Direct Energy, 2018a

260 11.6 Capacitor Inductor Example
(
L t, Q, dQ )
= 1
dt 2C Q2 − 1 ( ) 2 dQ
2 L (11.55)
dt
We can nd the path, charge on the capacitor as a function of time, by
solving for the least action
ˆ t2
(
δ
∣ L t, x, dx )
dt
dt ∣ =0 (11.56)
or by solving the Euler-Lagrange equation,
t 1
∂L
∂Q − d dt
∂L
) =0. (11.57)
∂ ( dQ
dt
In Eq. 11.56, δ indicates the rst variation as dened by Eq. 11.13. Solutions
depend on initial conditions such as the charge stored in the capacitor
and the current in the inductor at the initial time. We can use the Euler-
Lagrange equation to nd the equation ofmotion. The rst term ofEq.
11.57 is the generalized potential,
∂L
∂Q = Q C
(11.58)
which is the voltage v in volts.
generalized momentum.
The next term is the derivative ofthe
M =
∂L
∂ ( dQ
dt
) = −L dQ
dt
(11.59)
We can put the pieces together to nd an expression ofconservation ofthe
generalized potential.
Q
C + Q
Ld2 =0 (11.60)
dt2 This is a statement ofKirchho's voltage law. It looks more familiar ifit
is written in terms ofvoltage v = Q and current i C L = − dQ . dt
v − L di L
dt
=0 (11.61)
We can solve the equation ofmotion, Eq. 11.60, using appropriate initial
conditions, to nd the path. As in the mass spring example, Eq. 11.60 is
the wave equation, and its solutions are sinusoids. As expected, a circuit
made ofonly a capacitor and inductor is an oscillator.