Application of ODEs 6. Series RC Circuit PDF

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1/27/2020

Application of ODEs: 6. Series RC Circuit

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6. 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. (See the related section Series RL Circuit in the previous section.) In an RC circuit, the capacitor stores energy between a pair of plates. When voltage is applied to the capacitor, the charge builds up in the capacitor and the current drops off to zero.

Case 1: Constant Voltage The voltage across the resistor and capacitor are as follows:

V R = Ri and

1 ∫ i dt C

VC =

Kirchhoff's voltage law says the total voltages must be zero. So applying this law to a series RC circuit results in the equation:

Ri +

1 ∫ i dt = V C

One way to solve this equation is to turn it into a differential equation, by differentiating throughout with respect to t:

R

di i =0 + dt C

Solving the equation gives us:

i=

V −t/RC e R

Proof

We start with:

R

di i =0 + dt C

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Application of ODEs: 6. Series RC Circuit

Divide through by R:

di 1 )i = 0 +( RC dt We recognise this as a first order linear differential equation. Identify P and Q:

1 RC

P =

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 Applying the linear first order formula:

ie t/RC = K Since i =

V when t = 0 : R

K=

V R

Substituting this back in:

ie t/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. This equation does not apply if the voltage source is variable. 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. The current stops flowing as the capacitor becomes fully charged

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Application of ODEs: 6. Series RC Circuit

i V R

t τ

2τ

3τ

4τ

5τ

V − (t/RC ) Graph of i = , an exponential decay curve. e R

Applying our expressions from above, we have the following expressions for the voltage across the resistor and the capacitor:

V R = Ri = V e−t/RC VC =

1 ∫ i dt = V (1 − e−t/RC ) C

While the voltage over the resistor drops, the voltage over the capacitor rises as it is charged:

V V VC

VR

t τ

2τ

3τ

4τ

Graphs of V R = V e−t/RC (in green) and V C = V (1 − e

5τ −t/RC

) (in gray).

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. So we will make the substitutions:

i=

dq dt

and

q = ∫ i dt and so the equation in i involving an integral:

Ri +

1 ∫ i dt = V C

becomes the differential equation in q:

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Application of ODEs: 6. Series RC Circuit

dq 1 R + q =V dt C Example 1 A series RC circuit with R = 5 W and C = 0.02 F is connected with a battery of E = 100 V. At t = 0, the voltage across the capacitor is zero. (a) Obtain the subsequent voltage across the capacitor. (b) As t → ∞, find the charge in the capacitor. 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. Using the formulas V C = V (1 − e−t/RC ) and i =

V −t/RC . e R

Method 1 - Solving the DE in q 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 0.02

5

dq + 50q = 100 dt

dq + 10q = 20 dt We can solve this DE 2 ways, since it is variables separable or we could do it as a linear DE. The algebra is easier if we do it as a linear DE. Solving this differential equation as a linear DE, we have:

IF = e10t 10t

So qe

= ∫ (e10t )20 dt = 2e10t + K

So q = 2 + Ke−10t Now, since q (0) = 0, (that is, when t = 0 , q = 0 ) this gives: K = −2 . So q = 2(1 − e−10t ).

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Application of ODEs: 6. Series RC Circuit

q 2 1.5 1 0.5

t 0.1

0.2 −10t

Graph of q = 2(1 − e

0.3

0.4

0.5

0.6

), solution of a differential equation.

As t → ∞, q → 2C. Now,

VC =

1 ∫ i dt C

=

1 q C

=

1 2 (1 − e−10t ) 0.02

= 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.) Since t = 0 , q = 0 , we have K = − ln 2 . 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 Now

−t/RC

) and i =

V −t/RC e . R

1 1 = 10 = RC 5 × 0.02

So: https://www.intmath.com/differential-equations/6-rc-circuits.php

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V C

=

i= =

Application of ODEs: 6. Series RC Circuit

V (1 − e−t/RC ) = 100(1 − e−10t )

V −t/RC e R 100 −t/0.1 e 5

= 20e−10t Now

q = ∫ i dt = ∫ 20e−10t dt = −2e−10t + K 1 From here, we use q (0) = 0 and obtain: K 1 = 2. So q = 2(1 − e−10t ), as before. Also, as t → ∞, q → 2C .

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. Assume that when the switch is closed at t = 0, the charge on the capacitor is -0.05 C. Answer We will solve this 2 ways: 1. Solving in q. 2. Using Scientific Notebook. [NOTE: We cannot use the formulae V C = V (1 − e−t/RC ) and i =

V −t/RC e , since the voltage source is not r

constant.] From the formula: Ri +

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 = 8.5 cos 150t dt Now, we can solve this differential equation in q using the linear DE process as follows:

IF = e25t

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e25t q = ∫ e 25t8.5 cos 150t dt = 8.5 ∫ e 25t cos 150t dt Then we use the integration formula (found in our standard integral table):

∫ emt cos nt dt =

emt ( m cos nt + n sin nt) + n2

m2

We obtain:

e25t q = 8.5 ∫ e 25t cos 150t dt = 8.5

e25t ( 25 cos 150t+ 150e25t sin 150t) 23125

= 0.0092e25t cos 150t+ 0.055e25t sin 150t + K Dividing throughout by e

25t

gives:

q = 0.0092 cos 150t+ 0.055 sin 150t+ Ke−25t We now need to find K :

q (0) = −0.05 means K = −0.05 − 0.0092 = −0.0592 So this gives us:

q = 0.0092 cos 150t+ 0.055 sin 150t− 0.0592e−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 = 8.5 cos 150t dt q (0) = −0.05 Choosing Solve ODE - Exact from the Compute menu gives: Exact solution is:

q (t) = 0.0092 cos 150t+ 0.055 sin 150t− 0.059e−25t The graph for q (t) is as follows:

0.06

q

0.04 0.02

-0.02

t 0.05

0.1

0.15

0.2

-0.04 -0.06 -0.08 Graph of current q at time t. It's in steady state by around t = 0.12 .

We are also asked to find the current. We simply differentiate the expression for q:

i=

d (0.0092 cos 150t+ 0.055 sin 150t − 0.0592e−25t ) = −1.38 sin 150t+ 8.25 cos 150t+ 1.48e−25t dt

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Application of ODEs: 6. Series RC Circuit

The graph for i(t):

10

i

8 6 4 2 -2 -4

t 0.05

0.1

0.15

0.2

-6 -8 Graph of current i at time t. It's also in steady state by around t = 0.12 .

Example 3 In the RC circuit shown below, the switch is closed on position 1 at t = 0 and after 1 τ is moved to position 2. Find the complete current transient.

Answer At t = 0, the switch is at Position 1. We note that q (0) = 0 .

R

dq 1 1 + q1 = V dt C

500

dq 1 1 q 1 = 20 + dt 0.5 × 10−6

dq 1 + 4000q1 = 0.04 dt Using SNB to solve this differential equation, we have:

q1 (t) =

1 (1 − e−4,000t ) 100, 000

NOTE: By differentiating, this gives us:

i1 ( t) =

d 1 (1 − e −4000t ) = 0.04e−4000t dt 100000

We need to find τ :

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Application of ODEs: 6. Series RC Circuit −6

τ = RC = 500 × 0.5 × 10

= 0.00025.

Now, at t = 0.00025, the charge will be:

q1 (0.00025) =[

1 (1 − e−4,000t )] 100, 000 t=0.00025

= 6.3212 × 10−6 Position 2 At t = τ , switch at Position 2: Applying the formula R

500

dq 2 1 + q 2 = V again: dt C

dq 2 1 q 2 = −40 + 0.5 × 10−6 dt

NOTE: The negative voltage is because the current will flow in the opposite direction through the resistor and capacitor. Once again, we solve using Scientific Notebook:

500

dq 2 1 q 2 = −40 + 0.5 × 10−6 dt

q2 (0) = 6.3212 × 10−6 Exact solution is:

q2 ( t) = −0.00002 + 2.6321 × 10−5 e−4000t So the current transient will be:

i2 = =

dq 2 dt

d ( −0.00002 + 2.6321 × 10− 5e−4000t ) dt

= −0.10528e−4000t This expression assumes that time starts at t = 0 . However, we moved the switch to Position 2 att = 0.00025 , so we need:

i2 = −0.10528e−4000 (t−0.00025) = −0.10528e1−4000t So the complete current transient is:

i1 ( t) = 0.04e−4000t for 0 ≤ t ≤ 0.00025 i2 ( t) = −0.10528e1−4000t for t > 0.00025 The graph is quite interesting:

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Application of ODEs: 6. Series RC Circuit

i 0.04 0.02 -0.02

t 0.00025 0.0005 0.00075

0.001

-0.04 -0.06 -0.08 -0.1 -0.12 Graph of current i at time t.

Do not try this next one at home! Here's a great Java-based RLC simulator (on an external site). He is actually making a coil gun. You can play with each of R, L and C and see the effects. Play and learn :-) RLC Simulator Sorry, it won't work on a mobile device.

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