4.
Introduction
There are many reasons for studying initial and nal conditions. The most important reason is that the initial and nal conditions evaluate the arbitrary constants that appear in the general solution of a differential equation. In this chapter, we concentrate on nding the change in selected variables in a circuit when a switch is thrown open from closed position or vice versa. The time of throwing the switch is considered to be t = 0, and we want to determine the value of the variable at t = 0 and at + t = 0 , immediately before and after throwing the switch. Thus a switched circuit is an electrical circuit with one or more switches that open or close at time t = 0. We are very much interested in the change in currents and voltages of energy storing elements after the switch is thrown since these variables along with the sources will dictate the circuit behaviour for t > 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+ , (after switching). Past history will show up in the form of capacitor voltages and inductor currents. The computation of all voltages and currents and their derivatives at t = 0+ is the main aim of this chapter.
4.2
Initial and final conditions in elements
4.2.1 The inductor
The switch is closed at t = 0. Hence t = 0 corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed. The expression for current through the inductor is given by
i
1
L
Zt
vd
Figure 4.1 Circuit for explaining
switching action of an inductor
278
Network Theory 0 Z
vd
1
L
1
L
1
L
Zt
vd
Zt
0
i(t)
= i(0 ) +
vd
Putting t = 0+ on both sides, we get
i(0
) = i(0 ) + ) = i(0 )
1
L
0+ Z
vd
i(0
The above equation means that the current in an inductor cannot change instantaneously. Consequently, if i(0 ) = 0, we get i(0+ ) = 0. This means that at t = 0+ , inductor will act as an open circuit, irrespective of the voltage across the terminals. If i(0 ) = Io , then i(0+ ) = Io . In this case at t = 0+ , the inductor can be thought of as a current source of Io A. The equivalent circuits of an inductor at t = 0+ is shown in Fig. 4.2.
t = 0+
Figure 4.2 The initial-condition equivalent circuits of an inductor
The nal-condition equivalent circuit of an inductor is derived from the basic relationship
v di
=L
di dt
Under steady condition, = 0. This means, v = 0 and hence L acts as short at t = (nal dt or steady state). The nal-condition equivalent circuits of an inductor is shown in Fig.4.3.
t=
Io Io
Figure 4.3 The final-condition equivalent circuit of an inductor
Initial Conditions in Network
279
4.2.2 The capacitor
The switch is closed at t = 0. Hence, t = 0 corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed. The expression for voltage across the capacitor is given by
v
1
C
Zt
id
Figure 4.4 Circuit for explaining
switching action of a Capacitor
v (t)
1
C
0 Z
id
1
C
t
Zt
id
v (t)
= v (0 ) +
1
C
Z
0
id
Evaluating the expression at t =
v (0
0+ ,
we get 1
C
0+ Z
id
) = v (0 ) +
v (0
) = v (0 )
Thus the voltage across a capacitor cannot change instantaneously. If v (0 ) = 0, then v (0+ ) = 0. This means that at t = 0+ , capacitor C acts as short circuit. q0 q0 Conversely, if v (0 ) = then v (0+ ) = . These conclusions are summarized in Fig. 4.5.
C C
Figure 4.5 Initial-condition equivalent circuits of a capacitor
The nalcondition equivalent network is derived from the basic relationship
i dv
=C
dv dt
Under steady state condition, = 0. This is, at t = , i = 0. This means that t = dt or in steady state, capacitor C acts as an open circuit. The nal condition equivalent circuits of a capacitor is shown in Fig. 4.6.
280
Network Theory
Figure 4.6 Final-condition equivalent circuits of a capacitor
4.2.3 The resistor
The causeeffect relation for an ideal resistor is given by v = Ri. From this equation, we nd that the current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, voltage will change instantaneously if current changes instantaneously.
4.3
Procedure for evaluating initial conditions
There is no unique procedure that must be followed in solving for initial conditions. We usually solve for initial values of currents and voltages and then solve for the derivatives. For nding initial values of currents and voltages, an equivalent network of the original network at t = 0+ is constructed according to the following rules: (1) Replace all inductors with open circuit or with current sources having the value of current owing at t = 0+ . (2) Replace all capacitors with short circuits or with a voltage source of value vo = is an initial charge. (3) Resistors are left in the network without any changes.
EXAMPLE 4.1
i1 (0 q0 C
if there
Refer the circuit shown in Fig. 4.7(a). Find for t < 0.
+)
and
iL (0
+ ).
The circuit is in steady state
t=0
1W
1W
Figure 4.7(a)
Initial Conditions in Network
281
SOLUTION
The symbol for the switch implies that it is open at t = 0 and then closed at t = 0+ . The circuit is in steady state with the switch open. This means that at t = 0 , inductor L is short. Fig.4.7(b) shows the original circuit at t = 0 . Using the current division principle, 2 1 = 1A iL (0 ) = 1+1
1W
1W
Figure 4.7(b)
Since the current in an inductor cannot change instantaneously, we have
iL (0
) = iL (0 ) = 1A
At t = 0 , i1 (0 ) = 2 1 = 1A. Please note that the current in a resistor can change instantaneously. Since at t = 0+ , the switch is just closed, the voltage across R1 will be equal to zero because of the switch being short circuited and hence,
i1 (0
) = 0A
Thus, the current in the resistor changes abruptly form 1A to 0A.
EXAMPLE 4.2
Refer the circuit shown in Fig. 4.8. Find vC (0+ ). Assume that the switch was in closed state for a long time.
Figure 4.8
SOLUTION
The symbol for the switch implies that it is closed at t = 0 and then opens at t = 0+ . Since the circuit is in steady state with the switch closed, the capacitor is represented as an open circuit at t = 0 . The equivalent circuit at t = 0 is as shown in Fig. 4.9.
282
Network Theory
vC (0
) = i(0 )R2
Using the principle of voltage divider,
vC (0
)=
VS R1
+ R2
R2
5 1 = 2:5 V 1+1
Figure 4.9
Since the voltage across a capacitor cannot change instaneously, we have
vC (0
) = vC (0 ) = 2:5V That is, when the switch is opened at t = 0, and if the source is removed from the circuit, still + vC (0 ) remains at 2.5 V.
EXAMPLE 4.3
Refer the circuit shown in Fig 4.10. Find iL (0+ ) and vC (0+ ). The circuit is in steady state with the switch in closed condition.
Figure 4.10
SOLUTION
The symbol for the switch implies, it is closed at t = 0 and then opens at t = 0+ . In order to nd vC (0 ) and iL (0 ) we replace the capacitor by an open circuit and the inductor by a short circuit, as shown in Fig.4.11, because in the steady state L acts as a short circuit and C as an open circuit. 5 =1A 2+3 Using the voltage divider principle, we note that
iL (0
Figure 4.11
)=
vC (0
)=
+ +
5 3 =3V 3+2
Then we note that:
vC (0
) = vC (0 ) = 3 V ) = iL (0 ) = 2 A
iL (0
Initial Conditions in Network
283
EXAMPLE
4.4
In the given network, K is closed at t = 0 with zero current in the inductor. Find the values of i;
di dt
Refer the Fig. 4.12(a).
2 d i dt2
at t = 0+ if R = 8 and L = 0:2H.
SOLUTION
The symbol for the switch implies that it is open at t = 0 and then closes at t = 0+ . Since the current i through the inductor at t = 0 is zero, it implies that i(0+ ) = i(0 ) = 0. To nd
+ di(0 )
dt
Figure 4.12(a)
R = 8W
and
2 + d i(0 )
dt2
L = 0.2H
Applying KVL clockwise to the circuit shown in Fig. 4.12(b), we get
Ri
Figure 4.12(b)
+L
di dt di dt
= 12 = 12 (4.1)
8i + 0:2
At t = 0+ , the equation (4.1) becomes 8i(0+ ) + 0:2 8
di(0 di(0
+)
0+02
:
dt + dt + dt
= 12 = 12 12 0:2 = 60 A=sec =
) )
di(0
Differentiating equation (4.1) with respect to t, we get 8
+)
di dt
+ 0:2
2 d i dt2
=0
At t = 0+ , the above equation becomes 8
di(0 dt
+ 0:2
:
Hence
60 + 0 2
2 + d i(0 ) dt2
=0 =0 = 2400 A=sec2
2 + d i(0 ) dt2
2 + d i(0 ) dt2
284
Network Theory
EXAMPLE
4.5
di dt
In the network shown in Fig. 4.13, the switch is closed at t = 0. Determine i;
2 d i dt2
at t = 0+ .
Figure 4.13
SOLUTION
The symbol for the switch implies that it is open at t = 0 and then closes at t = 0+ . Since there is no current through the inductor at t = 0 , it implies that i(0+ ) = i(0 ) = 0.
R = 10W L = 1H
C = 1mF
Figure 4.14
Writing KVL clockwise for the circuit shown in Fig. 4.14, we get
Ri
+L
di dt
1
C
Zt
i( )d
= 10
(4.2) (4.2a)
Ri
Ri
+L 0+
0 di
dt
+ vC (t) = 10
Putting t = 0+ in equation (4.2a), we get 0
+
+L
R
di
0+
dt L
+ vC 0+ = 10
di
0+
dt +
+ 0 = 10 = 10
L
di
0
dt
= 10 A/sec
Differentiating equation (4.2) with respect to t, we get
R di dt
+L
2 d i dt2
i(t) C
=0
Initial Conditions in Network
285
At t = 0+ , the above equation becomes
R di
0+
dt
+L
2 d i
0+
Hence at
EXAMPLE
t
10 +
100 +
dt2 C 2 + d i 0 0 + L 2 dt C
0+
=0 =0
2 d i 2 d i
0+ 0
dt2 dt2
= 0+ ;
+
=0 = 100 A/sec2
4.6
Refer the circuit shown in Fig. 4.15. The switch K is changed from position 1 to position 2 at t = 0. Steady-state condition having been reached at position 1. Find the values of i, and
2 d i dt2
di dt
at t = 0+ .
Figure 4.15
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 condition, inductor acts as a short circuit. Hence at t = 0 , the circuit diagram is as shown in Fig. 4.16. 20 = 2A 10 Since the current through an inductor cannot change instantaneously, i 0+ = i (0 ) = 2A. Since there is no initial charge on the capacitor, vC (0 ) = 0. Since the voltage across a capacitor cannot change instanta+ neously, vC 0 = vC (0 ) = 0. Hence at t = 0+ the circuit diagram is as shown in Fig. 4.17(a). For t 0+ , the circuit diagram is as shown in Fig. 4.17(b).
i
Figure 4.16
Figure 4.17(a)
Figure 4.17(b)
286
Network Theory
Applying KVL clockwise to the circuit shown in Fig. 4.17(b), we get
Ri(t)
+L
di(t) dt
1
C
Zt
i( )d
=0
(4.3)
Ri
0+
Ri(t)
+L
di(t) dt
+ vC (t) = 0
(4.3a)
At t = 0+ equation (4.3a) becomes 0
+
+L
R
di
0+
dt
di
+ vC 0+ = 0 0+
dt di di
At t = 0+ , we get
R di
2+
+0=0 0+
dt dt
20 +
+
=0 = 20 A/sec
Differentiating equation (4.3) with respect to t, we get
R di dt
+L 0+
2 d i dt2
i C
=0
0+
dt
+L
2 d i
+
0+
=0 =0
Hence;
EXAMPLE 4.7
(
20) +
dt2 C 2 + d i 0 2 + L 2 dt C 2 + d i 0 dt2
2 106 A=sec2
In the network shown in Fig. 4.18, the switch is moved from position 1 to position 2 at t = 0. The steady-state has been reached before switching. Calculate i,
di dt
, and
2 d i dt2
at t = 0+ .
Figure 4.18
Initial Conditions in Network
287
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 condition, a capacitor acts as an open circuit. Hence at t = 0 , the circuit diagram is as shown in Fig. 4.18(a). We know that the voltage across a capacitor cannot change instantaneously. This means that Figure 4.18(a) + = v (0 ) = 40 V. vC 0 C At t = 0 , inductor is not energized. This means that i (0 ) = 0. Since current in an inductor cannot change instantaneously, i 0+ = i (0 ) = 0. Hence, the circuit diagram at t = 0+ is as shown in Fig. 4.18(b). The circuit diagram for t 0+ is as shown in Fig.4.18(c).
Figure 4.18(b)
Figure 4.18(c)
Applying KVL clockwise, we get
Ri
+L
di dt
1
C
Zt
i( )d
=0
(4.4)
At t = 0+ , we get
Ri(0
0+
Ri
+L
di dt
+ vC (t) = 0
)+L
di(0
+)
20
0+1
dt
+ vC (0+ ) = 0
+)
dt
di(0
+ 40 = 0
dt
+ di(0 )
40A/ sec
Diferentiating equation (4.4) with respect to t, we get
R di dt
+L
2 d i dt2
i C
=0
288
Network Theory
Putting t = 0+ in the above equation, we get
R
+ di(0 )
Hence
EXAMPLE 4.8
(
dt
+L
2 + d i(0 ) dt2
i(0
+)
40) + L
2 + d i(0 ) dt2
=0 =0 = 800A/ sec2
0
C +
2 d i(0 dt2
In the network shown in Fig. 4.19, v1 (t) = e is initially uncharged, determine the value of
for t
2
2 d v
0 and is zero for all
3 d v
dt3
t<
0. If the capacitor
dt2
and
at t = 0+ .
Figure 4.19
SOLUTION
Since the capacitor is initially uncharged, + v2 (0 ) = 0 Referring to Fig. 4.19(a) and applying KCL at node v2 (t):
v2 (t) v1 (t) R1
+C
dv2 (t) dt
1
R1
1
R2
v2 (t) R2 dt
=0 =
v1 (t) R1 dv2 dt
v2 (t)
+C
dv2 (t)
Figure 4.19(a)
Putting t = 0+ , we get
0:15v2 + 0:05
+) +) +)
= 0:1e
(4.5)
0:15v2 (0+ ) + 0:05
dv2 (0 dt dv2 (0 dt
= 0:1 = 0:1 = 0:1 = 2 Volts= sec 0:05
0:15
0 + 0 05
:
dv2 (0 dt
Initial Conditions in Network
289
Differentiating equation (4.5) with respect to t, we get 0:15
dv2 dt
+ 0:05
2 d v2 dt2
0:1e
(4.6)
Putting t = 0+ in equation (4.6), we nd that 0:1 0:3 = 8 Volts/ sec2 0:05 Again differentiating equation (4.6) with respect to t, we get = 0:15
2 d v2 dt2 2 + d v2 (0 ) dt2
+ 0:05
3 d v2 dt3
= 0:1e
(4.7) we nd that
Putting t = 0+ in equation (4.7) and solving for
3 + d v2 (0 ) dt3
3 d v2 + (0 ), dt3
0:1 + 1:2 = 26 Volts/ sec3 0:05
EXAMPLE
4.9
dvK dt
Refer the circuit shown in Fig. 4.20. The circuit is in steady state with switch K closed. At t = 0, the switch is opened. Determine the voltage across the switch, vK and at t = 0+ .
SOLUTION
Figure 4.20
The switch remains closed at t = 0 and open at t = 0+ . Under steady condition, inductor acts as a short circuit and hence the circuit diagram at t = 0 is as shown in Fig. 4.21(a). Therefore; For t
vK (0
) = vK (0 ) =0V
0
the circuit diagram is as shown in Fig. 4.21(b).
Figure 4.21(a)
Figure 4.21(b)
290
Network Theory
dvK dt
i(t)
=C
At (t) =
0+ ,
we get
i(0
)=C
dvK (0 dt
+)
Since the current through an inductor cannot change instantaneously, we get
i(0
Hence;
dvK (0 dt
) = i(0 ) = 2A + dvK (0 ) 2=C
dt
+)
2
C
2
1 2
= 4V/ sec
EXAMPLE
4.10
dv dt
In the given network, the switch K is opened at t = 0. At t = 0+ , solve for the values of v; and
2 d v dt2
if I = 2 A; R = 200 and L = 1 H
Figure 4.22
SOLUTION
The switch is opened at t = 0. This means that at t = 0 , it is closed and at t = 0+ , it is open. Since iL (0 ) = 0, we get iL (0+ ) = 0. The circuit at t = 0+ is as shown in Fig. 4.23(a).
Figure 4.23(a)
v (0
Figure 4.23(b)
+
) = IR =2
= 400 Volts
200
Initial Conditions in Network
291
Refer to the circuit shown in Fig. 4.23(b). For t 0+ , the KCL at node v (t) gives
v (t) R
1
L
Zt
v ( )d
(4.8)
0+
Differentiating both sides of equation (4.8) with respect to t, we get 1 dv (t) 1 0= + v (t)
R dt L
(4.8a)
At t = 0+ , we get 1
dv (0
+)
At t = 0+ , we get
1 dv (0 ) 1 + 200 dt 1
dt +
1
L
v (0
)=0
400 = 0
+)
dt
dv (0
8 104 V/ sec
Again differentiating equation (4.8a), we get
2 d v (t) R dt2
1
L
dv (t) dt
=0
EXAMPLE 4.11
1 d2 v (0+ ) 1 dv (0+ ) + =0 200 dt2 1 dt 2 + d v (0 ) = 200 2
dt
= 16 106 V/ sec2
8 10
In the circuit shown in Fig. 4.24, a steady state is reached with switch K open. At t = 0, the switch is closed. For element values given, determine the values of va (0 ) and va (0+ ).
Figure 4.24
292
Network Theory
SOLUTION
At t = 0 , the switch is open and at t = 0+ , the switch is closed. Under steady conditions, inductor L acts as a short circuit. Also the steady state is reached with switch K open. Hence, the circuit diagram at t = 0 is as shown in Fig.4.25(a).
iL (0
)=
5 5 2 + = A 30 10 3 5 20 10 = V 10 + 20 3 2 A: 3
Using the voltage divider principle:
va (0
)=
Since the current in an inductor cannot change instantaneously,
iL (0
) = iL (0 ) =
At t = 0+ , the circuit diagram is as shown in Fig. 4.25(b).
Figure 4.25(a)
Figure 4.25(b)
Refer the circuit in Fig. 4.25(b). KCL at node a:
va (0
+)
KCL at node b:
10 10 1 1 1 + + + va (0 ) 10 10 20
+) +)
va (0
+)
va (0
=0 20 1 5 + vb (0 ) = 20 10
+)
vb (0
+)
vb (0
5 2 + =0 10 3 1 5 1 1 + + + vb (0 ) = + va (0 ) 20 20 10 10 20 +
va (0
vb (0
+)
2 3
Initial Conditions in Network
293
Solving the above two nodal equations, we get,
va (0
)=
40 V 21
EXAMPLE
4.12
dvC (0 dt
Find iL (0+ ); vC (0+ );
+)
and
diL (0 dt
+)
for the circuit shown in Fig. 4.26.
Assume that switch 1 has been opened and switch 2 has been closed for a long time and steadystate conditions prevail at t = 0 .
SOLUTION
Figure 4.26
At t = 0 , switch 1 is open and switch 2 is closed, whereas at t = 0+ , switch 1 is closed and switch 2 is open. First, let us redraw the circuit at = 0 by replacing the inductor with a short circuit and the capacitor with an open circuit as shown in Fig. 4.27(a). From Fig. 4.27(b), we nd that iL (0 ) = 0
t
Figure 4.27(a)
Figure 4.27(b)
294
Network Theory
Applying KVL clockwise to the loop on the right, we get vC (0 ) 2+1 0=0 Hence, at
t
=0 :
vC (0 iL (0 vC (0
)=
2V 2V
Figure 4.27(c)
) = iL (0 ) = 0A
+
) = vC (0 ) =
The circuit diagram for Fig. 4.27(c).
0
is shown in
Applying KVL for righthand mesh, we get
vL vC
+ iL = 0
+
At t = 0+ , we get
vL (0
) = vC (0+ ) = 2
diL dt
iL (0
0=
2V
We know that At t = 0+ , we get
diL (0 dt
vL
=L
+)
vL (0 L
+)
2 = 1
2A/ sec
Applying KCL at node X ,
vC
10 2
+ iC + iL = 0
+)
Consequently, at t = 0+
iC (0
)=
10
vC (0
2
dvC dt +
iL (0
)=6
0=6A
Since We get,
EXAMPLE 4.13
dvC (0 dt
iC
=C =
+)
iC (0 C
6
1 2
= 12V/ sec
For the circuit shown in Fig. 4.28, nd: (a) i(0+ ) and v (0+ ) (b)
di(0
+)
(c) i( ) and v ( )
dt
and
dv (0
+)
dt
Initial Conditions in Network
295
SOLUTION
Figure 4.28
(a) From the symbol of switch, we nd that at t = 0 , the switch is closed and t = 0+ , it is open. At t = 0 , the circuit has reached steady state so that the equivalent circuit is as shown in Fig.4.29(a). 12 i(0 ) = = 2A 6 v (0 ) = 12 V Therefore, we have
v (0 i(0
) = i(0 ) = 2A
(b) For t
0
) = v (0 ) = 12V
+,
we have the equivalent circuit as shown in Fig.4.29(b).
Figure 4.29(a)
Figure 4.29(b)
Applying KVL anticlockwise to the mesh on the right, we get
vL (t) v (t)
+ 10i(t) = 0
Putting t = 0+ , we get
vL (0
v (0
) + 10i(0+ ) = 0
vL (0
vL (0
12 + 10
2=0
+
)=
8V
296
Network Theory
The voltage across the inductor is given by
vL
=L
di dt di(0
vL (0 di(0
)=L = = 1
L
+) +
+)
dt vL (0
dt
) 0.8A/ sec
1 ( 8) = 10 Similarly, the current through the capacitor is
iC
=C = =
dv dt i(0 C
or
dv (0 dt
+)
= 10
iC (0 C
+)
+)
2 10
0.2 106 V/ sec
(c) As t approaches innity, the switch is open and the circuit has attained steady state. The equivalent circuit at t = is shown in Fig.4.29(c).
i( v
) = 0 ( ) = 0
Figure 4.29(c)
EXAMPLE
4.14
Refer the circuit shown in Fig.4.30. Find the following: (a) (b) (c)
v (0
+)
and i(0+ ) and
i di(0
dv (0 dt
+)
+)
v(
) and ( )
dt
Figure 4.30
SOLUTION
From the denition of step function,
u(t)
=
1; 0;
0 t<0
t>
Initial Conditions in Network
297
From Fig.4.31(a), u(t) = 0 at t = 0 .
Similarly; or
u( u( t) t)
=
1; 0; 1; 0;
0 t<0
t>
0 t>0
t<
From Fig.4.31(b), we nd that u(
t)
= 1, at t = 0 .
t = 0+ t = 0
Figure 4.31(a)
Figure 4.31(b)
Due to the presence of u( t) and u(t) in the circuit of Fig.4.30, the circuit is an implicit switching circuit. We use the word implicit since there are no conventional switches in the circuit of Fig.4.30. The equivalent circuit at t = 0 is shown in Fig.4.31(c). Please note that at t = 0 , the independent current source is open because u(t) = 0 at t = 0 and the circuit is in steady state.
Figure 4.31(c)
40 = 5A 3+5 v (0 ) = 5i(0 ) = 25V
i(0
)=
Therefore
i(0 v (0
+ +
) = i(0 ) = 5A ) = v (0 ) = 25V
(b) For t 0+ ; u( t) = 0. This implies that the independent voltage source is zero and hence is represented by a short circuit in the circuit shown in Fig.4.31(d).
298
Network Theory
Figure 4.31(d)
Applying KVL at node a, we get 4+i=C At t = 0+ , We get 4 + i(0+ ) = C
dv dt
+
+)
5 +
v (0
dv (0 dt
+)
Evaluating at t = 0+ , we get
4 + 5 = 0:1
dv (0 dt
dv (0 dt
+)
+)
5 25 + 5
= 40V/ sec
Applying KVL to the leftmesh, we get 3i + 0:25
+)
di dt
+v =0
3i(0+ ) + 0:25
di(0
5 + 0 25
:
dt
+ v (0+ ) = 0
+)
di(0
di(0
dt +
+ 25 = 0
1 4
dt
40
160A/ sec
(c) As t approaches innity, again the circuit is in steady state. The equivalent circuit at t = is shown in Fig.4.31(e).
Figure 4.31(e)
Initial Conditions in Network
299
Using the principle of current divider, we get
4 5 = i( ) = 3+5 v ( ) = (i( ) + 4) 5
2.5A
= ( 2:5 + 4)5 = 7.5V
EXAMPLE 4.15
Refer the circuit shown in Fig.4.32. Find the following: (a) i(0+ ) and v (0+ ) (b)
di(0
+)
(c) i( ) and v ( )
Figure 4.32
SOLUTION
dt
and
dv (0
+)
dt
Here the function u(t) behaves like a switch. Mathematically,
u(t)
1; 0;
t> t<
0 0
The above expression means that the switch represented by u(t) is open for t < 0 and remains closed for t > 0. Hence, the circuit diagram of Fig.4.32 may be redrawn as shown in Fig.4.33(a).
Figure 4.33(a)
For t < 0, the circuit is not active because switch is in open state, This implies that all the initial conditions are zero. That is, iL (0 ) = 0 and vC (0 ) = 0 for t
0
+,
the equivalent circuit is as shown in Fig.4.33(b).
300
Network Theory
Figure 4.33(b)
From the circuit diagram of Fig.4.33(b), we nd that vC i= 5 + At t = 0 , we get + vC (0 ) vC (0 ) 0 + i(0 ) = = = = 0A 5 5 5 Also v = 15iL Evaluating at t = 0+ , we get + + v (0 ) = 15iL (0 ) = 15iL (0 ) = 15 (b) The equivalent circuit at t = We nd from Fig.4.33(c) that 0+
+
0 = 0V
is shown in Fig.4.33(c). ) = 5A
iC (0
Figure 4.33(c)
From Fig.4.33(b), we can write
vC dvC dt
= 5i =5
di dt di dt
Multiplying both sides by C , we get
C dvC dt
= 5C
Initial Conditions in Network
301
Putting t = 0+ , we get
di(0
iC
= 5C
di dt
+)
dt
1 + iC (0 ) 5C 1 5 = 5 1 4 = 4A/ sec =
Also
dv (0 dt
v dv dt dv dt dv dt
= 15iL = 15
diL dt
= 15 1 = 15vL
diL dt
At t = 0+ , we nd that
+)
= 15vL (0+ )
From Fig.4.33(b), we nd that vL (0+ ) = 0 + dv (0 ) = 15 0 Hence;
dt
= 0V/ sec
EXAMPLE
4.16
In the circuit shown in Fig. 4.34, steady state is reached with switch K open. The switch is closed at t = 0. di1 di2 Determine: i1 ; i2 ; and at t = 0+
dt dt
Figure 4.34
302
Network Theory
SOLUTION
At t = 0 , switch K is open and at t = 0+ , it is closed. At t = 0 , the circuit is in steady state and appears as shown in Fig.4.35(a). 20 = 1:33A 10 + 5 1:33 = 13:3V vC (0 ) = 10i2 (0 ) = 10
i2 (0
)=
Hence;
Since current through an inductor cannot change instantaneously, i2 (0+ ) = i2 (0 ) = 1.33 A. Also, vC (0+ ) = vC (0 ) = 13:3V. The equivalent circuit at t = 0+ is as shown in Fig.4.35(b).
i1 (0
)=
20
6:7 13:3 = = 0.67A 10 10
Figure 4.35(a)
Figure 4.35(b)
For t
0
+,
the circuit is as shown in Fig.4.35(c).
Writing KVL clockwise for the leftmesh, we get 10i1 + 1
C
Zt
i1 ( )d
= 20
0+
Differentiating with respect to t, we get 10
di1 dt
1
C
i1
=0
Figure 4.35(c)
i1 (0
Putting t = 0+ , we get 10
di1 (0 dt
+)
1
C
+ +
)=0 )= 0.67 105 A/ sec
di1 (0 dt
+)
10
1 10
i (0 6 1
Initial Conditions in Network
303
Writing KVL equation to the path made of 20V K 10 2di2 10i2 + = 20
dt
2H, we get
At t =
0+ ,
the above equation becomes 10i2 (0+ ) + 2di2 (0+ )
dt dt di2 (0 dt di2 (0
= 20 = 20 = 3.35A/ sec
EXAMPLE 4.17
10
1 33 + 2
:
+) +)
Refer the citcuit shown in Fig.4.36. The switch K is closed at t = 0. Find: (a) (b) (c) (d)
v1 v1
and v2 at t = 0+ and v2 at t = and
dv2 dt
dv1
dt 2 d v1 dt2
at t = 0+
at t = 0+
SOLUTION
Figure 4.36
(a) The circuit symbol for switch conveys that at t = 0 , the switch is open and t = 0+ , it is closed. At t = 0 , since the switch is open, the circuit is not activated. This implies that all initial conditions are zero. Hence, at t = 0+ , inductor is open and capactor is short. Fig 4.37(a) shows the equivalent circuit at t = 0+ .
Figure 4.37(a)
304
Network Theory
10 = 1A 10 + + v1 (0 ) = 0; i2 (0 ) = 0
i1 (0
)=
Applying KVL to the path, 10V source
(b) At = , switch conditions, capacitor circuit at = .
t t
10 + 10i1 (0+ ) + v1 (0+ ) + v2 (0+ ) = 0 10 + 10 + 0 + v2 (0+ ) = 0
v2 (0
10 10 2mH, we get
K
)=0
K C
remains closed and circuit is in steady state. Under steady state is open and inductor L is short. Fig. 4.37(b) shows the equivalent
i2 ( i1 v1 v2
) = 10 10 = 0.5A + 10 ( ) = 0 ( ) = 0 5 10 = 5V ( ) = 0
:
(c) For t
0
Figure 4.37(b)
+,
the circuit is as shown in Fig. 4.37(c).
Figure 4.37(c)
Initial Conditions in Network
305
i2
1
L
Zt
v2 ( )d
v1 (t) R2
0+
Differentiating with respect to t, we get
v2 L
dv1
R2 dt
Evaluating at t = 0+ we get
dv1 (0
+)
dt dv1 (0 dt
R2 L2
v2 (0
+)
= 0V/ sec
Applying KVL clockwise to the path 10 V source 10 + 10i + 1
C
10 4
K i2 ( )]d
F,
we get
Zt
[i( )
0+
=0
Differentiating with respect to t, we get 10 Evaluating at t = 0+ , we get
di(0 di dt
1
C
[i
i2 ]
=0
+)
dt
= = =
i2 (0
+)
10
i(0
+)
2
6
0 1 10 4 10
i(0+ ) = i1 (0+ ) + i2 (0+ ) 4 5 =1+0 = 1A
25000A/ sec
Applying KVL clockwise to the path 10 V source we get
10 10 2 mH,
K
10 + 10i + 10i2 + v2 = 0 10i + v1 + v2 = 10
Differentiating with respect to t, we get 10
di dt
dv1 dt
dv2 dt
=0
306
Network Theory
At t = 0+ , we get 10
di(0
+)
dt
dv1 (0 dt
+)
dv2 (0 dt dv2 (0 dt dv2 (0 dt
+) +) +)
=0 =0 = 25 104 V/ sec
10( 25000) + 0 +
(d) From part (c), we have 1
L
Zt
v2 ( )d
v1
0+
10
Differentiating with respect to t twice, we get 1 At t = 0+ , we get 1
L dv2 (0 dt dv2
L dt
1 d2 v1 10 dt2
+) +)
Hence;
EXAMPLE 4.18
2 d v
1 d2 v1 (0+ ) 10 dt2
1 (0 dt2
= 125 107 V/ sec2
Refer the network shown in Fig. 4.38. Switch K is changed from a to b at t = 0 (a steady state having been established at position a).
Figure 4.38
Show that at t = 0+ .
i1
= i2 =
V R1
+ R2 + R3
i3
=0
Initial Conditions in Network
307
SOLUTION
The symbol for switch indicates that at t = 0 , it is in position a and at t = 0+ , it is in position b. The circuit is in steady state at t = 0 . Fig 4.39(a) refers to the equivalent circuit at t = 0 . Please remember that at steady state C is open and L is short.
iL1 (0
) = 0;
iL2 (0
) = 0;
vC2 (0
) = 0;
vC1 (0
)=0
Applying KVL clockwise to the left-mesh, we get
+ vC3 (0 ) + 0
R2
+0=0 ) = V volts:
vC3 (0
Figure 4.39(a)
Since current in an inductor and voltage across a capacitor cannot change instantaneously, the equivalent circuit at t = 0+ is as shown in Fig. 4.39(b).
Figure 4.39(b)
i1 (0 i3 (0
+ +
) = i2 (0+ ) since ) = 0 since
R2 R3
iL1 (0
)=0
K
Applying KVL to the path v C3 (0+ )
V
R1
iL2 (0
)=0
we get,
+ R2 i1 (0+ ) + R3 i2 (0+ ) + R1 i1 (0+ ) = 0
308
Network Theory
Since i1 (0+ ) = i2 (0+ ), the above equation becomes = [R1 + R2 + R3 ] i1 (0+ ) V + + i1 (0 ) = i2 (0 ) = A R1 + R2 + R3
V
Hence;
EXAMPLE 4.19
di1 (0 dt
Refer the circuit shown in Fig. 4.40. The switch K is closed at t = 0. Find (a)
+)
and (b)
di2 (0 dt
+)
Figure 4.40
SOLUTION
The circuit symbol for the switch shows that at + t = 0 , it is open and at t = 0 , it is closed. Hence, at t = 0 , the circuit is not activated. This implies that all initial conditions are zero. That is, vC (0 ) = 0 and iL (0 ) = i2 (0 ) = 0. The equivalent circuit at t = 0+ keeping in mind that vC (0+ ) = vC (0 ) and iL (0+ ) = iL (0 ) is as shown in Fig. 4.41 (a).
i1 (0
Figure 4.41(a)
) = 0 and i2 (0+ ) = 0:
Figure. 4.41(b) shows the circuit diagram for t
Vo sin !t
0
1
C
+.
Zt
i1 ( )d
= i1 R +
0+
Differentiating with respect to t, we get
Vo ! cos !t
=R
di1 dt
i1 C
Initial Conditions in Network
309
At t = 0+ , we get
Vo !
=R =
di1 (0 dt
+)
i1 (0
+)
Also;
+ di1 (0 )
dt Vo sin !t
V0 A/ sec R
di2 dt
= i2 R + L
Evaluating at t = 0+ , we get 0 = i2 (0+ )R + L
di2 (0 dt
+)
EXAMPLE
di2 (0 dt
+)
= 0A/ sec
Figure 4.41(b)
4.20
t
In the network of the Fig. 4.42, the switch K is opened at steady state with the switch closed. (a) Find the expression for vK at t = 0+ .
= 0 after the network has attained
di(0
(b) If the parameters are adjusted such that i(0+ ) = 1, and the derivative of the voltage across the switch at t =
+)
0+ ,
dt
=
+
1, what is the value of
dvK dt
(0 ) ?
Figure 4.42
SOLUTION
At t = 0 , switch is in the closed state and at + t = 0 , it is open. Also at t = 0 , the circuit is in steady state. The equivalent circuit at t = 0 is as shown in Fig. 4.43(a).
i(0
)=
V R2
and vC (0 ) = 0
Figure 4.43(a)
310
Network Theory
For t 0+ , the equivalent circuit is as shown in Fig. 4.43(b). From Fig. 4.43 (b),
vK
= R1 i +
1
C
Zt
i( )d
At t = 0+ ,
vK (0
0+
vK
= R1 i + vC (t)
+)
= R1 i(0+ ) + vC (0+ )
vK (0
) = R1
V R2
+ vC (0 )
= R1
V volts R2
Figure 4.43(b)
(b)
vK
= R1 i + = R1
di dt
1
C
Zt
i( )d
0+
i C
dvK dt
Evaluating at t = 0+ , we get
dvK (0 dt
+)
= R1 = R1 1 = C
di(0
+)
(
dt
i(0
+)
1) +
1
C
R1 volts/ sec
Initial Conditions in Network
311
Reinforcement Problems
R.P
2 + d iL (0 ) dt2
4.1
t
Refer the circuit shown in Fig RP.4.1(a). If the switch is closed at at t = 0+ .
= 0, nd the value of
Figure R.P.4.1(a)
SOLUTION
The circuit at t = 0 is as shown in Fig RP 4.1(b). Since current through an inductor and voltage across a capacitor cannot change instantaneously, it implies that iL (0+ ) = 18A and vC (0+ ) = 180 V. The circuit for t 0+ is as shown in Fig. RP 4.1 (c).
Figure R.P.4.1(b)
Figure R.P.4.1(c)
Referring Fig RP 4.1 (c), we can write 2
10
3 diL
dt
+ 60iL + 288
10
Zt
iL (t)dt
=0
(4.9)
0+
312
Network Theory
At t = 0+ , we get
diL (0 dt
+)
= =
60 18 + 180 2 10 3 450 103 A= sec
Differentiating equation (4.9) with respect to t, we get 2 At t = 0+ , we get
2 + d iL (0 ) 2 dt
10
2 3 d iL dt2
+ 60
diL dt
+ 288
10
iL
=0
60(450)103 288 103 (18) 2 10 3 = 1:0908 1010 A= sec2 =
R.P
4.2
2 + d vC (0 ) dt2
For the circuit shown in Fig. RP 4.2, determine
and
3 + d vC (0 ) : dt3
Figure R.P.4.2
SOLUTION
Given
i(t)
= 2u(t) =
2; 0;
t t
0 0
=
Hence, at t = 0 , vC (0 ) = 0 and iL (0 ) = 0. For t 0+ , the circuit equations are
1 dvC (t) 1 + 64 dt 2
Zt
vL (t)dt
2 2
(4.10)
0+
1 dvC (t) + iL (t) = 64 dt
(4.11)
Initial Conditions in Network
313
[Note : At t =
iC
+ iL =
2 because of the capacitor polarity] 1 dvC (0+ ) + iL (0+ ) = 64 dt
0+ ,
equation (4.10) gives 2
Since, iL (0+ ) = iL (0 ) = 0, we get 1 dvC (0+ ) +0= 64 dt + dvC (0 ) =
dt
2 128 volts= sec
Differentiating equation (4.10) with respect to t we get 1 d2 vC (t) 1 + vL (t) = 0 64 dt 2 Also; At t = 0+ , we get
vC (0 vC vL
(4.12) (4.13)
24
1 = 2
Zt
vL dt
= iL
0+ +)
= iL (0+ ) 24 Since vC (0+ ) = 0 and iL (0+ ) = 0, we get vL (0+ ) = 0. At t = 0+ , equation (4.12) becomes 1 d2 vC (0+ ) 1 + vL (0+ ) = 0 64 dt2 2 1 d2 vC (0+ ) 1 + 0=0 64 dt2 2 2 + d vC (0 ) =0 2
+)
vL (0
dt
Differentiating equation (4.12) with respect to t we get 1 d3 vC 1 dvL + =0 3 64 dt 2 dt (4.14)
Differentiating equation (4.13) with respect to t, we get
dvC dt dvL dt
24
1 vL 2
314
Network Theory
At t = 0+ , we get
dvC (0 dt
+)
dvL (0 dt
+)
24 128
dvL (0 dt
=
+)
1 + vL (0 ) 2
24
dvL (0 dt
=0
+)
128 volts= sec
At t = 0+ , equation (4.14) becomes 1 d3 vC (0+ ) 1 dvL (0+ ) + =0 64 dt3 2 dt 3 + d vC (0 ) = 4096 volts= sec3 3
dt
R.P
4.3
In the network of Fig RP 4.3 (a), switch K is closed at t = 0. At t = 0 all the capacitor voltages and all the inductor currents are zero. Three node-to-datum voltages are identied as v1 , v2 and + v3 . Find at t = 0 : (i) (ii)
v1 , v2 dv1 dt
and v3
dt
dv2
and
dv3 dt
Figure R.P.4.3(a)
SOLUTION
The network at t = 0+ is as shown in Fig RP-4.3 (b). Since vC and iL cannot change instantaneously, we have from the network shown in Fig. RP-4.3 (b), + v1 (0 ) = 0
v2 (0 v3 (0
+ +
)=0 )=0
Initial Conditions in Network
315
Figure R.P.4.3(b)
For t
0
+,
the circuit equations are 1
C1
Zt
i1 dt
v C1
0+
v C2
1
C2
Zt
i2 dt
9 > > > > > > > > > > > > > = > > > > > > > > > > > > > ;
(4.15)
0+
v C3
1
C3
Zt
i3 dt
0+
From Fig. RP-4.3 (b), we can write
i1 (0 i2 (0
)= )=
v (0
+)
R1 v1 (0
; v2 (0 R2
+ +
+)
+)
and
i3 (0
)=0
Differentiating equation (4.15) with respect to t, we get
dvC1 dt
i1 C1
dvC2 dt
i2 C2
and
dvC3 dt
i3 C3
At t = 0+ , the above equations give
dv1 (0 dt dv2 (0 dt dv3 (0 dt
+) +) +)
= = =
i1 (0 i2 (0 i3 (0
+)
C1 + C2 + C3
= =
v (0
+)
) )
v1 (0
R 1 C1 +
v2 (0
+)
R 2 C2
=0
and
=0
316
Network Theory
R.P
4.4
For the network shown in Fig RP 4.4 (a) with switch K open, a steady-state is reached. The circuit paprameters are R1 = 10, R2 = 20, R3 = 20, L = 1 H and C = 1F. Take V = 100 volts. The switch is closed at t = 0. (a) Write the integro-differential equation after the switch is closed. (b) Find the voltage Vo across C before the switch is closed and give its polarity. (c) Find i1 and i2 at t = 0+ . (d) Find
di1 dt di2 dt
and
at t = 0+ .
di1 dt
(e) What is the value of t= ?
at
Figure R.P.4.4(a)
SOLUTION
The switch is in open state at t = 0 . The network at t = 0 is as shown in Fig RP 4.4 (b).
Figure R.P.4.4(b)
100 10 = A R1 + R2 30 3 10 200 VC (0 ) = i1 (0 )R2 = 20 = volts 3 3 Note that L is short and C is open under steady-state condition. For t 0+ (switch in closed state),
i1 (0
)=
we have and
20i1 + 20i2 + 10
6
di1 dt i2 dt
= 100 = 100
(4.16) (4.17)
Zt
0+
Initial Conditions in Network
317
10 A 3 200 + and VC (0 ) = VC (0 ) = Volts 3 From equation (4.16) at t = 0+ , Also
i1 (0
) = i1 (0 ) =
we have
di1 (0 dt
+)
= 100 =
20
100 A=sec 3
10 3
From equation (4.17), at t = 0+ , we have
+
1 100 i2 (0 ) = 20 Differentiating equation (4.17), we get 20 20di2 (0+ )
di2 dt
200 5 = A 3 3 (4.18)
+ 106 i2 = 0
From equation (4.18) at t = 0+ , we get
dt
+ 106 i2 (0+ ) = 0
di2 (0 dt
At t =
+)
,
di1 dt
106 5 3 20 106 = A= sec 12
i1 (
) = 100 =5A 20 ( ) = 0
2 + d i1 (0 ) . dt2
R.P
4.5
For the network shown in Fig RP 4.5 (a), nd The switch K is closed at t = 0.
Figure R.P.4.5(a)
318
Network Theory
SOLUTION
At t = 0 , we have vC (0 ) = 0 and i2 (0 ) = iL (0 ) = 0. Because of the switching property of L and C , we have vC (0+ ) = 0 and i2 (0+ ) = 0. The network at t = 0+ is as shown in Fig RP 4.5 (b).
Figure R.P.4.5(b)
Referring Fig RP 4.5 (b), we nd that
i1 (0
)=
v (0
+)
The circuit equations for t
0
R1
are
R1 i1
1
C
Zt
(i1
0+
i1 )dt +L di2 dt i2 )dt
= v (t)
(4.19)
and
R2 i2
1
C
Zt
(i2
0+
=0
(4.20)
vC (t)
{z
At t = 0+ , equation (4.20) becomes
R2 i2 (0
) + vC (0+ ) + L
di2 (0 dt di2 (0 dt
+)
=0 =0 (4.21)
Differentiating equation (4.19), we get
R1 di1 dt
+)
1
C
(i1
i2 )
dv (t) dt
(4.22)
Letting t = 0+ in equation (4.22), we get
R1 di1 (0 dt
+)
1
C
i1 (0
i2 (0
) = =
dv (0 dt
+)
dv (0 dt
di1 (0 dt
+)
1
R1
+)
v (0
+)
(4.23)
R1 C
Initial Conditions in Network
319
Differentiating equation (4.22) gives
R1
2 d i1 dt2
1
C
di1 dt
di2 dt
=
2 d v (t) dt2
Letting t = 0+ , we get
2 + d i1 (0 ) R1 dt2
1
C
di1 (0 dt
+)
di2 (0 dt
+)
= = =
2 + d v (0 ) dt2
2 + d i1 (0 ) R1 dt2 2 + d i1 (0 ) dt2
1
C
di1 (0 dt
+)
1
R1 C
2 + d v (0 ) dt2 +)
dt
1
R1
dv (0
2C R1
v (0
) +
2 + d v (0 ) dt2
R.P
4.6
Determine va (0 ) and va (0+ ) for the network shown in Fig RP 4.6 (a). Assume that the switch is closed at t = 0.
Figure R.P.4.6(a)
SOLUTION
Since L is short for DC at steady state, the network at t = 0 is as shown in Fig. RP 4.6 (b). Applying KCL at junction a, we get
va (0
) 10
va (0
) 20
vb (0
=0
Since vb (0 ) = 0, we get
va (0
) 10
) 0 =0 20 0:5 10 va (0 ) = = volts 0:1 + 0:05 3 +
Figure R.P.4.6(b)
va (0
320
Network Theory
Also;
iL (0
) = iL (0+ ) =
va (0
For t
0
20 2 = A 3
5 10
+,
we can write
va vb
10 20
va
and Simplifying at t = 0+ , we get
va
10
va
vb
vb
20
=0
10
+ iL = 0
and Solving we get,
va (0
1 + va (0 ) 4 1 + va (0 ) + 20
+)
1 1 + vb (0 ) = 20 2 3 1 + vb (0 ) = 20 6
40 = 1:905 volts 21
Exercise problems
E.P 4.1
di(0
Refer the circuit shown in Fig. E.P. 4.1 Switch K is closed at t = 0. Find i(0+ ),
+)
dt
and
2 + d i(0 ) . dt2
Figure E.P.4.1
Ans: i(0+ ) = 0.2A,
di(0+ ) = dt
2 103 A/ sec,
d2 i(0+ ) = 20 106 A/ sec2 dt2
Initial Conditions in Network
321
E.P
4.2
di dt
Refer the circuit shown in Fig. E.P. 4.2. Switch K is closed at t = 0. Find the values of i;
2 d i
dt2
and
at t = 0+ .
Figure E.P.4.2
Ans: i(0+ ) = 0,
E.P 4.3
di(0+ ) d2 i(0+ ) = = 10 A/ sec, dt dt2
1000 A/ sec2
Refering to the circuit shown in Fig. E.P. 4.3, switch is changed from position 1 to position 2 at
t
= 0. The circuit has attained steady state before switching. Determine i,
di dt
and
2 d i dt2
at t = 0+ .
Figure E.P.4.3
Ans: i(0+ ) = 0,
di(0+ ) = dt
40 A/ sec,
d2 i(0+ ) = 800 A/ sec2 dt2
322
Network Theory
E.P
4.4
dv1 dt C1 dv2 dt
In the network shown in Fig. E.P.4.4, the initial voltage on
v1 (0
is
Va
and on
C2
is
Vb
such that
) = Va and v2 (0 ) = Vb . Find the values of
and
at t = 0+ .
Figure E.P.4.4
Ans:
E.P
dv1 (0+ ) Vb Va = V/ sec, dt C1 R
4.5
2 d v2 dt2
dv2 (0+ ) Va Vb = V/ sec dt C2 R
In the network shown in Fig E.P. 4.5, switch K is closed at t = 0 with zero capacitor voltage and zero inductor current. Find at t = 0+ .
Figure E.P.4.5
Ans:
d2 v2 (0+ ) R2 Va = V/ sec2 dt2 R1 L1 C1
Initial Conditions in Network
323
E.P
4.6
2 d v1 dt2
In the network shown in Fig. E.P. 4.6, switch K is closed at t = 0. Find
at t = 0+ .
Figure E.P.4.6
Ans:
E.P
d2 v1 (0+ ) = 0 V/ sec2 dt2
4.7
t
The switch in Fig. E.P. 4.7 has been closed for a long time. It is open at
+ dv (0 )
dt
= 0. Find
+ di(0 )
dt
, i( ) and v ( ).
Figure E.P.4.7
Ans:
di(0+ ) = 0A/ sec, dt
dv (0+ ) = 20A/ sec, i( ) = 0A, v ( ) = 12V dt
324
Network Theory
E.P
4.8
diL (0 dt
In the circuit of Fig E.P. 4.8, calculate iL (0+ ),
+ ) dv (0+ ) C
dt
, vR ( ), vC ( ) and iL ( ).
Figure E.P.4.8
Ans: iL (0+ ) = 0 A,
dvC (0+ ) = 2 V/ sec, dt
E.P 4.9
diL (0+ ) = 0 A/ sec dt
vR ( ) = 4V, vC ( ) =
20V, iL ( ) = 1A
Refer the circuit shown in Fig. E.P. 4.9. Assume that the switch was closed for a long time for + diL (0 ) and iL (0+ ). Take v (0+ ) = 8 V. t < 0. Find
dt
Figure E.P.4.9
Ans: iL (0+ ) = 4 A,
E.P 4.10
diL (0+ ) = 0 A/ sec dt
Refer the network shown in Fig. E.P. 4.10. A steady state is reached with the switch K closed and + dv2 (0 ) . with i = 10A. At t = 0, switch K is opened. Find v2 (0+ ) and
dt
Initial Conditions in Network
325
Figure E.P.4.10
Ans: v2 (0+ ) = 0,
E.P 4.11
10Ra Rc dv2 (0+ ) = V= sec. dt Ca (Ra + Rb )(Ra + Rc )
Refer the network shown in Fig. E.P. 4.11. The network is in steady state with switch K closed. + dvk (0 ) . The switch is opened at t = 0. Find vk (0+ ) and
dt
Figure E.P.4.11
Ans: vk (0+ ) =
Va Rc Volts, Ra + Rb + Rc
dvk (0+ ) Va (Ca + Cb ) = V/ sec dt (Ra + Rb + Rc )(Ca Cd + Cb Ca + Cb Cd )
E.P 4.12
2 + d i1 (0 ) . dt2
Refer the network shown in Fig. E.P. 4.12. Find
326
Network Theory
d2 i1 (0+ ) 1 Ans: = 2 dt Ra
E.P 4.13
Figure E.P.4.12
10 10 + 2 2 A/ sec2 Ra Ca
Refer the circuit shown in Fig. E.P. 4.13. Find steady state at t = 0 .
di1 (0 dt
+)
. Assume that the circuit has attained
Figure E.P.4.13
di1 (0+ ) 10 Ans: A/ sec = dt RA
E.P 4.14
Refer the network shown in Fig. E.P.4.14. The circuit reaches steady state with switch K closed. + 2 + dv1 (0 ) d v2 (0 ) At a new reference time, t = 0, the switch K is opened. Find . and 2
dt dt
Figure E.P.4.14
Initial Conditions in Network
327
Ans:
E.P
dv1 (0+ ) 10 = V/ sec, dt Ca (Ra + Rb )
4.15
10Rb d2 v2 (0+ ) = V/ sec2 dt2 La Ca (Ra + Rb )
The switch shown in Fig. E.P. 4.15 has been open for a long time before closing at t = 0. Find: + + i0 (0 ), iL (0 ) i0 (0 ), iL (0 ), i0 ( ), iL ( ) and vL ( ).
Figure E.P.4.15
Ans: i(0 ) = 0, iL (0 ) = 160mA, i0 (0+ ) = 65mA, iL (0+ ) = 160mA, i0 ( ) = 225mA, iL ( ) = 0, vL ( ) = 0
E.P 4.16
The switch shown in Fig. E.P. 4.16 has been closed for a long time before opeing at t = 0. Find: i1 (0 ), i2 (0 ), i1 (0+ ), i2 (0+ ). Explain why i2 (0 ) = i2 (0+ ).
Figure E.P.4.16
Ans: i1 (0 ) = i2 (0 ) = 0.2mA, i2 (0+ ) =
i1 (0+ ) =
0.2mA
328
Network Theory
E.P
4.17
The switch in the circuit of Fig E.P.4.17 is closed at t = 0 after being open for a long time. Find: (a) i1 (0 ) and i2 (0 ) (b) i1 (0+ ) and i2 (0+ ) (c) Explain why i1 (0 ) = i1 (0+ ) (d) Explain why i2 (0 ) = i2 (0+ )
Figure E.P.4.17
Ans: i1 (0 ) = i2 (0 ) = 0.2 mA, i1 (0+ ) = 0.2 mA, i2 (0+ ) =
0.2mA
