# Homework Help: Electric Circuits Problem

1. Aug 31, 2011

### VitaX

1. The problem statement, all variables and given/known data

[PLAIN]http://img535.imageshack.us/img535/7201/electriccircuits113.png [Broken]

2. Relevant equations

p = dW/dt; p = vi; W = Integral (p(t)) dt

3. The attempt at a solution

I think for the most part I got parts a and b correct. What I'm having significant trouble with is part c. The answer in the back of the book is 42.678 W. I believe I have to find power first, then find the energy from integrating that power and lastly take the derivative with respect to time in order to find the instantaneous power. The only thing is, what values of t are even present to plug in at the end to find the answer to part c? I'm getting confused on this one, badly.

Last edited by a moderator: May 5, 2017
2. Aug 31, 2011

### tiny-tim

Hi VitaX!
You should get a formula for power flow, as a function of t …

then just find the maximum.

But if you're still having difficulty, show us your full calculations so far

3. Aug 31, 2011

### VitaX

Well I'm looking at this http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powerac.html

Instantaneous Power = Vm*Im*Cos(theta)*Sin^2(wt) - Vm*Im*Sin(theta)*Sin(wt)*Cos(wt)

I know to find t I would have to plug in my values and then proceed to take the derivative and set equation equal to zero, but I'm having a lot of difficulty even attempting to take the derivative of this equation. Isn't there a way to simply this down somewhat for ease of use?

Edit: Nevermind with part C, I believe the teacher just wanted us to find the max using a graphing method.

Last edited: Sep 1, 2011
4. Sep 1, 2011

### tiny-tim

Hi VitaX!

(just got up :zzz: …)
For future reference, it would have been easier just to use P = VI and the original formulas,

V = 10sin(2π103t), I = 10sin(2π103t - 45°) = (10/√2)(sin(2π103t) - cos(2π103t))

Also that formula for P from hyperphysics isn't very informative …

an easier formula (from the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=303") is

$$P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)$$
$$=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2$$​
(because $2cosAcosB = cos(A-B) + cos(A+B))$)
$$=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)$$​

which clearly separates the constant part and the variable part

Last edited by a moderator: Apr 26, 2017