * 0.6}{2} [ \cos ( 2 * 100 \pi t + \frac{\pi}{4}) + \cos ( \frac{\pi}{4}) \)

I am a little unsure how to progress from here??
 
Mar 2016
14
0
England
The maxmum of the

\(\displaystyle [ \cos ( 2 * 100 \pi t + \frac{\pi}{4}) = 1 \)

So this makes

\(\displaystyle p = \frac{0.02 * 0.6}{2} [ \cos ( 1 + \cos \frac{\pi}{4}) \)

Is this correct?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$P = 0.006\bigg[\cos\left(-\dfrac{\pi}{4}\right)-\cos\left(200\pi t + \dfrac{\pi}{4}\right)\bigg]$

$P = 0.006\bigg[\dfrac{\sqrt{2}}{2}-\cos\left(200\pi t + \dfrac{\pi}{4}\right)\bigg]$

max power will occur when $\cos\left(200\pi t + \dfrac{\pi}{4}\right)=-1$

$P_{max} = 0.006\bigg[\dfrac{\sqrt{2}}{2}+1\bigg]$
 
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Mar 2016
14
0
England
Thanks Skeeter

So this works out to a value of about 0.01W.

How would i go about working out the first time that the power reaches a maximum value?

Thanks again for your assistance!
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
max power will occur when $\cos\left(200\pi t + \dfrac{\pi}{4}\right)=-1$ and note that $\cos{\pi} = -1$

what now?
 
Mar 2016
14
0
England
does that make

\(\displaystyle \cos (200 \pi t + \frac{\pi}{4} ) = \cos \pi \)

And a trigonometric equation?

\(\displaystyle \cos (200 \pi t + \frac{\pi}{4}) - \cos \pi = 0 \)

Is this correct?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
how about ...

$200\pi t + \dfrac{\pi}{4} = \pi$

... solve for $t$
 
Mar 2016
14
0
England
Thanks Skeeter

\(\displaystyle t = \frac {\pi - \frac {\pi}{4}} {200 \pi} \)

t = 3.75 * 10^-3

Thanks for your help Skeeter!!