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Electrical Engineering
Principles & Applications

Chapter 4- Transients

Slide 1

Outline
1
...
Understand the concepts of transient response
and steady-state response
3
...
Second-order circuits

Slide 2

Transients
1
...
By writing circuit equations, we obtain
integrodifferential equations

Slide 3

Transients

Slide 4

Discharge of a Capacitance through a
Resistance
Which can be written as:

dv C (t )
1
=−
RC
v C (t ) …(2)
dt
RC
Write KCL at top node with
switch closed:

Solution of (2) is a function
with same form as its first
Derivative
an exponential

ic (t ) + iR (t ) = 0
C
RC
Slide 5

v C (t ) = Ke

dv

K and s are constants

C

dt
dv

C

dt

(t ) +

(t ) +

v C (t )
= 0
R

v C (t ) = 0 … (1)

st

…(2)

to be determined
Substitute (2) in (1):

RCKse + Ke = 0
st

st

Discharge of a Capacitance Through a
Resistance

RCKse + Ke = 0
st

( RCs + 1) Ke
RCs + 1 = 0
−1
s=
RC

vC (t ) = Ke

st

st

=0
Voltage immediately after
the switch closes

v C (0 + ) = V i

− t RC

Since voltage across the capacitor
cannot change instantaneously
(current would be infinite)
Slide 6

( )

v C 0 + = V i = Ke 0 = K
Thus,

vC (t ) = Vi e

−t RC

Time Constant

The time interval = RC is called the time constant of the circuit
...
Assume that the
switch will be closed at t = 0
...
Replace capacitances with open circuits
2
...
Solve the remaining circuit
Slide 10

DC Steady State
Compute the current ix and the voltage vx when the switch is closed

ix =

10
R1 + R 2

=1A

v x = R2 i x = 5 V
Slide 11

DC Steady State
Compute the unknown currents and the unknown voltages when the
switches are closed

v a = 50 V, i a = 2 A

i1 = 2 A, i 2 = 1A, i 3 = 1 A
Slide 12

Summary of Solving RC or RL Circuits
•

The steps involved in solving simple circuits containing dc sources,
resistances, and one energy-storage element (inductance or
capacitance) are:
1
...
If the equation contains integrals, differentiate each term in
the equation to produce a pure differential equation
3
...
Substitute the solution into the differential equation to
determine the values of K1 and s
5
...
Write the final solution
Slide 13

Example: RL Transient Analysis
:

V s
= 2
R
R
s = −
L

Write KVL equation:

K

di
Ri ( t ) + L
= Vs
dt
Solution:

i(t ) = K1 + K 2e

st

RK1 + ( RK2 + sLK2 )e st = Vs
Slide 14

1

=

Using initial conditions

K 2 = −2
i (t ) = 2 − 2 e

−t /τ

iL (0 + ) = 0

L
, τ =
R

Example (cont
...

0
Slide 16

K1 + K2est

Then the solution is of the form
−t /τ

i ( t ) = Ke

τ = L / R2
i ( 0 + ) = V s / R1 = Ke − 0 = K
 Vs


R1
i (t ) = 
V
 s e −t /τ
 R1


t<0
t>0

Example: RL Transient Analysis

di ( t )
dt
− LV s
e
=
R 1τ

v (t ) = L

− t /τ

− R2
V se
=
R1

and
Slide 17

v (t ) = 0

− t /τ

t<0

, τ = L / R
t > 0

2

Second-Order Circuits
Circuits with two energy–storage elements
Write KVL:

di(t )
1
L
+ Ri(t ) + ∫ i(t )dt + vC (0) = vs (t )
dt
C0
t

dvs (t )
di(t ) 1
d i(t )
L 2 +R
+ i(t ) =
dt C
dt
dt
d 2i(t ) R di(t ) 1
1 dvs (t )
i(t ) =
+
+
2
dt
L dt LC
L dt
2

Define:
R
α =
2L
ω

0

=

f (t) =

1
LC

1 dvs (t)
L dt

d 2 i (t )
di (t )
+ 2α
+ ω 02 i (t ) = f (t )
dt 2
dt
Slide 18

Second-order DE

(Damping coefficient)
(Undamped resonant
frequency)
(forcing function)

Solution to the DE
2

d

i (t )

dt

2

+ 2α

di (t )
+ ω
dt

2
0

i (t ) =

x (t ) = x p (t ) + x c (t )

Voltage or current

Particular solution

f

(t )
Complementary solution

Particular Solution: (also known as forced response)
Due to forcing function
...
Over-damped case ( > 1):
If > 1 (or equivalently, if > 0), the roots of the
characteristic equation are real and distinct
...
Critically damped case ( = 1):
If = 1 (or equivalently, if = 0 ), the roots are real and
equal
...
Under-damped case ( < 1):
Finally, if < 1 (or equivalently, if < 0), the roots are
complex
...
) In other words, the
roots are of the form

s1 = −α + jωn and s2 = −α − jωn
in which j is the square root of -1 and the natural frequency is given by

ω n = ω 02 − α 2
Then the complementary solution is

xc (t ) = K1e

Slide 23

−αt

cos(ω n t ) + K 2 e

−αt

sin(ω n t )

Example
Find the vc(t) for R = 300, 200, 100

The solution has the form:
Particular solution:
xp = vcp = Vs = 10 V

Slide 24

x (t ) = x p (t ) + x c (t )

Example (cont
...
)
α = 15000 > ω 0 (ζ > 1) Over-damped

R = 300
Recall that:

s1 = −α − α 2 − ω 02 = − 2
...
382 × 10 4

x

c

(t ) =

K

1

e

s1t

+ K

2

e

s2t

v c ( t ) = 10 + K 1 e s1 t + K 2 e s 2 t
vc (0) = 0 = 10 + K1 + K2
dvc (t)
i(t) = C
dt
dv (0)
i(0) = 0 = c = s1K1 + s2 K2
dt

K1 = 1
...
708

v c ( t ) = 10 + 1
...
708 e s 2 t
Slide 26

Example (cont
...
)
R = 100

α = 5000 < ω 0 (ζ = 0
...
774
v c ( t ) = 10 − 10 e − α t cos( ω n t ) − 5

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