Multivector Review and Training Center

TRANSIENTS

Electrical Transient – is the study of characteristics of current, potential drop,

power charged, energy dissipation and energy stored across different load
parameters as energized by a D.C. or A.C. source by means of the activation of a
switch.

In general, transient disturbances are produced whenever:

a. an apparatus or circuit is suddenly connected to or disconnected from the
supply.
b. a circuit is shorted.
c. there is a sudden change in the applied voltage from one finite value to
another.

Transient Disturbances are classified as follows:

1. Initiation Transients – these are produced when a circuit which is
originally dead is energized.
2. Subsidence Transients – these are produced when an energized circuit is
rapidly de-energized and reaches an eventual steady-state of zero current
or voltage.
3. Transition Transients – these are due to sudden but energetic changes
from one steady state to another.
4. Complex Transients – these are produced in a circuit which is
simultaneously subjected to two transients due to two independent
disturbances.
5. Relaxation Transients – transition occurs cyclically towards states,
which when reached, become unstable themselves.

DC TRANSIENT
Series RL Circuit
Case 1 Initiation Transient:
Given initial condition: at t = 0, i = 0

Model Circuit:

SW

R VR
E
L VL
i

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When the switch is closed, then by KVL;

VR  VL  E
di
Ri  L  E  L.D.E. of the first order
dt

By variable separable
di 1
 dt
E  Ri L

lnE  Ri   t  k 1
1 1

R L
R
lnE  Ri   t  Rk 1 where: Rk1 = k2
L
R
lnE  Ri   t  k2
L

e   k e
R R R
 t  t  t
E  Ri  e L
 k2  e L k2
3
L

k 3
let k 
R
R
E  t
therefore, i   ke L  general solution
R
R
E k  t
i   3 e L ; k 3  ek2
R R

where: k = constant of integration

To find k, at t = 0, i = 0
E
0   ke0
R
E
k , substitute back in general solution
R
Hence,
R
E E Lt
i  e
R R
E
where: = steady state current component
R
R
E Lt
 e = transient current component
R
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Finally:
E   t
R
i 1e L   particular solution
R  

E
current growth equation from i  0 to i  I max 
R
Note: For inductor with no initial current,
Open at t = 0  i = 0
E
Shorted at t =   I = Imax =
R
Voltage Across R (VR):
  t
R
* VR  Ri  E 1  e L 
 
 

Voltage Across L (VL):

di
VL  L
dt
E   t  R  
R
 L  0  e L   
 R   L  
R
 t
VL  Ee L

OR
VL  E  VR
  t
R
 E  E 1  e L 
 
 
R
 t
* VL  Ee L

Power Across R (PR):

VR2
PR  VR  i  i 2 R 
R
Using:
PR  VR  i
  t
R
E  t
R
 E 1  e L    1  e L 
  R 
   

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2
E2   t
R
* PR  1  e L 
R  
 
when t = 0, PR = 0
E2
t = , PR =
R
Note: t =  (steady state)

Power Across L (PL):

E   t
R R
 t
PL  VL  i  Ee L
 1e L 
R  

E 2   L t t
R 2R

PL  e e L 
R  

When t = 0, PL = 0
t = , PL = 0

Total Power (PT):

PT  PR  PL  Ei
E   t
R
PT  E  1e L 
R  

E2   t
R
PT  1  e L 
R  
 

Energy Stored in the Magnetic Field (WL):

di
dWL  PL dt  VL idt  L  idt  Lidi
dt
when: t = 0, i = 0, WL = 0
t = t, i = I, WL = WL
Hence

 
W1 t
dWL  L idi
0 0

1 2
WL  LI
2
At maximum condition (t = ), I = Imax
1 2
WL  LI max
2

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CASE 2 Subsidence Transient:

Given initial condition: when t = 0, i = Io

Model Circuit:
1 2

R
E
L
i Io

When the switch is in position 1

E   t
R
i 1e L 
R  


When the switch is in position 2

R
i
L
Io

V R  VL  0
di
Ri  L 0
dt
di R
  dt
i L

At t = 0, i = IO and At t = t, i = I
di R t
    dt
i

IO i L 0
R
 t
i  I Oe L

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Or from general solution

R R
E  t  t
i  ke L  ke L
R

At t = 0, i = IO
IO = ke0, therefore k = IO
R
 t
i  I Oe L
 current decay equation i  I O to i  0
E
Note: When the switch transfer from 1 to 2 during steady-state condition, I O 
R

Time Constant (TC) :

The time constant of any equation dealing with transient is the value of time (t,
sec) which makes the magnitude or absolute value of the exponent of the nos. e
equivalent to unity or 1.
In both current growth and current decay curves the current tends to approach their
steady-state values after 5TC.
Thus, the transition interval is: 0  t  5TC

For RL circuit:
L
TC 
R

For RC circuit:
TC = RC

Time constant for RL circuits:

Time constant at Initiation: The time for the current to reach 63.2% of its final
value.
Proof:

E   t
R
i 1e L 
R  

L
When t  1 TC 
R
i
E
R
  E
1  e 1  0.632  63.2% I max
R

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E
R
E
0.632 E
(1 – e – t
) exponential rise
R
R i= L
R

0 t (sec)
1 TC 5 TC

Time constant at Subsidiency: The time elapsed for the current to reach 36.8% of
its initial value.

R
 t
Proof: i  Io e L

L
when t  1 TC 
R
i  I o e 1  0.368I o  36.8% I o

i = Io e – t
) exponential decay
R
L

Io

0.368 Io
t (sec)
0 1 TC 5 TC

Transient in Series RC Circuit (D. C.):

R VR
E
i C VC

dq q 1
C C
VR  Ri  R VC   i dt
dt

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When the switch is closed:

q
E  VR  VC  Ri 
C
dq q
ER 
dt C
q dq
E R
C dt
dq
CE  q  RC
dt
dq 1
 dt
CE  q RC

 lnCE  q  
t
 k1
RC
t
lnCE  q    k1
RC
t t
 k 1 
CE  q  e RC
e RC
 e k 1 where: e  k1  k 2
t

q  CE  k 2e RC

t

q  CE  ke RC
 general soluton

To find i,
dq  
t
 1 
i  CE0  e RC   
dt   RC 
t
E  RC
i e
R
E
But  I O , therefore
R
t

i  I Oe RC

Voltage Across R:
t
E  RC
VR  Ri  R  e
R
t

VR  Ee RC

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Voltage Across C:
 
t

CE 1  e RC 
 
q
VC    
C C
t

VC  E(1  e RC

OR
t

VC  E  VR  E  Ee RC

 
t

VC  E 1  e RC 
 
 

CASE 1 (Charging Process)

Given Condition: at t = 0, q =0
t

q  CE  ke RC
0  CE  k; k   CE
 
t
  
t

q  CE 1  e RC   Qmax 1  e RC 
   
   
t t
 E  RC
i  Io e RC
 e
R

Note: If the capacitor is initially uncharged,

E
shorted at t = 0, I o 
R
open at t = , Io = 
Proof:
t t
 E  RC
i  Io e RC
e 
R
q
E  VR  VC  Ri 
C
At t = 0, i = Io, q = 0
0
E  RI o 
C
E
Io 
R

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Power Across R:
VR2
PR  VR  i  i 2 R 
R
t t
 E  RC
PR  Ee RC
 e
R
2t
E 2  RC
PR  e
R
E2
At t = 0, PR 
R
At t = , PR = 0

Power Across C:
PC  VC i
 
t
E  t
PC  E 1  e RC  e RC
 R
 
At t = 0, PC = 0
At t = , PC = 0

CASE 2: if at t = 0 , q = Qo (charging with Qo)

R
E ++++
Qo C
i

From,
t

q  CE  ke RC

At t = 0, q = Qo

Q o  CE  ke0
k  Q o  CE
t
q  CE  Q o  CE  e

RC

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Qo

t E   t 
i  Io e RC
 C  e RC 
R  
 
Proof:
q
E  VR  VC  Ri 
C
At t = 0, q = Qo, i = Io
Qo
E  RI o 
C
Qo
E
Io  C
R
Hence,
Qo

t E t
i  Io e RC
 C e  RC
R

Energy stored in C due to charge Q in “t” seconds:

 q  dq q
dWC  PCdt  VC idt    dt  dq
 C  dt C
Q
1  1  q2 

Q
WC  qdq    
C 0
 C  2 0
1  Q2  1 1
WC     QVC  CVC2
2  C  2 2
Q
Since VC  ; Q  CVC
C
At maximum condition (t = ), Q = Qmax = CE

Hence,
1  Qmax
2
 1 1
WC     QmaxE  CE 2
2  C  2 2

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CASE 3: (discharging with Qo

R
E
–Qo C
i ++++

t

From q  CE  ke RC

At t = 0, q = -Qo
-Q o = CE + ke0
k = -(Qo + CE)
Hence,
t
q  CE  Q o  CE  e

RC

Qo

t E t
i  io e RC
 C e  RC
R
Proof:
q
E  VR  VC  Ri 
C
At t = 0
Qo
E
E  RI o 
 Qo  ; Io  C
C R

CASE 4 (For discharging of initially charged capacitor):

2 ++++
Qo C
V C R
R i
i

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t

q  Qo e RC

Qo

t  t

i  Io e RC
 C e  RC
R

CASE 5:

2
++++ R
C1 R V C1 Qo i
C2 C2

Qo
t  t
 C1  R CT
i  Io e RC
 e
R
C1C2
CT 
C1  C2
 Q C  C 
t

q   o T  1  1  e RCT 
 C1  CT 


RLC Circuit (Double Energy Transient)

Series RLC Circuit (DC):

R VR
E L VL
i
C VC

By KVL,
VR  VL  VC  C

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di q
dt C 
Ri  L  idt  E

Differentiate “i” with respect to t:

 di d 2i i 1
R  L   0
 dt dt C L

 R  di
d 2i i
   0
dt 2
 
L dt LC
(L.D.E. of the second order with three possible solutions)
OR
 2 R 1 
D  D  i 0
 L LC 

Auxiliary Equation:
R 1
m2  m 0
L LC

By Quadratic Equation Formula:

2
R R 4
   
2L  
L LC
m
21
2
R R 4
   
2L  
L LC
m
4
2
 R  1
  
R  2L  LC
m 
2L 4

2
R  R  1
Let  ,    
2L  2L  LC

therefore,
m=+
m1 =  + 
m2 =  – 

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Damping Cases of RLC in Series:

CASE I: Over-damped Case:

- when the roots are real and unequal.
2
R 4
Test:    0
L LC
Or
b 2  4ac  0
2
 R   1 
Or if    
 2L   LC 
 is a real number
m1  m 2
m 1    , m 2    
D    D     i  0
General Solution:

i  k 1em1t  k 2em2t amperes

OR

i  k 1e    t  k 2e    t
i  k 1e t  et  k 2e t  e t

i  e t k 1et  k 2e t 
CASE II: Critically-damped case
- when the roots are real and equal/repeated.

2
R 4
Test:    0
 
L LC
OR b 2  4ac  0
2
 R   1 
OR If    
 2L   LC 
m    ;   0
m  m1  m 2  
D   D    i  0

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General Solution:
i  k 1em1t  k 2 tem2t  k 1e t  k 2 tet
i  e t k 1  k 2 t 

CASE III: Underdamped Case: (Oscillatory Case)

- when the roots contain a real and imaginary part.
2
R 4
Test:    0
 
L LC
OR b 2  4ac  0
2
 R   1 
OR If    
 2L   LC 
 is imaginary ,   j
m
2
R  R  1
m    
2L  2L  LC

R  1  R  
2

m     
2L  LC  2L  
2
R 1  R 
m j  
2L LC  2L 
R

2L
2
1  R 
let    
LC  2L 
m    j
OR
m 1    j
m 2    j
D    jD    ji  0
General Solution:
i  k A em1t  k B em2t
i  k A e  jt  k Be   jt

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i  k A e t  e jt  k Be t  e  jt

i  e t k A e jt  k Be  jt 
From Euler’s Formula:
e+jt = cos t + j sint
i = et [kA(cost + j sint) + kB (cost – j sint)]
= et [(kA+ kB)cost + (jkA – jkB) sint]
i = et (k1cost + k2 sint)

where:  = angular velocity of oscillation, rad/sec

f = frequency of oscillation, cps or hertz
2
1  R 
  2f   
LC  2L 
2
1 1  R 
f  
2 LC  2L 

AC Transients

Series RL Circuit

R
e ~
L
i

e = Em sin(t + )
 = position of sinusoidal voltage at the instant of switching/application
For example:
1
At t  0, e  0 At t  0, e  Em At t  0, e  Em
2
0  Em sin0   
1
1  sin  E m  E m sin 
2
0   90   30

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after the switch is closed,

e  eR  eL

Em sint     Ri  L
di
dt
i  i T  i SS

where: i = resultant
iT = transient component
iSS = steady-state component

R
sint     Z 
 t Em
i  Ce L

ZL
L
Z L  R  jL 
R
X L L
tan  Z  
R R
X L
 Z  tan 1 L  tan 1
R R

Series RC Circuit

R
e ~
C
i

t
sint     Z 
 Em
i  Ce RC

ZC
1
Z C  R  jX C  R  j
C
XC 1
tan  Z  
R CR
X 1
 Z  tan 1 C  tan 1
R CR

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