Transcript 6. Transient cct

Transient Analysis
The Growth of current in
R-L Circuit
a S
b
i
R
vR
+
vL
V
L
-
When switch S just switches to a : t=0 , i=0 , VL =0
Then current start flowing with the rate of di/dt thus a
voltage develop in inductor given by
di
vL  L  e
dt
a S
R
b
i
vR
+
vL
V
-
Using Voltage Kirchoff’s law
V  vR  vL
di
 iR  L
dt
Divided by R
V
L di
i 
R
R dt
di
I i 
dt
Where I = V/R = steady current
 = L/R = time constant
L
1
Rearrange the equation
1
dt 
di

Ii
Integrate both sides we have
 
Then the solution is
t
  ln I  i   A

1
dt 
(*)
A  ln I
Apply the initial state t=0 and i=0
t
I
Substitute in (*) , we have
 ln
I  i 

Voltage
1
di
Ii

t /
i

I
1

e
Or


di
d
IL t /
t / 
vL  L  L I 1  e
 e
dt
dt

Substitute  = L/R
vL  IR e
t /
 Ve
t /

vL, i
V
vL,
V/R
i
0
t
Decay analysis for R-L
circuit
a S
b
R
vR
+
V
-
vL
L
i
When switch S is switched to b iL will begin to decay
in a reverse direction with the rate of di/dt thus a
voltage develop in inductor given by
di
vL  L   e
dt
Using Voltage Kirchoff’s law
vR  vL  0
Or
Rearrange
1
1
 di  dt
i

Integrate

The solution

iR  L
di
0
dt
where  = L/R

1
1
di 
dt
i

t
  ln i  A

Apply initial condition t=0 , i=I
Or we have
t
 ln I  ln i

i  Ie
t / 
I
 ln  
i
Voltage
LI  t / 
d t / 
 LI e
 e
dt

di
vL  L
dt
Substitude  = L/R and IR = V ; we have
vL = - V e-t/
vL, i
V/R
i
+
0
t
-
vL
V
Example 1
1
2
r
10
R
10V
L
0.1H
15
vL
For network as in figure
(a)Determine the mathematical expressions for the variation of the
current in the inductor following the closure of the switch at t=0
on to position 1
(b)When the switch is closed on to position 2 at t=100ms,
determine the new expression for the inductor current and also
for the voltage across R;
(c)Plot the current waveforms for t=0 to t=200ms.
(a) For the switch in position 1, the time constant is
L 0.1
T1 

 10ms
R1 10

Therefore i1  I 1  e
 Tt
1

 t 3
 t 3
10 


 1  e 1 01 0   1  e 1 01 0 
 

10 
(b) For the switch in position 2, the time constant is
L
0.1
T2 

 4ms
R2 10  15
Therefore
i2  Ie  t / T2
 t 3
 t 3
10

e 41 0  0.4e 41 0
10  15
The voltage drop
vR  i2 R  0.4 15e

t
4103
 6e

t
4103
i(A)
1.0
0.5
20
40
60
80
t
100 120 140 160 180
At time t=0 till t=40ms i follows the equation
i1  1  e


t
10103
At time t=40ms till t=100ms i is saturated which is equaled to 1
At time t>100ms i follows the equation
i2  0.4e

t
41 03


Example 2
For the network in example 1, the switch is closed on to
position 1 as before but it is closed on to position 2 when
t=10ms. Determine the current expressions and hence plot
the current waveforms
When switch is switched on to
position 1 , i follows
i1  1  e


t
10103
 amperes



 10103 

At t=10ms the magnitude of i is i1  1  e 1010   1  e 1  0.632 A


3
When switch is switched on to
position 2 , i
i2  Ie  t / T2  0.632e

t
4103
A
Example 3
For network in the Figure , the switch is closed on to position 1
when t=0 and then moved to position 2 when t=1.5ms. Determine
the current in the inductor when t=2.5ms.
20
1
1A
2
10mH
10
10
20

First we simplify the supply of the circuit using Norton
Theorem
10  2010
20
Re 
1A
10
10
is / c
The simplified circuit is
10  20  10
 7.5 
10

1  0.33 A
10  20
1
0.33A
2
10mH
7.5
20

Switch at position 1
7.5
Current Maximum magnitude I 
 0.33  0.2 A
5  7.5
At time t=1.5ms
L 10 10
T1  
R
5
i1  0.21  e

 T1
t
1
Switch at position 2
3
 2.0ms
  0.21  e




 12.5
  0.106 A

L 10 10 3
T2  
 0.4ms
R
5  20
t2=2.5ms- t1=2.5ms-1.5ms=1ms
i2  0.106e
 T2
t
2
 0.106e
 01.4
 0.009 A  9mA
Energy stored
To neutralize the induced e.m.f which represents the power
absorbed by magnetic field a voltage source is needed. The
energy produced by the voltage and current developed is
stored in the form of magnetic field. This is given by
di
vL  L
p  ivL
and
dt
di
 p  iL
dt
Average energy absorbed
Total energy absorbed
di
w  pdt  iL dt  iLdi
dt
 
1 2
W   iLdi  Li
0
2
I
L
0
1 2
 LI
2
The Growth of current in
R-C Circuit
a S
b
+
R
i
vR
vC
V
C
-
When switch S just switches to a : t=0 , q=0 , VC =0
Then voltage start develops across the capacitor with the rate
of dVC/dt thus a current flowing in the circuit is given by
dv C
iC  C
dt
a S
b
+
R
i
vR
vC
V
C
-
Using Voltage Kirchoff’s law
Or
V = vR + vC
V – vC = vR = iR
Substitute i we have
Or
dvC
V  vC  RC
dt
dvC
V  vC  
dt
Where  =RC
Rearrange we have
dvC
dt

V  vC 
dv C
dt
 V  vC   
Integrate
t
  ln( V  vC )  A

Then we have
At t=0 we have vC=0 thus A=ln V
Substitute A we have
Or
t
V
 ln

V  vC
vC
 V1  e 
t / 
Substitute VC in the current formulation
d
CV  t 
 t /
i  CV (1  e ) 
e
dt

V t / 
i e
R
dvC
iC
dt
vC, i
V
vC
V
R
i
t
Voltage development and current decay
in serial RC circuit
Decay analysis for R-C
circuit
a S
b
+
V
R
i
vR
vC
C
-
When the switch is connected to b, vC = V. The accumulated
charges in C now start to discharge via R. The initial current iI is
given by:
V
iI  
R
In the process of discharging, voltage across C, vC is
decaying. When charges are full discharged, vC = 0
and the final current iF = 0.
Using voltage‘s Kirchoff law
vC = -vR
i
Therefore
But i  C
vR  vC

R
R
dvC
dt
Rearrange
Substitute we have
dt
dvC

RC
vC
Then we have
C
dvC
v
 R
dt
R
And substitute RC = 
dt
dvC


vC
Integrate
We get
 
dt


dvC
vC
t
  ln vC  A

If t=0 ; vC=V , then we have
Substitute A we have
We get
A = ln V
t
V
  ln v C  ln V  ln

vC
V
 et / 
vC
or
vC  Ve t / 
For current we have
vC V  t 
i
 e
R R
vC , i
V
vC
0
-V/R
t
i
Decaying of Voltage and
currrent in RC circuit
Example 4
A capacitor is made of two pieces of aluminium foil separated
by a piece of paper of 0.2mm thick and having a relative
permittivity of r = 5. Determine the area of aluminium
required to produce a capacitance of 1200 pF? Assume that
the permittivity of the free space is o ,8.854 x 10-12 F/m.
C = 1200 pF = 1200 x 10-12 F
d = 0.2 mm = 2 x 10-4 m
r = 5; dan o = 8.854 x 10-12 F/m
A = ?
C
 r o A
d
1200 1012  2 10 4
2
A


54
.
2
cm
 r 0
5  8.854 1012
Cd
Example 5
Two capacitors, C1 = 1200 pF and C2 = 2200 pF are connected
in parallel. The combination is connected to an a.c voltage
source VS = 155 V having a series resistor RS. After the
voltage across the capacitor reach a steady state, the
connection of capacitor to the battery is disconnected and
discharged via a resistor RP.
1. Draw the circuit.
2. Calculate Rs so that the energy stored in capacitor reaches 25
mJ in 250ms during capacitor charging
3. Calculate Rp , after 1ms discharging, the voltage drop is 20%
from its steady state.
4. Calculate the energy stored in capacitor after 1.5ms
discharging.
(1)
RS
VS
155 V
RP
C1
1200 pF
C2
2200 pF
(2)
Total capacitance
CT  C1  C2  1200  2200  3400 pF
W  12 CvC  25 106 J
Given
2
Voltage across cap.
During charging
rewrite
e
then
 t / RS C
2W
2  25 10 6
vC 

 121 V
12
CT
3400 10

vC  VS 1  e
t / R S C
vC VS  vC
 1

VS
VS
t
RS C
 ln

vC
 1  et / R
VS
or
or
e
t / R SC
S
C
VS

VS  vC
VS
155
 ln
 1.517
VS  v C
155  121
t
250 10 6
RS 

 48.5k
12
1.517  C 1.517  3400 10
(3)
vC  VSe  t / R
During discharging
P
CT
Given that vc drop 20% from its steady voltage , vC= 0.8VS
0.8VS  VSe t / R
et / R
t
R PC
P
P
CT
CT
or
0.8  e t / R
P
CT
 1.25
 ln 1.25  0.223
t
110 3
RP 

 1.32M
12
0.223  CT 0.223  3400 10
(4)
After 1.5ms discharge
1.5 10 3

 0.334
6
12
RPC 1.32 10  3400 10
t
v C  1.55e
 t / RP CT
 1.55e
0.334
 111V
W  CvC  12  3400 1012 1112  21mJ
1
2
2