Energy Conservation in RLC circuit

  • Thread starter Opus_723
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  • #1
Opus_723
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Homework Statement



Assume the current in a series RLC circuit is given by I = ACω(sin(ωt) + [itex]\frac{α}{ω}[/itex]cos(ωt))e[itex]^{-αt}[/itex].

Calculate the energy stored in the circuit at t=0. Then calculate the energy stored in the circuit one-quarter cycle later, at t=[itex]\frac{\pi}{2ω}[/itex].

Verfiy that the difference is equal to the energy dissipated in the resistor R during this interval.

For this problem, assume the damping is slight, that is, that [itex]\frac{α}{ω}[/itex] << 1, and neglect quantities proportional to α[itex]^{2}[/itex].

The Attempt at a Solution



I'm pretty sure I have the first part right, since it seems like a straightforward use of [itex]\frac{CV^{2}}{2}[/itex] + [itex]\frac{LI^{2}}{2}[/itex]. A couple of e[itex]^{-αt}[/itex]'s simplify to 1 when you plug in 0 and [itex]\frac{\pi}{2ω}[/itex] for t since [itex]\frac{α}{ω}[/itex] << 1.

But for the next part, I think they're asking me to integrate I[itex]^{2}[/itex]Rdt over the interval, and I am not seeing how to approach that integral. Expanding the equation for I to get I[itex]^{2}[/itex] just makes a mess no matter what small terms I ignore. I don't know if I'm just being clumsy with my math or if I'm approaching it wrong. I even tried using complex numbers to represent the power but I ended up with a nonsensical answer. Any advice on tackling this part of the problem?
 

Answers and Replies

  • #2
voko
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I am not sure what mess you are getting. After the expansion, you should get three terms, one of which gets neglected, the other two are readily integrated. What is the difficulty you are having?
 
  • #3
Opus_723
178
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Well, after expanding, I get I[itex]^{2}[/itex]R = A[itex]^{2}[/itex]C[itex]^{2}[/itex]ω[itex]^{2}[/itex]R(e[itex]^{-2αt}[/itex]sin[itex]^{2}[/itex](ωt) + [itex]\frac{2α}{ω}[/itex]e[itex]^{-2αt}[/itex]sin(ωt)cos(ωt) + [itex]\frac{α^{2}}{ω^{2}}[/itex]e[itex]^{-2αt}[/itex]cos[itex]^{2}[/itex](ωt))

Of course the last term in the parentheses can be dropped, but I don't know how to integrate the other two. It's probably simple, but I'm not seeing it.
 
Last edited:
  • #4
Opus_723
178
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EDIT: Sorry for the double post. Computer froze. Is there any way to delete these?
 
  • #5
voko
6,054
391
Convert the trigonometric products to functions of double angle, then integrate the exp-trig products by parts - if you have to, otherwise those products are frequently given in tables of integrals.
 
  • #6
Opus_723
178
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Ah, thank you. I always forget my trig identities, and then I feel silly.
 

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