106
3.7
CHAPTER 3. SECOND ORDER ODE
Application-LRC Circuits
The charge in an LRC-circuit can also be modeled by a second order linear
DE with constant coefficients. The DE is
Lq 00 + Rq 0 +
1
q = 0.
C
Here q(t) is charge at some specified point in the circuit, and the terms in
the DE are voltages.
We briefly describe the derivation from physics. For a capacitor,
q(t) = Cv(t),
where C is the capacitance. So
vcapacitor (t) =
1
q(t).
C
The voltage across an inductor is equal to
vinductor (t) = L
d2 q
,
dt2
where L is the inductance. The voltage across a resistor is equal to
vresistor (t) = R
dq
,
dt
where L is the resistance (this is Ohm’s Law).
From Kirchhoff’s Law, these voltages all sum to zero, so we obtain
Lq 00 + Rq 0 +
1
q = 0.
C
If an external voltage E(t) source is added to the system, then the sum
will not be zero, but will need to be E(t). So we obtain
1
q = E(t).
C
Note that if voltage in this DE is measured in volts, then q(t) is measured
in Coulombs, L are Henries, R is in Ohms, C is in Farads, and q 0 (t) is amperes
(amps).
Lq 00 + Rq 0 +
3.7. APPLICATION-LRC CIRCUITS
107
Example 3.27 An LRC-circuit is attached to a 9 volt battery where L = 1
Henry, R = 2 Ohms, and C is 31 Farads. Find the steady state for q(t).
Solution: Using undetermined coefficients, we set q(t) = A and plug this
into the DE
q 00 + 2q 0 + 3q = 9.
We obtain q(t) = 3, so the charge limits to 3 coulombs. Note that this would
imply that the capacitor would eventually have a voltage difference of 9 volts
across it.
¤
Example 3.28 An LRC-circuit is attached to a 12 volt battery is turned on
and off periodically so that E(t) = 6 sin t + 6. If the circuit has L = 1 Henry,
R = 2 Ohms, and C is 31 Farads. Find the steady state for q(t).
Solution: Using undetermined coefficients, we set q(t) = A sin t+B cos t+C.
From plugging to the DE:
−A sin t + −B cos t − 2B sin t + 2A cos t + 3A sin t + 3B cos t + 3C = 6 sin t + 6
We obtain 2A − 2B = 1 and 2A + 2B = 0 and C = 2 so A = 14 , B = − 41
¤
and C = 2. Thus, q(t) = 14 sin t − 14 cos t + 2.
As one can see, LRC circuits behave in exactly the same manner as spring
mass systems. Hence, just as spring mass systems, LRC circuits can be
characterized as overdamped, underdamped, or critically damped, and can
exhibit resonance if the resistance is low enough.
Exercises
1. An LRC circuit has inductor with inductance f rac12 Henry, resistor
with resistance 4 Ohms, and capacitor with capacitance 2 Farads. If
the initial charge on the capacitor is 5 coulombs, with initial current of
2 amps, find the charge on the capacitor at any time t.
2. An LRC circuit has inductor with inductance 3 Henries, resistor with
1
resistance 1 Ohm, and capacitor with capacitance 10
Farads. It is
attached to an 10 volt battery. Find the steady state solution.
108
CHAPTER 3. SECOND ORDER ODE
3. LRC circuits come in 3 varieties, Find conditions on L, R, and C so
that the circuit is (a) overdamped (b) critically damped (c) underdamped (d) lightly underdamped. Find the resonance frequency for a
sinusoidally driven lightly underdamped LRC circuit.
3.8
Wronskians and Variation of Parameters
In this section, we will need the concept of the Wronskian defined below:
3.8.1
The Wronskian
Wronskian of 2 functions
Define the Wronskian of two differentiable functions f (t) and g(t) to be
¯
¯
¯ f (t) g(t) ¯
¯
¯ = f (t)g 0 (t) − g(t)f 0 (t)
W [f, g](t) = ¯ 0
f (t) g 0 (t) ¯
Example 3.29 Compute the Wronksian of f (t) = t2 and g(t) = sin(6t)
Solution:
¯
¯
¯ f (t) g(t) ¯
¯
¯ = f (t)g 0 (t) − g(t)f 0 (t),
W [f, g](t) = ¯ 0
f (t) g 0 (t) ¯
so
¯ 2
¯ t
sin(6t)
W [t , sin(6t)] = ¯¯
2t 6 cos(6t)
2
¯
¯
¯
¯
= 6t2 cos(6t) − 2t sin(6t). ¤
√
Example 3.30 Compute the Wronksian of f (t) = t and g(t) = e2t
Solution:
¯
¯
¯ f (t) g(t) ¯
¯ = f (t)g 0 (t) − g(t)f 0 (t),
W [f, g](t) = ¯¯ 0
f (t) g 0 (t) ¯
so
¯ √
√ 2t
¯ t
W [ t, e ] = ¯¯ √
1
2 t
¯
e2t ¯¯
2e2t ¯