LRC Circuits
In this presentation we will go over the derivation of the current
in a circuit with a resistor, capacitor, and inductor being
powered by an alternating current signal.
Anthony Ruth
February 24, 2012
The System
We begin with a picture of an LRC circuit. Using Kircho's loop
rule we get equation 1.
Vs + VR + VL + VC
=0
(1)
Voltages
We need to substitute in expressions for the voltages. Equation 2
has the form of each voltage.
VS
= Vo sin(wt )
VR
= −IR
VL = −L
dI
dt
VC
=−
Q
(2)
C
The source is an alternating voltage with amplitude V0 oscillating
at angular frequency w . The voltage across the resistor is given by
ohm's law. The voltage on the inductor is the rate of change of
magnetic ux which is equal to the inductance times the rate of
change of current. The voltage on the capacitor is the charge on
the capacitor divided by the capacitance.
Charge Dierential Equation
Substituting in the expressions for the voltages we get equation 3.
V0 sin(wt ) − IR − L
dI
dt
−
Q
C
=0
(3)
Now we need to get the equation into a single variable. Recall that
the current is the rate of change of charge on the capacitor as
given by equation 4.
I
=
dQ
dt
(4)
I will substitute equation 4 into equation 3 and rearrange to obtain
equation 5.
Q
LC
+
R dQ
L dt
+
d 2Q
dt 2
=
V0
sin(wt )
L
(5)
Charge Dierential Equation
To simplify the math to come we will use the substitutions in
equation 6.
1
LC
= w02
R
L
= 2a (6)
This gives us a dierential equation that we can solve for the
current. Equation 7 is what we must solve.
w02 Q + 2a
dQ
dt
+
d 2Q
dt 2
=
V0
sin(wt )
L
(7)
Equation 7 is a second order linear inhomogenous dierential
equation. To solve it we start with the homogenous part given by
equation 8.
w02 Q + 2a
dQ
dt
+
d 2Q
dt 2
=0
(8)
Solving The Homogenous Dierential Equation
Linear homogenous dierential equations have solutions of the form
given by equation 9. Since this is the homogenous part of the
equation, I will label it as QG for general.
QG
= De kt
(9)
Substituting this back into equation 8 gives us equation 10.
w02 + 2ak + k 2 = 0
(10)
Equation 10 is easily solved for k by using the quadratic formula. k
is given by equation 11.
q
k = −a + a2 − w02
(11)
We can solve for D by noting that when t = 0, Q is the charge
intially on the capacitor which we will call Q0 . The solution to the
homogenous equation is given by equation 12.
QG
√2 2
= Q0 e −at e t a −w0
(12)
Solving The Inhomogenous Dierential Equation
We now return to equation 7.
w02 Q + 2a
dQ
dt
+
d 2Q
dt 2
=
V0
sin(wt )
L
Consider a solution of the form given by equation 13. I labeled this
part of the solution as QP for particular.
QP
= Ecos (wt ) + Fsin(wt )
(13)
Plugging this into equation 7 gives us equation 14.
w02 Ecos (wt ) + w02 Fsin(wt ) − 2aEwsin(wt ) + 2aFwcos (wt )
V0
− Ew 2 cos (wt ) − Fw 2 sin(wt ) =
sin(wt ) (14)
L
Solving The Inhomogenous Dierential Equation
We can seperate the terms involving cosine and the terms involving
sine as in equation 15
w02 Ecos (wt ) + 2aFwcos (wt ) − Ew 2 cos (wt ) = 0
w02 Fsin(wt ) − 2aEwsin(wt ) − Fw 2 sin(wt ) =
V0
sin(wt ) (15)
L
These two equations can be solved for E and F to give equation 16.
E
=−
L((w
2
2awV0
− w02 )2 − 2aw )
F
=−
L((w
2
−w
This gives us Qp in equation 17
QP
=−
L((w
2
2awV0
−w
2 2
0)
− 2aw )
cos (wt )−
L((w
V
0
2aw
2) −
)
0
(w 2 −w02 )
2
−w
V
0
2) −
2aw
2
(16)
2
)
sin(wt )