Response of RLC networks to general class of excitation functions'

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• Response of R-L-C networks to general class of
  excitation functions.
• Let us first consider simple R-L and R-C (1st
  order) networks subjected to step input
  excitation (Fig.2.17).

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• i (0) 0

t 0
R
VR
L
VL
V
Fig. 2.17
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• The coil is initially having no stored magnetic
  energy and the switch is closed at t 0. The
  voltages across R and L are VR(t) and VL(t) and
  the dynamic equation will be

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(1)
5

• whence
• At t 0, i 0

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(2)
(3)
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• where

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• Fig. 2.18 shows these variables against time.

V
T
VR
V
V/R
Ist.
VR
iR iL
Fig. 2.18
VL
i
t
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(No Transcript)
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• If we maintain this slope of the current, then
  i.e., if we maintain
  the same initial slope, then after a lapse of
  T seconds meets the steady
  value of the current

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• If the value of R in the circuit is increased,
  will reduce and hence the voltage VR and VL
  and the current iR will attain the steady value
  after a time much less than T.

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• Once again,
• At t T

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• T can be defined as the time after which, the
  current comes to 63 of the final value.

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• R-C Circuit Consider now a simple R-C as shown
  in Fig. 2.19, excited by a DC voltage V.

t 0
R
Fig. 2.19
C
V
i
15

(4)
16

(5)
(6)
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i0 V/R
i(t)
t
Fig. 2.20
(a)
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V
vc
vR
t
i
Fig. 2.20
(b)
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• Since T R C, if R increases, T increases and
  the system response becomes sluggish.
• If we now have a circuit with R-L-C elements as
  shown in Fig 2.21, excited by a voltage V.

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R
C
L

• Differentiating we get,

V
Fig. 2.21
(7)
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• is the characteristic equation and the roots of
  p are

(7a)
(8)
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• We now convert equation (7) to a standard form
  and determine the initial value of R such that
  the radical term in equation (8) becomes zero.

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• We define the two terms ? and ?n as.

(9)
(10)
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• where ? dimensionless damping ratio and ?n
  natural frequency of oscillation of the network
  when
• R 0. Thus,
• Equation (7) can be rewritten as

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• If we use the term Q the quality factor of the
  coil , then equation (12) can be also
  written as
• 
  (13)

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• Let us once again refer to eqn. (8)
• and

(14)
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• Three cases now arise, for the nature of the
  response.
• Case I If ? gt1, the roots are real.
• Case II If ? 1, the roots are equal and
  real.
• Case III If ? lt1, the roots are complex
  conjugate.

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• If we trace the loci of the roots in the complex
  plane of p as ? changes, they appear as shown
  in Fig. 2.22.

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j?
??n

• The location of roots
• shown for ? lt 1

• The loci of roots as
• ? varies from o to ?.

?n
?
Real ?
? 0
j?
j?n
?
? ?
? gt1
? ??
? 1
Fig. 2.22
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• At ? 0 , p1, p2 ? j?n R 0
• At 0 lt ? lt 1, p1, p2 - ? ? j?
• ? ? ?n,
• ? ?2 ?2 ?n2
• And Cos ? ?

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• Case I when ? gt 1 , Q lt 1/2
• Since both the roots are real and negative, the
  response is a monotonically decreasing one
  (Fig.2.23.).

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i
t
Fig. 2.23
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• Case II When ? 1, the roots are equal
  and with repeated roots the solution is of the
  form
• At t ? 0, i (0) ?
• you can calculate this by applying lHospitals
  rule

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• Case III ? lt 1 or Q gt 1/2
• which can be finally written as

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• Let us now consider a situation with non-zero
  initial condition.
• Example 2.5
• Consider the network given in Fig.2.24a where V
  100V, R1 20?, R2 30 ?, L 1 h.

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t 0
K

R1
L
R2
V
Fig. 2.24 (a)
5
Before t 0, there was a steady initial current

3
vc
i 2
T1
t (sec)
Fig. 2.24 (b)
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• The switch K is closed at t 0. Determine the
  expression for the current.

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• After the switch is closed, the steady value of
  the current i (?) will be
  5A.
• So the current i (t) starts from an initial
  value of 2A and increases to 5A and the time
  constant involved is
• 0.05
  sec (Fig.2.24b)

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• We can write the general expression for the
  current as

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• If after the steady value of 5A is obtained, we
  put the switch k off the current will decrease
  from 5A to 2A with an exponential decay with a
  time constant

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43

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• The fall in the current is shown in Fig.2.24 C

5
3
i(t)
T2 50
0
t
Fig. 2.24 (c)
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• Initial Conditions
• The current cannot change instantaneously in an
  inductor and the voltage can not change
  instantaneously across a capacitor.

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• If an uncharged capacitor is connected to an
  energy source, an instantaneous current with flow
  and the capacitor behaves as a short circuit
  and zero charge means zero voltage or
  short circuit.

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• With an initial charge qo, the capacitor is
  equivalent to a voltage source of value V0 qo
  /C.
• Thus we have the equivalent circuit
  representations at t 0 for the different
  elements as follows.

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At t 0
At t ?
Elements

0.C
S.cct
i(0)
i(0)
-

-
-
V0 q0 /C
V0
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• will have zero values at steady state( at t ?)
  and hence they are shown as given in the right
  column Fig 2.25

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• Exercise 2
• 1.(a) Determine the initial values of voltages
  across the different capacitors in the following
  circuit Fig 2.26. When the switch is put on ( t
  0)

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C2
R1
C1
R2
-
V0
C3
C4
Fig. 2.26
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• R1 100 ? , R2 100 ? , C1 20 ?F, C2 ?F,
  C3 100 ?F, C4 40 ?F ,V0 100 V.
• (b) What would be the voltages across the
  network elements for the circuit shown,

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• if the switch k is kept open initially and at t
  0, it is suddenly closed ? Determine i1(0), i2
  (0), i3 (0), Vc1(0), Vc2 (0).

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50?
100V
C1 40 ?F
50?
i2
i1
L 1h
i3
C2 20 ?F
Fig. 2.27