
#1
Feb2711, 01:05 PM

P: 10

The energy of an RLC circuit decreases by 1.00% during each oscillation when R=2.00 ohms. If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1.00 kHz. Find the values of inductance and capacitance.




#2
Feb2711, 01:38 PM

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P: 11,415

Do you have a strategy? What do you know? What equations are relevant? Where's your attempt?




#3
Feb2711, 05:28 PM

P: 10

w=1/(LC)^(1/2)
f=w/2pi = 1/(2pi(LC)^(1/2)) = 1.00kHz some how this is supposed to relate to the equation for a damped object on a spring. L(d^2Q/dt^2) + R(dQ/dt) + Q/C = 0 <> m(d^2x/dt^2) + b(dx/dt) + kx = 0 Other than this I really have no idea... 



#4
Feb2711, 06:13 PM

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P: 11,415

RLC Circuit Find inductance and capacitance
Okay, it may be a bit simpler than you think.
From what you have written you can determine the value of ω_{o}. Next determine the Q of the circuit. You're told that the energy decreases by 1% each cycle, so what is the Q? (hint: Q is energy stored / energy dissipated per cycle). 



#5
Feb2711, 06:46 PM

P: 10

So ωo = 2pif = (2pi)1.0 khz
and the resistance would have something to do with Q? 



#6
Feb2711, 06:57 PM

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P: 11,415





#7
Feb2711, 07:32 PM

P: 10

Q= 2pif x (energy stored / energy dissipated per cycle)
= ωo(energy stored / energy dissipated per cycle) = ωo(0.01) ? 



#8
Feb2711, 07:47 PM

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P: 7,395

The Quality Factor, Q_{0}, is the ratio: (energy stored)/(energy dissipated) for each cycle. What is (energy stored)/(1.00% of energy stored) ? 



#9
Feb2711, 07:53 PM

P: 10

oh, so Qo= 100




#10
Feb2711, 07:56 PM

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P: 11,415

Now, there are expressions for the natural frequency ω_{o} and Q for RLC circuits. These involve the circuit components R, L, and C (naturally). Since you have R, with the expressions for ω_{o} and Q you can solve for L and C. The tricky thing is trying to decide whether its a parallel RLC circuit or a series RLC circuit, because the expression for Q is different for each. What formulas have you learned for ω_{o} and Q for RLC circuits? 



#11
Feb2711, 08:19 PM

P: 10

ωo = 1/(LC)^1/2 > L = ((1/ωo)^2)/C
Q = (1/R)(L/C)^(1/2) > L = C(QR)^2 > C=1/QRωo = 7.96x10^(9) Farads > L=Q^2(R^2)(C) = 3.184 I think this looks right! Thanks a bunch, really appreciate it! 



#12
Feb2711, 08:36 PM

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P: 11,415

Watch your orders of magnitude. I put the capacitance in the ~1μF range, and the inductance around 30 mH.



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