Energy Conservation in RLC circuit

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Homework Help Overview

The problem involves calculating energy stored in a series RLC circuit at specific time intervals, as well as determining the energy dissipated in the resistor during that time. The context includes assumptions about slight damping and neglecting higher-order terms.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial calculation of energy using standard formulas and the challenges of integrating the expression for power dissipation. There are questions about the integration process and the handling of terms in the expansion of the current equation.

Discussion Status

Some participants have provided suggestions for integrating the terms resulting from the expansion, while others express uncertainty about their approach. There is an ongoing exploration of different methods to tackle the integration, with no explicit consensus reached yet.

Contextual Notes

Participants are operating under the assumption of slight damping and are neglecting terms proportional to α², which may affect their calculations and reasoning.

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?
 
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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?
 
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:
EDIT: Sorry for the double post. Computer froze. Is there any way to delete these?
 
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.
 
Ah, thank you. I always forget my trig identities, and then I feel silly.
 

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