ELECTRICAL ENGINEERING DEPARTMENT

EE241: ELECTRIC CIRCUITS I

SPRING 2023

LECTURES 19-21
CH7- FIRST-ORDER CIRCUITS
(TRANSIENT RESPONSE OF 1ST ORDER CIRCUITS)
Dr. Mohammed A. Hassan
m.hassan@kfu.edu.sa
1
Last Lectures: Capacitors
Physical structure Parallel Capacitors

εA
C=
d
Ceq = C1 + C2 + C3 + .... + C N
Electrical Characteristics

q = Cv Series Capacitors

dv
i=C O.C. in DC
dt
1
Energy Stored w = C v2 1 1 1 1 1
2 = + + + ... +
Ceq C1 C2 C3 CN 2
Last Lectures: Inductors
Physical structure Series Inductors

N 2 A
L=
l
Leq = L1 + L2 + L3 + ... + LN
Electrical Characteristics Parallel Inductors
di
v=L S.C. in DC
dt
1 2
Energy Stored w(t ) = Li (t ) 1 1 1
= + ++
1
2 Leq L1 L2 LN
3
Remember: Important characteristics of basic elements

4
First-Order Circuits

Source-Free RC Circuit
Source-Free RL Circuit
Unit-step Function
Step Response of an RC Circuit
Step Response of an RL Circuit

5
Source-Free RC Circuit
 A first-order circuit is characterized by a first-order differential equation.
By KCL
v dv dv 1
iR + iC = 0 +C =0 =− dt
R dt v RC
Ohms
law Capacitor law

• Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations.

• Apply the laws to RC and RL circuits produces differential equations.

Solution of the 1st order D.E

v(t ) = V0 e −t /  where =RC

• Vo is the initial voltage V(0) across the capacitor.
•  = RC is the time constant.
6
Source-Free RC Circuit
 The natural response of a circuit refers to the behavior (in terms of voltages and
currents) of the circuit itself, with no external sources of excitation.

Time constant =RC

Decays more slowly

Decays faster

• The time constant  of a circuit is the time required for the response to decay by a factor
of 1/e or 36.8% of its initial value.
• 𝑣 decays faster for small 𝜏 and slower for large 𝜏.
7
Example 1
Refer to the circuit below, determine vC, vx, and io for t ≥ 0.
Assume that vC(0) = 30 V.
Solution:
𝑣𝑐 𝑡 = 𝑉𝑜 𝑒 −𝑡/𝜏
𝑉𝑜 = 𝑣𝑐 0 = 30𝑉
1
𝜏 = 𝑅𝑒𝑞 𝐶 = 8 + 12\\6 ∗ = 4 𝑠𝑒𝑐
3
𝑣𝑐 𝑡 = 30 𝑒 −0.25𝑡 V
4
Using voltage divider 𝑣𝑥 𝑡 = 𝑣𝑐 𝑡 𝑣𝑥 𝑡 = 10 𝑒 −0.25𝑡 V
4+8
𝑣𝑥 𝑡 − 𝑣𝑐 𝑡
Ohm’s low 𝑖𝑜 𝑡 = 𝑖𝑜 𝑡 = −2.5 𝑒 −0.25𝑡 A
8
8
Practice Exercise 1
Refer to the circuit in the following Figure, Let 𝑣𝑐 0 = 60 𝑉. Determine 𝑣𝑐 ,
𝑣𝑥 , and 𝑖𝑜 for 𝑡 ≥ 0.

9
Example 2
The switch in circuit below is opened at 𝑡 = 0, find 𝑣(𝑡) for 𝑡 ≥ 0.
Solution:
For 𝑡 < 0 C acts as O.C.
Using voltage divider
3
𝑉0 = 𝑣𝑐 0 = 24 = 8𝑉
3+6
For 𝑡 ≥ 0
𝑣𝑐 𝑡 = 𝑉𝑜 𝑒 −𝑡/𝜏
𝑉𝑜 = 𝑣𝑐 0 = 8𝑉
1
𝜏 = 𝑅𝑒𝑞 𝐶 = 12\\4 ∗ = 0.5 𝑠𝑒𝑐
6
𝑣 𝑡 = 𝑣𝑐 𝑡 = 8 𝑒 −2𝑡 V, 𝑡≥0

10
Practice Exercise 2
If the switch in the circuit opens at 𝑡 = 0, find 𝑣(𝑡) for or 𝑡 ≥ 0 and 𝑤𝑐 (0).

11
Source-Free RL Circuit
 A first-order RL circuit consists of an inductor L and a resistor
By KVL
di di R
vL + vR = 0 L + iR = 0 = − dt
dt i L

Inductors Ohms
law law

Solution of the 1st order D.E

− t / L
i (t ) = I 0 e where =
R
• Io is the initial current I(0) through the Inductor.
𝐿
•  = is the time constant
𝑅
12
Source-Free RL Circuit

A general form representing a RL

i (t ) = I 0 e − t / 
L
where =
R

• The time constant  of a circuit is the time required for the response to decay
by a factor of 1/e or 36.8% of its initial value.
• i(t) decays faster for small  and slower for large .
• The general form is very similar to a RC source-free circuit.
13
Source-Free RC and RL Circuits
Comparison between a RL and RC circuit

RL source-free circuit RC source-free circuit

L
i (t ) = I 0 e − t /  where = v(t ) = V0 e−t / where  = RC
R

The key to working with a source-free RL (or RC) circuit is finding:

(a) Initial value Io (or Vo), (b) time constant 𝜏 14
Example 3
The switch in the circuit of the following Figure has been closed for a long
time. At 𝑡 = 0, the switch is opened. Calculate 𝑖(𝑡) for 𝑡 > 0.
Solution:
For 𝑡 < 0 ( L acts as S.C.)
40𝑉 12
Io = 𝑖 0 = ∗ = 6𝐴
(2 + 12\\4)Ω 16
For 𝑡 ≥ 0

𝑅𝑒𝑞 = (12 + 4)\\16 = 8Ω

𝐿 2 1
𝜏= = = 𝑆
𝑅𝑒𝑞 8 4

𝑡
−
𝑖 𝑡 = 𝐼𝑜 𝑒 𝜏 = 6 𝑒 −4𝑡 𝐴
15
Practice Exercise 3
For the following circuit, find 𝑖(𝑡) for or 𝑡 > 0.

16
Example 4
In the following circuit, 𝑓𝑖𝑛𝑑 𝑖𝑜 , 𝑣𝑜 , and 𝑖 for all time, assuming that the
switch was open for a long time.
Solution:
For 𝑡 < 0 ( L acts as S.C.)

For 𝑡 ≥ 0 𝑡
𝐿 2 𝑖 𝑡 = 𝐼𝑜 𝑒 −
𝜏 = 2 𝑒 −𝑡 𝐴, 𝑡>0
𝑅𝑒𝑞 = 3\\6 = 2Ω, 𝜏= = = 1 𝑆,
𝑅𝑒𝑞 2
𝑑𝑖
𝑣𝑜 𝑡 = −𝑣𝐿 = −𝐿 = −2 −2𝑒 −2 = 4𝑒 −𝑡 𝑉, 𝑡>0
𝑑𝑡
𝑣𝐿 2
𝑖𝑜 𝑡 = = − 𝑒 −𝑡 𝐴, 𝑡>0
6 3
17
Example 4
Solution (Cont.):
For all time

A plot of 𝑖 and 𝑖𝑜

18
Practice Exercise 4
For the following circuit, find 𝑖, 𝑖𝑜 and 𝑣𝑜 for all 𝑡. Assume that the switch was
closed for a long time.

19
Summary: Source-Free RC and RL Circuits
Comparison between a RL and RC circuit

RL source-free circuit RC source-free circuit

L
i (t ) = I 0 e − t /  where = v(t ) = V0 e−t / where  = RC
R

The key to working with a source-free RL (or RC) circuit is finding:

(a) Initial value Io (or Vo), (b) time constant 𝜏 20
First-Order Circuits

Source-Free (Natural Response) RC Circuit

Source-Free (Natural Response) RL Circuit
Unit-step Function
Step Response of an RC Circuit
Step Response of an RL Circuit

21
Unit-Step Function
 The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

 0, t0
u(t ) = 
1, t0

 u(t) is used to represent an abrupt change for:

1- voltage source. 2- for current source:

22
Step-Response of a RC Circuit
 The step response of a circuit is its behavior when the excitation is the step
function, which may be a voltage or a current source.

• Initial condition: v(0-) = v(0+) = V0

dv v − Vs u (t )
• Applying KCL, c + =0
dt R
dv v − Vs
=−
dt RC
• Where u(t) is the unit-step function

•Solving the 1st order D.E. using separation of parameters:

dv 1
=− dt
v − Vs RC 23
Step-Response of a RC Circuit
 Integrating both sides and considering the initial conditions, the solution of the
equation is: 𝑣 𝑡 𝑡 𝑣 − 𝑉𝑠 𝑡
ln 𝑣 − 𝑉𝑠 ቤ𝑉 = − ฬ 𝑙𝑛 =−
𝑜 𝑅𝐶 0 𝑉𝑜 − 𝑉𝑠 𝑅𝐶
𝑡
𝑣−𝑉𝑠 −𝑅𝐶
∴ 𝑉𝑜 −𝑉𝑠
=𝑒

V0 t0
v(t ) =  − t /
Vs + (V0 − Vs ) e t 0

Final value Initial value Source-free

at t -> ∞ at t = 0 Response

Complete Response = Natural response + Forced Response

(stored energy) (independent source)

= V0e–t/τ + Vs(1–e–t/τ)
24
Step-Response of a RC Circuit
Three steps to find out the step response of an RC circuit:
1. The initial capacitor voltage v(0).
2. The final capacitor voltage v() —
DC voltage across C (after long time, C acts as OC).
3. The time constant .

− t /
v (t ) = v () + [v (0+) − v ()] e
Important Note: If the circuit connected to the capacitor at t>0 is more complex,
use Thevenin equivalent circuit. V(∞)=VTh and R=RTh

25
 Example 5
Find 𝑣(𝑡) for 𝑡 > 0 in the circuit in below. Assume the switch has been open for a long
time and is closed at 𝑡 = 0. Then, calculate 𝑣(𝑡) at 𝑡 = 1, and 𝑡 = 4.
Solution:

26
 Practice Exercise 5
Find 𝑣(𝑡) for 𝑡 > 0 in the circuit in below. Assume the switch has been open for a long
time and is closed at t = 0. Calculate 𝑣(𝑡) at 𝑡 = 0.5.

27
 Example 6
In the following circuit, the switch has been closed for a long time and is
opened at 𝑡 = 0. Find 𝑖 and 𝑣 for all time.
Solution:

20
𝑣 ∞ = 𝑉𝑇ℎ = 30 = 20𝑉
20 + 10

28
Example 6
30 − 10
Solution (Cont.): 𝑖 0− = = 2A
10
For 𝑡 > 0

2A

Practice: Plot 𝑣 and 𝑖 29

 Practice Exercise 6
Find 𝑖(𝑡) and 𝑣(𝑡) for 𝑡 > 0 in the circuit in below. Assume the switch has been open for
a long time and is closed at t = 0.

60

60 V, 𝑡<0
𝑣(𝑡) = ቊ
50 + 10𝑒 −1.5𝑡 V, 𝑡>0

60 − 𝑣(𝑡) 0 A, 𝑡<0
𝑖 𝑡 = =ቊ
5 2(1 − 𝑒 −1.5𝑡 ) A, 𝑡>0
30
 Step-Response of an RL Circuit
 The step response of a circuit is its behavior when the excitation is the step
function, which may be a voltage or a current source.

i (t ) = i () + [i (0) − i ()] e −t / L

 =
R
Three steps to find out the step response
of an RL circuit:
1. The initial inductor current 𝑖(0) at 𝑡 = 0− .
2. The final inductor current 𝑖  = 𝑉𝑠 /𝑅 (after long time, L acts as SC).
3. The time constant =L/R.

Note: If the circuit connected to the inductor at t>0 is more complex, use
Thevenin equivalent circuit. I(∞)=VTh/RTh and R=RTh
31
 Example 7
The switch in the circuit shown below has been closed for a long time. It opens at
𝑡 = 0. Find 𝑖(𝑡) for 𝑡 > 0.
Solution:
I o = i (0− ) = 3 A
10
i ( ) = 3 = 2A
10 + 5
L 1.5
= = = 0.1 S
RTh 15

i (t ) = i () + [i (0) − i ()] e −t /

i (t ) = 2 + e −10t
32
 Example 8
The switch in the circuit shown below has been closed for a long time. It opens at 𝑡 = 0.
Find 𝑖(𝑡) for 𝑡 > 0.
Solution:

33
 Practice Exercise 7
The switch in the circuit shown below has been open for a long time. It closes at 𝑡 = 0.
Find 𝑖(𝑡) for 𝑡 > 0.

i (t ) = 5 − 3e −6t
34
Summary of Transient-Response for RC and RL circuits
Step Response (Forced Response):
v (t ) = v () + [v (0) − v ()] e−t /

 = RC

i (t ) = i () + [i (0) − i ()] e −t /

L
Three steps to solve any transient-response problem:  =
R
1. The initial Values.
2. The final value (after long time, C acts as OC and L acts as SC).
3. The time constant .
Note: Natural Response (source free) is a special case where final value =0 35
End of Ch7

Questions?

36