RLC Circuit Find inductance and capacitance


by dfs730
Tags: capacitance, circuit, inductance, rlc
dfs730
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#1
Feb27-11, 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.
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gneill
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#2
Feb27-11, 01:38 PM
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Do you have a strategy? What do you know? What equations are relevant? Where's your attempt?
dfs730
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#3
Feb27-11, 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...

gneill
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#4
Feb27-11, 06:13 PM
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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 ωo. 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).
dfs730
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#5
Feb27-11, 06:46 PM
P: 10
So ωo = 2pif = (2pi)1.0 khz

and the resistance would have something to do with Q?
gneill
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#6
Feb27-11, 06:57 PM
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Quote Quote by dfs730 View Post
So ωo = 2pif = (2pi)1.0 khz

and the resistance would have something to do with Q?
Yes, the resistance is where energy is dissipated. But in this case you're given specific information about how the energy is lost (per cycle of oscillation). You can determine the Q from that.
dfs730
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#7
Feb27-11, 07:32 PM
P: 10
Q= 2pif x (energy stored / energy dissipated per cycle)
= ωo(energy stored / energy dissipated per cycle)
= ωo(0.01)
?
SammyS
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#8
Feb27-11, 07:47 PM
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Quote Quote by dfs730 View Post
Q= 2pif x (energy stored / energy dissipated per cycle)
= ωo(energy stored / energy dissipated per cycle)
= ωo(0.01)
?
Problem statement: The circuit loses (dissipates) 1.00% of its energy during each cycle.

The Quality Factor, Q0, is the ratio: (energy stored)/(energy dissipated) for each cycle.

What is (energy stored)/(1.00% of energy stored) ?
dfs730
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#9
Feb27-11, 07:53 PM
P: 10
oh, so Qo= 100
gneill
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#10
Feb27-11, 07:56 PM
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Quote Quote by dfs730 View Post
Q= 2pif x (energy stored / energy dissipated per cycle)
= ωo(energy stored / energy dissipated per cycle)
= ωo(0.01)
?
It's simply energy stored/energy lost for a given cycle. You're told that 1% of the energy is lost per cycle. Imagine that there happens to be 100 units of energy (you don't care what the units are) that begin a cycle. A 1% loss represents 1 unit of energy. So Q = 100/1 = 100.

Now, there are expressions for the natural frequency ωo and Q for RLC circuits. These involve the circuit components R, L, and C (naturally). Since you have R, with the expressions for ωo 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 ωo and Q for RLC circuits?
dfs730
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#11
Feb27-11, 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!
gneill
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#12
Feb27-11, 08:36 PM
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Watch your orders of magnitude. I put the capacitance in the ~1μF range, and the inductance around 30 mH.


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