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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
III. Practice Problem 1: R-C DC Circuit
Task: Work on solving for the charge for the given circuit; indicated links give (partial)
solutions.
R=2
C=1/6
EMF=100
An R-C circuit consists of a 100 volt DC battery connected in series with a 2 ohm resistor
and a 1/6 farad capacitor; there is no initial charge on the capacitor and no current flows
until the switch is closed. The associated Initial Value Problem is
2 Q 0 (t) +
Q(t)
= 100,
1/6
Q(0) = 0
Method of Separation of Variables:
[a] Put the ODE in differential form.
[b] Use algebra to separate variables.
[c] Integrate both sides of the preceding equation to get a general solution to the ODE.
[d] Use the initial condition to compute the value of the arbitrary constant and determine
the implicit solution to the IVP.
[e] Use the rules of logs and exponentials in order to solve the implicit solution to the
ODE for the charge Q(t) for this circuit.
[f] Rework the solution when the capacitor in this circuit is charged to 30 coulombs at
the time when the switch is closed.
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
[a]
III. Practice Problem 1: R-C DC Circuit
Put the ODE in differential form.
The ODE
2 Q 0 (t) +
becomes
Q(t)
= 100
1/6
Q 0 (t) + 3 Q(t) = 50
or
dQ
= 50 − 3 Q
dt
which, after multiplying through by dt , becomes
dQ = (50 − 3 Q) dt
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
[b] Use algebra to separate variables.
The differential form
dQ = (50 − 3 Q) dt
can be written
dQ
= dt
50 − 3 Q
so that all terms involving Q are on one side of the equation and all terms involving t on
the other side.
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
[c] Integrate both sides of the preceding equation to get a general solution to the ODE.
Integrating both sides of
dQ
= dt
50 − 3 Q
gives
Z
dQ
=
50 − 3 Q
We use the substitution
dt
x = 50 − 3 Q
dx
= −3
dQ
1
dQ = − dx
3
to get
or
so that
Z
¶
µ
1
1
− dx
x
3
Z
1
1
=−
dx
3
x
1
= − ln |x| + C
3
1
= − ln |50 − 3 Q| + C
3
Hence the general solution to the ODE in implicit form is
Z
1
dQ =
50 − 3 Q
Z
1
− ln |50 − 3 Q| = t + C
3
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
[d] Use the initial condition to compute the value of the arbitrary constant and determine
the implicit solution to the IVP.
Since Q = 0 when t = 0 , we put t = 0 into
1
− ln |50 − 3 Q| = t + C
3
to get
1
1
− ln |50 − 0| = 0 + C =⇒ C = − ln 50
3
3
This leads us to the solution to the IVP in implicit form:
1
1
− ln |50 − 3 Q| = t − ln 50
3
3
or, multiplying through by −3 ,
ln |50 − 3 Q| = ln 50 − 3t
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
[e] Use the rules of logs and exponentials in order to solve the implicit solution to the
ODE for the charge Q(t) for this circuit.
Take the antilogarithm of both sides of the solution to the IVP
ln |50 − 3 Q| = ln 50 − 3t
to get
eln|50−3 Q| = eln 50−3t
which, when simplified, yields
|50 − 3 Q| = 50 e−3t
or
50 − 3 Q = ±50 e−3t
Since Q = 0 when t = 0 , the sign must be + , allowing us to solve for Q :
50 − 3 Q = +50 e−3t
or
Q(t) =
whose graph is plotted below.
¢
50 ¡
1 − e−3t
3
R-C Circuit: charge Q(t)
EMF=100 R=2 C=1/6
16
14
12
10
8
6
4
2
0
ODEs and Electric Circuits
0.5
1
1.5 t
6
2
2.5
3
3.5
III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
III. Practice Problem 1: R-C DC Circuit
[f] Rework the solution when the capacitor in this circuit is charged to 30 coulombs at
the time when the switch is closed.
Start with the general solution
1
− ln |50 − 3 Q| = t + C
3
from part [c] and substitute Q = 30 and t = 0 :
1
− ln |50 − 3(30)| = 0 + C
3
which yields
1
C = − ln(40)
3
Hence
1
1
− ln |50 − 3 Q| = t − ln(40)
3
3
which we can solve as before.
Multiply by −3 :
ln |50 − 3 Q| = −3t + ln(40)
Take antilogs:
|50 − 3 Q| = e−3t+ln(40)
Simplify exponentials:
e−3t+ln(40) = e−3t × eln(40) = e−3t × 40
to arrive at:
|50 − 3 Q| = 40 e−3t
or
50 − 3 Q = ±40 e−3t
Now Q = 30 when t = 0 means
50 − 3(30) = ±40 e0
or
−40 = ±40
So the sign must be − and the solution to the IVP is
50 − 3 Q = −40 e−3t
ODEs and Electric Circuits
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III. Practice Problem 1: R-C DC Circuit
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ODEs and Electric Circuits
or
Q(t) =
III. Practice Problem 1: R-C DC Circuit
¢
10 ¡
5 + 4e−3t
3
Charge (below left) and current (below right) are plotted for Q(0) = 0 (thinner curve)
and Q(0) = 30 (thicker curve).
R-C Circuit: charge Q(t)
EMF=100 R=2 C=1/6
R-C Circuit: current I(t)
EMF=100 R=2 C=1/6
30
25
40
20
20
15
10
0
5
0
0.2
0.4
0.6
0.8
t
1
1.2
1.4
1.6
–20
0.2
0.4
0.6
0.8
t
ODEs and Electric Circuits
1
1.2
1.4
1.6
8
–40
III. Practice Problem 1: R-C DC Circuit