Transient ransient Analysis of AC Circuits
Transients The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients.
Transient Tr ansient of AC Circuits
Dr H.H. Hanafy
2
Transients The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients.
Transient Tr ansient of AC Circuits
Dr H.H. Hanafy
2
R-L Circuit Current
Buildup Transients
For the simple R-L circuit the voltage differential equation at the transient period is given by:
v(t) v R( t) v L( t) v(t) Ri( R i(t) t) L
di(t) d t
v(t) 2 V sin( t v ) Transient Tr ansient of AC Circuits
Dr. H.H. Hanafy
3
This Equation is a first order differential equation and has a solution given by:
i(t ) i ss t i t ( ) ( ) ss t r Where: i ss(t) = the steady state current. itr (t)= the transient current which is a decaying current current with time. Transient Tr ansient of AC Circuits
Dr. H.H. Hanafy
4
To get i ss(t) we solve the circuit in steady state conditions: V I ( R JX L ) I Z Z Impedance ( R JX L ) Z Z
I ss
2
R X L
V v
2
V
Z Z
i ss (t ) 2 Transient of AC Circuits
V Z
1
t an
X L R
( v )
sin( t v ) Dr H.H. Hanafy
5
The itr (t) could be assumed in the following form:
itr A e
t /
Where : A is a constant
is the time constant and given by :
L R
t /
i(t ) i ss (t ) Ae Transient of AC Circuits
Dr H.H. Hanafy
6
To get the constant ‘A’ we use the initial conditions Let at t = 0 the current i(t) has an initial value of Io
I o 2
V Z
sin( v ) A
A I o 2
V Z
sin( v )
Hence: i (t )
V V 2 sin( t v ) { I o 2 sin( v )}e t / Z Z
v R (t ) R i(t ) Transient of AC Circuits
v L (t ) v(t ) v R (t ) Dr H.H. Hanafy
7
If Io= 0
A 2
V Z
sin( v )
Hence: i (t )
V V 2 sin( t v ) { 2 sin( v )}e t / Z Z
v R (t ) R i(t ) Transient of AC Circuits
v L (t ) v(t ) v R (t ) Dr H.H. Hanafy
8
If (θ v-φ) = 90 and τ >> T
If (θ v-φ) = 0 then itr = 0 Transient of AC Circuits
Dr H.H. Hanafy
9
De-energizing Transients For the simple R-L circuit the voltage differential equation at the transient period is given by:
0 V R (t ) V L (t ) 0 Ri(t ) L
di(t ) dt
This Equation is a first order differential equation and has a solution given by:
t /
i(t ) i ss (t ) Ae Transient of AC Circuits
Dr H.H. Hanafy
10
i ss(t) =0
t /
i(t ) Ae
To get the constant ‘A’ we use the initial conditions Let at t = 0 the current i(t) has an initial value of Io t / Hence:
I o A
V R (t ) R i (t )
V L (t ) V R (t ) Transient of AC Circuits
i(t ) I oe
V R (t ) RI oe
t /
V L (t ) RI oe Dr H.H. Hanafy
t / 11
R-C Circuit Capacitor
Charging Equations
For the simple R-C circuit the voltage differential equation at the transient period is given by:
v(t ) v R (t ) vc (t ) v(t ) Ric (t ) vc (t ) v(t ) RC
dvc (t ) dt
vc (t )
v(t) 2 V sin( t v ) Transient of AC Circuits
Dr H.H. Hanafy
12
This Equation is a first order differential equation and has a solution given by:
vc (t ) v ss (t ) vtr (t ) Where: v ss(t) = the steady state voltage. V tr (t)= the transient voltage which is a decaying voltage with time. Transient of AC Circuits
Dr. H.H. Hanafy
13
To get v ss(t) we solve the circuit in steady state conditions:
V I c ( R JX C ) I c Z Z Impedance ( R JX C ) Z 1 X C
2 C
Z R X 2
I css
V v
tan
V
Z Z
R
v
icss (t ) 2 I sin( t v ) Transient of AC Circuits
Dr H.H. Hanafy
14
V ss J I css X C
v ss (t ) 2 I X c sin( t v 90) v ss (t ) 2
Transient of AC Circuits
VX c Z
sin( t v 90)
Dr H.H. Hanafy
15
The vtr (t) could be assumed in the following form:
vtr (t ) A e
t /
Where : A is a constant is thetimeconstantand given by : RC
t /
vc (t ) v ss (t ) Ae Transient of AC Circuits
Dr H.H. Hanafy
16
To get the constant ‘A’ we use the initial conditions Let at t = 0 the voltage vc(t) has an initial value of Vo
V o
2
A V o
Transient of AC Circuits
VX c Z
2
sin( v
VX c Z
90) A
sin( v
Dr H.H. Hanafy
90)
17
vc (t ) 2
VX c Z
sin( t v 90)
{V o 2
VX c Z
ic (t ) C
sin( v 90)}e
t /
dvc (t ) dt
v R (t ) v(t ) vC (t ) ic (t ) R Transient of AC Circuits
Dr H.H. Hanafy
18
If Vo = 0
If (θ v+φ) = 0 and τ >> T Transient of AC Circuits
Dr H.H. Hanafy
19
Capacitor
Discharging Equations
For the simple R-C circuit the voltage differential equation at the transient period is given by:
0 v R (t ) vc (t ) 0 Ric (t ) vc (t ) 0 RC
dvc (t ) dt
vc (t )
This Equation is a first order differential equation and has a solution given by:
t /
vc (t ) v ss (t ) Ae Transient of AC Circuits
Dr H.H. Hanafy
20
v ss (t ) 0 t /
vc (t ) Ae
To get the constant ‘A’ we use the initial conditions Let at t = 0 the voltage vc(t) has an initial value of Vo
V o A Hence:
vc (t ) V oe
Transient of AC Circuits
t /
Dr H.H. Hanafy
21
ic (t ) C
ic (t )
dV c (t ) dt
V o R
e
t /
v R (t ) V C (t ) ic (t ) R
v R (t ) V o e Transient of AC Circuits
t /
Dr H.H. Hanafy
22
R-L-C Circuit For the simple R-L-C circuit the voltage differential equation at the transient period is given by:
v(t ) v R (t ) v L (t ) vc (t ) v(t ) Ri(t ) L
di(t ) dt
1
i (t ) dt C
v(t) 2 V sin( t v ) Transient of AC Circuits
Dr H.H. Hanafy
23
v(t ) Ri(t ) L v (t ) R \
di(t ) dt
di(t ) dt
1
i (t ) dt C
2
L
d i (t ) 2
dt
1 C
i (t )
This Equation is a second order differential equation and has a solution given by:
i(t ) i ss (t ) itr (t )
Where: i ss(t) = the steady state current. itr (t)= the transient current which is a decaying current with time. Transient of AC Circuits
Dr H.H. Hanafy
24
To get i ss(t) we solve the circuit in steady state conditions: V I R J I X L J I X C V I ( R JX L JX C ) I Z Z R J ( X L X C ) Z Z R ( X L X C ) 2
I
V v
V
Z Z
i ss (t ) 2 Transient of AC Circuits
1 X L
t an
2
X C R
v V
R ( X L X C ) 2
Dr. H.H. Hanafy
2
sin( t v ) 25
To get itr (t) we solve the homogeneous differential equation. The homogeneous differential equation is obtained by setting the forcing function to zero. •
•
0 R
ditr (t ) dt
2
d itr (t ) 2
2
L
d itr (t ) 2
dt
R ditr (t )
1
1 C
itr (t )
itr (t ) 0
dt L dt LC This equation has a general solution with the form:
Transient of AC Circuits
Dr H.H. Hanafy
26
K 1 , K 2 are constants determined by initial conditions (capacitor voltage and inductor current). •
•
S1 and S2 are the roots of the characteristic equation:
S 2
R L
S
1 LC
0
2 S 2 n S n 0 2
n
1 LC
R
C
2
L
Transient of AC Circuits
naturalfrequency dampingratio Dr H.H. Hanafy
27
S 1, 2
2 n 4 4 2 n
2
2 n
2
S 1 n n 1 2
S 2 n n 1 2
Transient of AC Circuits
Dr H.H. Hanafy
28
1. Overdamped case
If ζ > 1, the roots of the characteristic equation are real and distinct. Then the solution is
itr t K 1e K 2e s1t
s2t
In this case, we say that the circuit is overdamped.
Transient of AC Circuits
Dr H.H. Hanafy
29
2. Critically damped case
If ζ = 1, the roots of the characteristic equation are real and equal. Then the solution is
itr t K 1e K 2te s1t
s1t
In this case, we say that the circuit is critically damped.
Transient of AC Circuits
Dr H.H. Hanafy
30
3. Underdamped case
if ζ < 1, the roots of the characteristic equation are complex. In other words, the roots are of the form
S 1 n J d S 2 n J d d n 1 Transient of AC Circuits
Dr H.H. Hanafy
2
31
For complex roots, the solution is of the form:
itr t K 1e
nt
cos d t K 2e
nt
sin d t
In this case, we say that the circuit is underdamped.
Transient of AC Circuits
Dr H.H. Hanafy
32
4. Undamped case
if ζ = 0 (R=0), the roots of the characteristic equation are imaginary. In other words, the roots are of the form
S 1 J n S 2 J n Transient of AC Circuits
Dr H.H. Hanafy
33
For imaginary roots, the solution is of the form:
itr t K 1 cos nt K 2 sin nt In this case, we say that the circuit is undamped or oscillatory.
Transient of AC Circuits
Dr H.H. Hanafy
34
Transient of AC Circuits
Dr H.H. Hanafy
35
R-L-C Circuit For the simple R-L-C circuit the voltage differential equation at the transient period is given by:
i (t ) c
dv c (t ) dt
v(t ) v R (t ) v L (t ) vc (t ) v(t ) RC
dvc (t ) dt
2
LC
d vc (t ) dt
vc (t )
v(t) 2 V sin( t v ) Transient of AC Circuits
Dr H.H. Hanafy
36
v(t ) RC
dvc (t ) dt
2
LC
d vc (t ) dt
vc (t )
This Equation is a second order differential equation and has a solution given by:
vc (t ) v ss
(t ) vtr (t )
Where: v ss(t) = the steady state voltage. vtr (t)= the transient voltage which is a decaying voltage with time. Transient of AC Circuits
Dr H.H. Hanafy
37
To get v ss(t) we solve the circuit in steady state conditions: V I R J I X L J I X C V I ( R JX L JX C ) I Z Z R J ( X L X C ) Z Z R ( X L X C ) 2
I
V v
V
Z Z
i ss (t ) 2 Transient of AC Circuits
1 X L
t an
2
X C R
v V
R ( X L X C ) 2
Dr. H.H. Hanafy
2
sin( t v ) 38
V ss J I ss X C
v ss (t ) 2 I X c sin( t v 90) v ss (t ) 2
Transient of AC Circuits
VX c Z
sin( t v 90)
Dr H.H. Hanafy
39
To get vtr (t) we solve the homogeneous differential equation. The homogeneous differential equation is obtained by setting the forcing function to zero. •
•
0 RC
dvtr (t ) dt
2
d vtr (t ) dt
2
LC
R dvtr (t ) L
dt
d vtr (t ) dt
1 LC
vtr (t )
vtr (t ) 0
This equation has a general solution with the form:
Transient of AC Circuits
Dr H.H. Hanafy
40
K 1 , K 2 are constants determined by initial conditions (capacitor voltage and inductor current). •
•
S1 and S2 are the roots of the characteristic equation:
S 2
R L
S
1 LC
0
2 S 2 n S n 0 2
n
1 LC
R
C
2
L
Transient of AC Circuits
naturalfrequency dampingratio Dr H.H. Hanafy
41
S 1, 2
2 n 4 4 2 n
2
2 n
2
S 1 n n 1 2
S 2 n n 1 2
Transient of AC Circuits
Dr H.H. Hanafy
42
1. Overdamped case
If ζ > 1, the roots of the characteristic equation are real and distinct. Then the solution is
vtr t K 1e K 2e s1t
s2t
In this case, we say that the circuit is overdamped.
Transient of AC Circuits
Dr H.H. Hanafy
43
2. Critically damped case
If ζ = 1, the roots of the characteristic equation are real and equal. Then the solution is
vtr t K 1e K 2te s1t
s1t
In this case, we say that the circuit is critically damped.
Transient of AC Circuits
Dr H.H. Hanafy
44
3. Underdamped case
if ζ < 1, the roots of the characteristic equation are complex. In other words, the roots are of the form
S 1 n J d S 2 n J d d n 1 Transient of AC Circuits
Dr H.H. Hanafy
2
45
For complex roots, the solution is of the form:
vtr t K 1e
nt
cos d t K 2e
nt
sin d t
In this case, we say that the circuit is underdamped.
Transient of AC Circuits
Dr H.H. Hanafy
46
4. Undamped case
if ζ = 0 (R=0), the roots of the characteristic equation are imaginary. In other words, the roots are of the form
S 1 J n S 2 J n Transient of AC Circuits
Dr H.H. Hanafy
47
For imaginary roots, the solution is of the form:
vtr t K 1 cos nt K 2 sin nt In this case, we say that the circuit is undamped or oscillatory.
Transient of AC Circuits
Dr H.H. Hanafy
48
