Solving an Oscillatory Circuit

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Homework Help Overview

The discussion revolves around solving an oscillatory circuit, specifically focusing on the behavior of charge in an RLC circuit described by a second-order differential equation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants share their attempts at finding the charge Q, with some consistently arriving at the same expression. Others question the steps taken in deriving the characteristic equation and the interpretation of results.

Discussion Status

There is an ongoing exploration of the characteristic equation and the roots derived from it. Some participants have provided insights into the use of the quadratic formula, while others express confusion about the calculations and assumptions made in the process.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the application of the quadratic formula and the interpretation of coefficients in the characteristic equation. There is also mention of homework constraints that may influence the approach taken.

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Homework Statement



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Homework Equations





The Attempt at a Solution



Is the correct answer a choice?

I get Q= .17e-6cos(4000t), every single time.
 

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1MileCrash said:

Homework Statement



attached

Homework Equations





The Attempt at a Solution



Is the correct answer a choice?

I get Q= .17e-6cos(4000t), every single time.

Well, how do you get that?
 
Q'' + 4e6Q = 0 =>

r^2 + 4e6 = 0 (characteristic equation) => r = 0 +/- 4000i

=> Q = c1cos(4000t) + c2sin(4000t)
=> Q' = -4000c1sin(4000t) + 4000c2cos(4000t)

=> Q(0) = c1 = .17e-6
=> Q'(0) = 4000c2 = 0 => c2 = 0

So

Q = .17e-6cos(4000t)
 
1MileCrash said:
Q'' + 4e6Q = 0 =>

r^2 + 4e6 = 0 (characteristic equation) => r = 0 +/- 4000i

=> Q = c1cos(4000t) + c2sin(4000t)
=> Q' = -4000c1sin(4000t) + 4000c2cos(4000t)

=> Q(0) = c1 = .17e-6
=> Q'(0) = 4000c2 = 0 => c2 = 0

So

Q = .17e-6cos(4000t)

The square root of 4e6 is 2000, not 4000.
 
Dick said:
The square root of 4e6 is 2000, not 4000.

I'm confused. Why would I be taking the square root of 4e6?

Shouldn't I be taking the square root of b^2 - 4(a)(c)?

Which is 0 - 4(1)(4e6) = 16000000, and the square root of that is 4000.


Thanks
 
1MileCrash said:
I'm confused. Why would I be taking the square root of 4e6?

Shouldn't I be taking the square root of b^2 - 4(a)(c)?

Which is 0 - 4(1)(4e6) = 16000000, and the square root of that is 4000.


Thanks

You don't NEED to use the quadratic formula to solve r^2+4e6=0, but if you do you shouldn't forget there is a 2a in the denominator.
 
Dick said:
You don't NEED to use the quadratic formula to solve r^2+4e6=0, but if you do you shouldn't forget there is a 2a in the denominator.

Wow, I was thinking a = 1 and ignored it.

And I know I don't need to, it just like to for complex since it spits it out in the right form already.

That's what I get for rushing.

Thanks again
 

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