320 Week 9: Alternating Current Circuits
+Q
C
S
L
R
Figure 122: UndrivenLRCcircuitwhere
I=−dQ
dt
(713)
If we substitute this relation in for theI’s and divide byL, we get the following second order,
linear, homogeneous ordinary differential equation:
d^2 Q
dt^2 +R
L
dQ
dt+Q
LC= 0 (714)
We recognize this as the differential equation for adamped harmonic oscillator. To solve it, we
“guess”^92 :
Q(t) =Q 0 eαt (715)
and substitute this into the ODE to get the characteristic:
α^2 +R
L
α+1
LC
= 0 (716)
We solve for:α = −R
2 L
±
√(
R
L) 2
−LC^4
2
= −
R
2 L
±iω 0√
1 −
R^2 C
4 L
= −
R
2 L
±iω 0√
1 −
τL
4 τR= −R
2 L
±iω′
(717)whereτL=R/L τC= 1/RC,ω′= 0
√
1 − 4 ττLR, and our final solution looks like:Q(t) =Q 0 e−Rt 2 L
cos(ω′t) (718)(after we choose the real part of the complex exponential and use the initial conditions).
Note well the analogy with the damped, undriven harmonic oscillator.
From this we can easily find the current through and voltage acrossall of the elements of the
circuit. Finally, given the current and voltages it is easy to show thatenergy is conserved, that the
initial energy stored in the capacitor exactly balances the energy consumed in the resistor ast→∞.
This is again left as an exercise – the more of this that you work out onyour own (it is quite easy
- compute thepowerdelivered to each circuit element over time and integrate over time tofind the
total energy consumed by the resistor or residual in the capacitor or inductor) the better you will
learn it.
(^92) Not really.