Initial conditions

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• Initial conditions
• R.V. Srinivasa Murthy
• Assistant Professor
• Dept. of E C
• A.P.S. College of Engineering
• Bangalore

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Initial conditions

• The reason for studying initial and final
  conditions in a network is to evaluate the
  arbitrary constants that appear in the general
  solution of the differential equations written
  for the network.

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In this chapter we concentrate on finding the
change in selected variables in a circuit when a
switch is thrown from open to closed or vice
versa position.

• Note that
• t 0 indicates the time of throwing the
  switch
• t 0- indicates time immediately before throwing
  the switch and
• t 0 indicates time immediately after throwing
  the switch

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We are very much interested in the change in
currents and voltages of energy storage elements
(inductor and capacitor) after the switch is
thrown since these variables along with the
sources will dictate the circuit behavior for t gt
0.

• Initial conditions in a network depend on the
  past history of the circuit (before t 0- ) and
  structure of the network at t 0
• Past history will show up in the form of
  capacitor voltages and inductor currents.

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Initial and final conditions in elementsi) The
resistor

• The cause effect relation for the ideal resistor
  is given by
• V R i
• From this equation we find that the current
  through a resistor will change instantaneously,
  if the voltage changes instantaneously.
• Similarly voltage will change instantaneously if
  current changes instantaneously.

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ii) The inductor

• Initial condition
• The switch is closed at t 0 The
  expression for current through the inductor is
  given by

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• The above equation indicates that the current in
  an inductor can not change instantaneously.
• Hence if i(0-) 0, then i(0) 0.
• This means that at t 0 inductor will act as an
  open circuit, independent of voltage across the
  terminals.

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If i (0 -) I 0 (i.e., if a residual
current is present) then i (0) I 0 , meaning
that an inductor at t 0 can be thought of as
a current source of I0 which is as shown
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Final (or steady state) condition

• The final - condition equivalent circuit of an
  inductor is derived from the basic relationship
  V L di/ dt
• Under steady state condition di 0 which means
  v 0 and hence L acts as a short dt
  at t 8 (final or steady state)

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The capacitor

• The switch is closed at t 0 .
• The expression for voltage across the capacitor
  is given by

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V(0) V(0-) means that the voltage across the
capacitor can not change instantaneously. If
V(0-) 0 then V (0) 0 indicating that
the Capacitor acts as a short at t 0
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Final (or steady state ) condition

• The final-condition equivalent network is
  determined from the basic relationship
  i C dv/dt
• Under steady state condition dv / dt 0 which
  means at t 8 the Capacitor acts as a open
  circuit.

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Example 1 In the given circuit k isclosed
at t0 with zero current in the inductorFind
the values of i , di/dt ,d2i/dt2 at t0

• Solution The symbol for the switch implies
  that it is open at t 0- and then closes at t
  0. From the data given it is also clear that i
  (0-) 0. Hence from the circuit i(0) i(0-)
  0

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To find di (0) and d2i (0) dt
dt2 Applying KVL to the circuit we get R
i (t) L di (t) 12
dt8 i (t) 0.2 di (t) 12 .(1)
dtAt t 0 8 i (0) 0.2 di
(0) 12 dt8 x 0
0.2 di (0) 12
dt di (0) 12 60 A/secdt 0.2
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Differentiating equation (1) with respect to
time8 di 0.2 d2i 0 dt
dt2At t 0 the above equation
becomes8 di (0-) 0.2 d2i (0)
0 dt dt2 8x 60 0.2
d2 i(0) 0 d t2
Hence d2 i (0) - 2400 A / Sec2
dt2
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Example 2
In the
network shown in figure the switch is closed at t
0 determine i, di and d2 i at t 0 dt
dt2

• Solution At t 0 the circuit appears as
  shown in figure.
• From the circuit i(0) 0 i(0-)

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Writing KVL clock wise for the circuit when the
switch is closed, we get

• Putting t 0 in equation (2), we get

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Example 3 Refer to the circuit shown in
figure. The switch K is changed from portion 1 to
position 2 at t 0. Steady state condition
having been reached in position 1, find the
values of i, di and d2i at t 0
dt dt2

• Solution The symbol for switch K implies that
  it is in position 1 at t 0 - and in position 2
  at t 0. Under steady state conditions,
  inductor acts as a short circuit.  
• Hence at t 0 - the circuit diagram appears as
  shown in figure

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i(0-) 20 2A 20V 10Since
the current through an inductor can not change
instantaneously i(0-) 20 2 A
10Since the current through an inductor can
not change instantaneously i(0) i(0-) 2
ASince there is no initial charge on capacitor
Vc (0-) 0Since the voltage across a
capacitor can not change instantaneously Vc (0)
Vc (0-) 0Hence at t 0 the circuit
appears as shown
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Example 4 In the network the switch is moved
from position 1 to position 2 at t 0.The steady
state having reached before switching. Calculate
i, di/dt, d2i/dt2 at t 0
SolutionThe symbol for switch K implies it is in
position 1 at t 0- and position 2 at t 0 .
Under steady state condition a capacitor acts as
an open circuit. Hence at t 0-, the circuit
diagram is as shown in figure.
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We know that the voltage across a capacitor
cannot change VC(0-) 40V instantaneously. This
means that VC(0) VC(0-) 40 Volts. Also i(0-)
0 Hence i(0)0

• The circuit diagram for t 0 is as shown

For t 0 the circuit diagram appears as shown.
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Example 5 The Network shown in the figure has
the switch K opened at t 0. Solve for v, dv
and d2v at t 0 dt dt2

• Solution Initial condition for Inductor is open
  circuit and hence the circuit at t 0 reduces
  to the form shown in figure.
• V(0) I R 1 x 100 100 Volts

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Consider the given circuit. Let V be the node
Voltage
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Example 6 In the network shown the switch K
is opened at t 0. At t 0, solve for the
values of V, dv and d2v if I 10 A ,R 100
ohms and C 1µF.dt dt2

• Solution At t 0, switch K is opened.
  Equivalent circuit at t 0 is as shown. Since
  capacitor is a short at t 0 we get V(0) 0

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Consider the given network. Let V be the node
voltage. Applying KCL we get, V C dv I
..(1)R dt
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