EE 201 ELECTRIC CIRCUITS
LECTURE 23
The material covered in this lecture will be as follows:
⇒ First Order Circuits
⇒ Review of First Order Differential Equations and Solution
⇒ Natural Response of the RC Circuit
⇒ The Time Constant of the RC Circuit
At the end of this lecture you should be able to:
⇒ Understand the term “first order circuits”.
⇒ Understand the term “natural response of a circuit”.
⇒ Recognize first order differential equations.
⇒ Find the general solution of first order differential equations
⇒ Recognize that circuits with a natural response lead to homogeneous differential equations
⇒ Derive a differential equation for the RC circuit.
⇒ Solve the differential equation for the RC circuit.
⇒ Apply the initial condition in the solution of the differential equation
⇒ Find the time constant of the RC circuit
⇒ Understand the meaning of the term “time constant”
First Order Circuits:
Circuits with only one energy storage element are called first order circuits.
Reason ⇒ these circuits are generally described by first order differential equations.
In general if the circuit has:
N energy storage elements ⇒ N th order circuits ⇒ described by N th order differential
equations.
All circuits shown are first order circuits:
For circuit c) C1 C2 ⇒ Circuit effectively has only one capacitor C = C1 + C2
For circuit d) L1 & L2 (in series) ⇒ Circuit effectively has only one inductor L = L1 + L2
R3
L
C
is
R1
Vs
R
R2
(a)
(b)
L1
Vs
R3
L2
R2
R2
R1
R1
C1
C2
Vs
(c)
(d)
Figure 1
Review of First Order Differential Equation and Solution:
The differential equation:
dx(t )
+ ax(t ) = f (t )
dt
(1)
Is usually encountered in the analysis of first order circuits.
x(t ) is an unknown function of time.
t is the time variable.
a = constant
f (t ) = a given function of time
The coefficient of the first derivative
dx(t )
is one ⇒ Normalized differential equation.
dt
When f (t ) ≠ 0 ⇒ The differential equation is called inhomogeneous
When f (t ) = 0 ⇒ The differential equation is called homogeneous
Eqn. (1) has the following general solution:
x(t ) = e − at ∫ e at f (t ) dt + Ae − at
(2) [Substitute eqn. (2) into eqn. (1) to prove this fact].
A = an arbitrary constant
Special Case:
If f (t ) = b , where b is some constant:
x(t ) = e − at ∫ e at bdt + Ae − at ⇒
∴ the solution of
x(t ) = e − at
dx (t )
+ ax(t ) = b
dt
e at
b
b + Ae − at = + Ae− at
a
a
(3)
⇒
x (t ) =
b
+ Ae − at
a
(4)
This is an important special case, which we will encounter very often in the coming lectures.
Natural Response of First Order Circuits:
When a circuit does not contain an independent source ⇒ its response is called natural response
Circuits exhibiting a natural response give rise to homogeneous differential equations.
In this lecture and the next lecture we will see some examples of circuits exhibiting a natural
response.
Circuit has no
independent
sources
Homogenous
Differential
equation
Natural
Response
Figure 2
Natural Response of the RC circuit:
The RC circuit consists only of a resistor in parallel with a capacitor.
The response of the RC circuit is natural, because no independent sources exist in the circuit.
Figure 3
Example 1:
The RC circuit shown has an initial voltage vc (0) across the capacitor.
Should vc (t ) :
1- decrease with t ?
2- increase with t ?
3- remain unchanged?
Figure 4
Solution:
The capacitor has an initial voltage ⇒ it has an initial energy [because wc (t ) =
1 2
Cvc (t ) ]
2
C R ⇒ vR = vc
pR =
vR2 vc 2
=
>0
R
R
⇒ R absorbs electric power
[This means that as long as the capacitor has voltage across it, the resistor continues to absorb power]
The power absorbed by R must be supplied by C
Since wc (t ) =
1 2
Cvc (t )
2
⇒
⇒ Energy stored in C must decrease with t
vc (t ) must decrease with t
This also means the voltages and current in the circuit are functions of time.
Figure 5
We can also answer this question by considering the charge q stored in the capacitor.
Capacitor has initial voltage ⇒ it also has initial charge [because q (0) = Cvc (0) ]
The resistor provides a path for the mobile negative charges to recombine with the positive charges
on the positive capacitor plate.
This results in the reduction of charge on the positive and negative plates.
∴ q decreases with time
This results in a decrease in the voltage vc across the capacitor [because q (t ) = Cvc (t ) ]
[Note the actual direction of current is opposite to the direction of flow of negative charges]
Figure 6
Example 2:
The voltage across the 2F capacitor is 6V at t = 0 . [i.e. vc (0) = 6 ]
a) Find v c (t ) for t ≥ 0 .
b) Plot v c (t ) for t ≥ 0 .
Figure 7
Solution:
a) To find vc (t ) :
KCL ⇒
ic (t ) + iR (t ) = 0
⇒
2
dvc (t ) vR (t )
+
=0
dt
5
⇒
2
dvc (t ) vc (t )
+
=0
5
dt
(1)
Eqn. (1) is a differential equation, which has vc (t ) as the unknown.
We must normalize eqn. (1) before obtaining its solution.
2
dvc (t ) vc (t )
+
= 0 (Not normalized) ⇒
dt
5
dvc (t ) vc (t )
+
= 0 (Normalized by dividing by 2)
dt
10
∴
dvc (t )
+ 0.1vc (t ) = 0 [the differential equation is homogeneous. Why?]
dt
Since
∴
dx (t )
+ ax(t ) = b
dt
dvc (t )
+ 0.1vc (t ) = 0
dt
⇒
x (t ) =
⇒
vc (t ) =
b
+ Ae − at
a
0
+ Ae −0.1t
0.1
⇒
vc (t ) = Ae −0.1t (2)
Where we used a = 0.1 & b = 0
To determine the arbitrary constant A in Eqn. (2) ⇒ Use the given initial voltage vc (0) = 6
Eqn. 2
vc (t ) = Ae −0.1t
∴ vc (t ) = 6e −0.1t
⇒
vc (0) = Ae −0.1×0 = Ae 0 = A = 6
for t ≥ 0 [Required solution]
Figure 8
b) The graph of vc (t ) = 6e −0.1t for t ≥ 0 is shown.
As expected, vc (t ) decreases with t . It actually decreases exponentially with time.
Figure 9
Example 3:
In the R-C circuit shown below, vc (t0 ) is known.
a) Develop an expression for vc (t ) for t ≥ t0
b) Plot vc (t ) for t ≥ t0
Figure 10
Solution:
a) We will use a similar method as in example 1.
KCL ⇒
∴
ic (t ) + iR (t ) = 0
⇒
C
dvc (t ) vR (t )
+
=0
dt
R
⇒
C
dvc (t ) vc (t )
+
=0
dt
R
dvc (t ) vc (t )
+
= 0 (Normalized)
dt
RC
∴a =
1
RC
∴ vc (t ) =
&
b=0
[Using
1
−
t
0
+ Ae RC = Ae − t / RC
1
RC
dvc (t )
b
+ avc (t ) = b ⇒ vc (t ) = + Ae − at ]
dt
a
(1)
From Eqn. (1) ⇒ vc (t ) = Ae − t / RC ⇒
vc (t0 ) = Ae − t0 / RC
⇒
Substituting Eqn. (2) into Eqn. (1)
vc (t ) = [vc (t0 )e + t0 / RC ]e − t / RC
∴ vc (t ) = vc (t0 )e − ( t −t0 ) / RC
⇒
for
vc (t ) = vc (t0 )e − ( t −t0 ) / RC
t ≥ t0
is the required solution.
A = vc (t0 )e + t0 / RC (2)
Figure 11
b) The graph of vc (t ) for t ≥ t0 is shown:
Vc (t) [V]
Vc (to)
t [s]
to
Figure 12
Special Case of the Natural Response of the RC Circuit:
The result of the previous example:
vc (t ) = vc (t0 )e − ( t −t0 ) / RC
for
t ≥ t0
describes the natural response of an RC circuit in the most general form.
It assumes that we know vc (t0 ) and we are interested in vc (t ) for t ≥ t0
When the initial time t0 = 0 , the natural response of the RC circuit is simplified.
vc (t ) = vc (t0 )e − ( t −t0 ) / RC
for
t ≥ t0
∴ vc (t ) = vc (0)e − t / RC for t ≥ 0
t0 = 0
⇒
vc (t ) = vc (0)e − ( t − 0) / RC
for
t≥0
Figure 13
The Time Constant of the RC Circuit:
The natural response of the RC circuit for t0 = 0 :
vc (t ) = vc (0)e − t / RC for t ≥ 0 ,
vc (t ) = vc (0)e − t /τ for t ≥ 0 ,
can be rewritten as:
Where τ = RC
τ = RC is called the time constant of the RC circuit [in seconds].
The Meaning of the Time Constant:
The figure shows a graph of vc (t ) = vc (0)e − t /τ for t ≥ 0
When t = τ
⇒
1
1
vc (τ ) = vc (0)e −τ /τ = vc (0)e −1 = vc (0) =
vc (0) 0.37vc (0)
e
2.7183
It takes τ seconds for the voltage across the capacitor to drop by a factor of
Figure 14
The figure shows a graph of vc (t ) = vc (0)e − t /τ for τ = 0.5, 1, 3, 6 and ∞
Higher values of τ ⇒ slower rate of voltage decrease
1
0.37
e
When τ → ∞
⇒ the voltage becomes constant with time.
When τ → 0
⇒ the voltage goes to zero instantaneously.
In practice, RC circuits have a very wide range of values of the time constant τ from as low as picoseconds [ 10 −12 s ] to several seconds.
RC circuits have various applications such as their use in timers and filters.
Figure 15