Week 9: Alternating Current Circuits 319
In the figure above, the capacitorCon the left is initially charged up to chargeQ 0. At time
t= 0 the switch is closed and current begins to flow. If we apply Kirchhoff’s voltage/loop rule to
the circuit, we get:
Q
C
−L
dI
dt= 0 (702)
where
I=−dQ
dt(703)
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+
Q
LC
= 0 (704)
We recognize this as the differential equation for aharmonic oscillator!To solve it, we “guess”^91 :
Q(t) =Q 0 eαt (705)and substitute this into the ODE to get the characteristic:
α^2 +1
LC
= 0 (706)
We solve for:
α=±i√
1
LC
=±iω 0 (707)and get:
Q(t) =Q0+e+iω^0 t+Q 0 −e−iω^0 t (708)
or (taking the real part and using the initial conditions):
Q(t) =Q 0 cos(ω 0 t) (709)Note well that this overall solution methodology isidenticalto that used for the simple harmonic
oscillator, with spring constantkeff=C^1 and massm=L.
One can, of course, analyze energy in this circuit. At any instant oftime, the energy in the
circuit is clearly all the energy stored in the capacitor:
UC(t) =Q(t)2
2 C(710)
This energyover timeoscillates between the capacitor and the energy in the inductor:
UL(t) =^1
2LI(t)^2 (711)Show that the sum of these two energies is a constant, and that the constant equals the initial
energy in the capacitor! This is precisely analogous to what happensto the conserved total energy
as it oscillates between potential energy in a spring and kinetic energy of motion of the mass in a
harmonic oscillator.
Non-driven LRC circuit
In the figure above, the capacitorCon the left is initially charged up to chargeQ 0. At timet= 0
the switch is closed and current begins to flow. If we apply Kirchhoff’svoltage/loop rule to the
circuit, we get:
Q
C−L
dI
dt−IR= 0 (712)(^91) Not really.