Power Dissipation by R at t=3.7ms:

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Homework Statement

The current through a resistor is i(t) = 2sin(30t + 169°), and R = 50Ω
What is the power dissipated by R at t = 3.7ms?

Homework Equations

p = i²R
or
p = .5(V_max)(I_max)*(cos([tex]\phi[/tex]-[tex]\theta[/tex])+cos(2*[tex]\omega[/tex]*t+[tex]\phi[/tex]+[tex]\theta[/tex]))

The Attempt at a Solution

I feel stupid for asking this but I just can't get the numbers right. I think this is a situation where the problem is so easy that it is difficult... anyway

i(t) = 2*sin(30t + 169°) = 2*cos(30t + 169°-90°)= 2*cos(30t + 79°)

so

p = i²R = (2*cos(30t + 79°))²*50= 4*50*(cos(30t + 79°))²

if t = 3.7ms = .0037s

4*50*(cos(30*.0037 + 79°))² = 7.1372 Watts

the actual answer is 1.31 Watts and I can't get this right. I must be making a stupid mistake somewhere that is so simple that I just can't see it. I also get my answer if I use the second formula I listed. Where is my mistake?

Thanks!

 

[SirAskalot]

You need to use either radians or degrees. You can't add apples and bananas.

 

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You need to use either radians or degrees. You can't add apples and bananas.

Ah... such a simple mistake I made. Now it comes out right if I multiply the 79 by pi/180 (aka put everything into radians). Thanks for that.