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05/01/2019

Application of ODEs: 6
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Application: Series RC Circuit

An RC series circuit

In this section we see how to solve the differential equation arising from a circuit consisting of a resistor and a capacitor
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)
In an RC circuit, the capacitor stores energy between a pair of plates
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Case 1: Constant Voltage
The voltage across the resistor and capacitor are as follows:

V R = Ri
and

VC =

1
∫ i dt
C

Kirchhoff's voltage law says the total voltages must be zero
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Application of ODEs: 6
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Identify P and Q:

P =

1
RC

Q=0
Find the integrating factor (our independent variable is t and the dependent variable is i):

∫ P dt = ∫

1
1
dt =
t
RC
RC

So

IF = et/RC
Now for the right hand integral of the 1st order linear solution:

∫ Qe∫ P dt dt = ∫ 0 dt = K
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Series RC Circuit

Applying the linear first order formula:

iet/RC = K
Since i =

V
when t = 0:
R

K=

V
R

Substituting this back in:

iet/RC =

V
R

Solving for i gives us the required expression:

V −t/RC
e
R

i=

Important note: We are assuming that the circuit has a constant voltage source, V
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The time constant in the case of an RC circuit is:

τ = RC
The function

i=

V −t/RC
e
R

has an exponential decay shape as shown in the graph
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Series RC Circuit

i
V
R

t
2τ

τ
Graph of i =

3τ

4τ

5τ

V −(t/RC )
e
, an exponential decay curve
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intmath
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2τ

3τ

4τ

5τ

(in green) and V C = V (1 − e

−t/RC

) (in gray)
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Series RC Circuit

Case 2: Variable Voltage and 2-mesh Circuits
We need to solve variable voltage cases in q, rather than in i, since we have an integral to deal with if we use i
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02 F is connected with a battery of E = 100 V
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(a) Obtain the subsequent voltage across the capacitor
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Answer
We will solve this 3 ways, since it has a constant voltage source:
1 and 2: Solving the DE in q, as:
a linear DE and
variables separable
3
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R

Method 1 - Solving the DE in q

1

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Series RC Circuit

From the formula: Ri +

R

1
∫ i dt = V , we obtain:
C

dq
1
+ q=V
dt
C

On substituting, we have:

5

dq
1
+
q = 100
dt
0
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The algebra is easier if we do it as a linear DE
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So q = 2(1 − e−10t )
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Application of ODEs: 6
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5
1
0
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1

0
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3

0
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5

0
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As t → ∞, q → 2 C
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02

= 100(1 − e10t )
For comparison, here is the solution of the DE using variables separable:

dq
= 10(2 − q )
dt
dq
= 10 dt
2−q
−ln∣2 − q ∣ = 10t + K
(We could continue and get the same expression as above
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Series RC Circuit

So

−ln∣2 − q ∣ = 10t − ln 2
− ln 2 + ln∣2 − q ∣ = −10t
2−q
= e−10t
2
2 − q = 2e−10t
q = 2(1 − e−10t )
Method 2:
We use the formulae V C = V (1 − e

−t/RC

Now

) and i =

V −t/RC
e

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02

So:

V C = V (1 − e−t/RC ) = 100(1 − e−10t )
i=
=

V −t/RC
e
R
100 −t/0
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Series RC Circuit

−10t

So q = 2(1 − e

), as before
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Example 2
Find the charge and the current for t > 0 in a series RC circuit where R = 10 W, C = 4 × 10 -3 F and E = 85 cos 150t V
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05 C
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Solving in q
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Using Scientific Notebook
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]
r

1
∫ i dt = V , we obtain:
C

dq
1
+ q=V
dt
C

Since R = 10, C = 4 × 10−3 , and V = 85 cos 150t, we have:

10

dq
1
+
q = 85 cos 150t
dt
4 × 10−3

10

dq
+ 250q = 85 cos 150t
dt

dq
+ 25q = 8
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5 cos 150t dt = 8
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Application of ODEs: 6
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5 ∫ e25t cos 150t dt
= 8
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0092e25t cos 150t+ 0
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0092 cos 150t+ 0
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05 means K = −0
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0092 = −0
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0092 cos 150t+ 0
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0592e−25t
Method Using Scientific Notebook
We set up the differential equation and the initial conditions in a matrix (not a table) as follows:

dq
+ 25q = 8
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05
Choosing Solve ODE - Exact from the Compute menu gives:
Exact solution is:

q (t) = 0
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055 sin 150t− 0
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Application of ODEs: 6
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06
0
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02

t
0
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02

0
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15

0
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04
-0
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08
Graph of current q at time t
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12
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We simply differentiate the expression for q:

i=

d
(0
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055 sin 150t − 0
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38 sin 150t+ 8
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48e−25t
dt

The graph for i(t):

10
8
6
4
2
-2
-4
-6
-8

i

t
0
...
1

0
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2

Graph of current i at time t
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12
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Find the complete current transient
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Application of ODEs: 6
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We note that q (0) = 0
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5 × 10−6

dq 1
+ 4000q 1 = 0
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04e−4000t
dt 100000

We need to find τ :

τ = RC = 500 × 0
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00025
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Application of ODEs: 6
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00025, the charge will be:

q 1 (0
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00025

= 6
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5 × 10−6

NOTE: The negative voltage is because the current will flow in the opposite direction through the resistor and capacitor
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5 × 10−6

q 2 (0) = 6
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00002 + 2
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00002 + 2
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10528e−4000t
This expression assumes that time starts at t = 0
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00025, so we need:
`i_2=-0
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00025))` `=-0
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Application of ODEs: 6
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04e^(-4000t)` for `0 <= t <= 0
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10528e^(1-4000t)` for `t>0
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04

i

0
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02

t
0
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0005 0
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001

-0
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06
-0
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1
-0
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Do not try this next one at home!
Here's a great Java-based RLC simulator (on an external site)
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You can play with each of V, R, L and C and see the effects
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