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Power in a circuit with an inductor and integrals

  • Thread starter bokonon
  • Start date
  • #1
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1. The power into a circuit element in the product of the voltage across the element and the current through the element. Assuming a voltage v(t) = V_p cos(omega*t) across inductor L, integrate the power over one cycle and show that the net energy into the inductor is zero.

V_p is the peak voltage

2. Homework Equations :
P=IV
I=(V_p/(omega*L))sin(omega*t)



Ok, so I'm really bad at calculus. I combined the three above equations to solve for Power, P and got: P=((V_p)^2/(omega*L))cos(omega*t)sin(omega*t). I assume I have to take the integral of this, but I'm not even sure what one cycle means. Can anybody point me in the right direction?

My best guess of the integral would be something like (since integral(sinxcosx)=.5sinx^2):

.5sin(omega*t)^2*V_p/(omega*L)

Thanks
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
In one complete cycle omega*t changes from zero to 2*pi. Take the integration between these limits.
 
  • #3
20
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Can you help me out a little bit more about how to take that integral?
 
  • #4
rl.bhat
Homework Helper
4,433
5
P= Vp^2/wL*sinwt*coswt = (Vp^2/2wL)*sin2wt.
Now find the integration with limits.
 

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