Solution of the RC circuit differential equation
ODE:
ansatz:
dq(t) q(t)
Em
+
=
sin(ωt)
dt
RC
R
q(t) = Q cos(ωt − φ)
(1)
(2)
where the Q and φ are adjustable parameters that will depend (in ways to be uncovered) upon R, C, ω, and
Em .
Plug the ansatz into the ODE, giving
−Qω sin(ωt − φ) +
Q
Em
cos(ωt − φ) =
sin(ωt).
RC
R
(3)
If this equation is to hold for all times, it must hold for the time t = 0:
−Qω sin(−φ) +
Define XC ≡
Q
cos(−φ)
RC
=
0
tan φ
=
−
(4)
1
ωRC
(5)
1
and call it the “reactance” or “impedance” of the capacitor (dimensions: ohm). Then
ωC
tan φ = −
XC
.
R
(6)
R
XXXφ
−XC
XXX
p
XX
X
R2 + XC2
XXX
If the ansatz is to hold for all times, then it must also hold for the time when ωt = φ, at which time
equation (3) becomes
Q
RC
=
Em
sin φ
R
Q = Em C sin φ = Em C
(7)
−p
XC
R2 + XC2
!
(8)
So we’ve shown that if ansatz (2) is a solution to ODE (1), then the parameters Q and φ must take on
the values given in equations (6) and (8). But we’ve not yet shown that ansatz (2) is really a solution. To
do so, plug equations (6) and (8) into (3)
Em C sin φ
cos(ωt − φ) =
RC
−ωRC sin φ sin(ωt − φ) + sin φ cos(ωt − φ) =
−ωEm C sin φ sin(ωt − φ) +
Em
sin ωt
R
sin ωt
(9)
(10)
and use the dreaded (at least, dreaded by me) sum angle formulas
sin(ωt − φ)
=
sin ωt cos φ − cos ωt sin φ
(11)
cos(ωt − φ)
=
cos ωt cos φ + sin ωt sin φ.
(12)
This gives (using ωRC = R/XC )
[−
R
R
sin φ cos φ + sin2 φ] sin ωt + [
sin2 φ + sin φ cos φ] cos ωt = sin ωt.
XC
XC
(13)
But from equation (6)
sin φ = − p
sin2 φ =
XC
+ XC2
XC2
R2 + XC2
R2
;
;
cos φ = p
R
+ XC2
RXC
sin φ cos φ = − 2
R + XC2
R2
(14)
(15)
whence (13) becomes
[1] sin ωt + [0] cos ωt = sin ωt.
Which is manifestly true!
(16)