8 .3 1 Consider the circuit in Fig. 8.79. Find VL and Vc @ t >0.
@ t< 0
40
(50) = 40 �
40 + 10
50 �
�=
= 1�
50 �ℎ��
�0 =
@ t>0
Apply KCL:
2 = I0+ i1= 1 +i1 =1
i1 = 1 A
Apply KVL :
-VL+ 40i1 + V0 = 0
VL= 40 (1) + 40 = 80 V
ANSW ER:
VC = 40 V
VL = 80 V
8 .3 3 Find for in the circuit of Fig. 8.81.
@ t<0
v(0) =
i(0) =
@t > 0
30
(5) = 10 V
15
V 30
=
= 2A
R 15
R
5
=
= 2.5
2L
2(1)
1
1
ωo =
=
= 0.5
LC
1(4)
α =
α > ωo , OVERDAMPED
s = − 2.5 ± 6.25 − 0.25 = − 4.95, 0.0505
v(t) = Vs + [A1 e−4.95t + A2 e−0.0505t ]
Vs= 20
v(0) = 10 = 20 + A1 + A2
A2 =− 10 − A1
Cdv(0)
2
1
i(0) = dt = 4 = 2
0.5= -4.95A1 − 0.0505A2
0.5= -4.95A1 − 0.0505(10 + A1 ) = − 0.005 A1 = 0.0001125, A2 =− 10.001
ANSW ER:
�(�) = [�� + �. ��������−�.��� + ��. ��� �−�.����� V
8 .3 5 Using Fig. 8.83, design a problem to help other students better understand the step
response of series RLC circuits.
R =2 L=1H C=1/5F
V1 = 8V V2= 12V
@ t<0
i L (0)= 0
V c(0) = 8 V
@ t>0
R
2
=
=1
α =
2L
2(1)
α < ωo , UNDERDAMPED
ωo =
s = − 1 ± 1 − 0.199 = − 1, 2
v(t) = Vs + [(A1 cos2t + A2sin2t)e−t ]
1
LC
=
1
1(1/5)
= 2
Vs= 12
v(0) = 8 = 12 + A1 + A2
A1 =− 4
Cdv(0)
i(0) = dt = 0
dv
= [ − (A1 cos 2 t + A2 cos 2t)e−t + [2(−A1 cos 2 t + A2 cos 2t)e−t
dt
0 =
dv(0)
= − A1 + 2A2
dt
ANSW ER:
�(�) = �� − (���� � � + ��� ��)�−� V
i(t) = [(5 s in 0 .6))e−0.8t ] A
8 .3 7 For the network in Fig. 8.85, solve for for i(t) t 7 0.
@ t<0
18�2 − 6�1 = 0 �� �1 = 3�2
−30 + 6 (i1 − i2) + 10 = 0
10
�1 − �2 =
3
5
3
�(0) = �1, = 5A
-10 -6 �2 + �(0) = 0
�(0) = �1, �2 =
V(0) = 10 + 6(5/3) = 20
@ t>0
R = 6 || 12 = 4
R
4
=
=4
α =
2L
2(1/2)
;
ωo =
α = ωo , CRITICALLY DAMPED
v(t) = Vs + [(A1 + A2 t)e−4t ]
;
1
LC
=
1
1/2(1/8)
= 4
Vs= 10
v(0) = 20 = 10 + A1
A1 = 10
i(c) =
Cdv(0)
dt
ANSW ER:
= C [ − 4 (10 + A2 t)e−4t ] + C[(A2 )e−4t ]
�(�) =− ��(�) = ��−��
8 .3 9 Determine for in the circuit of Fig. 8.87. v(t) t 7 0
@ t<0
60u(-t) = 60
30u(t) = 0
20
v(0) = 20+30 (60) = 24V
i(0) = 0 A
@ t>0
R = 20|| 30 = 12
α =
R
12
=
= 24
2L
2(0.25)
;
ωo =
1
LC
=
α > ωo , OVERDAMPED
s = − 24 ± 576 − 7.95 = − 47.833, − 0.167
v(t) = Vs + [A1 e−47.833t + A2 e−0.167t ]
;
1
1/2(1/4)
= 2.82
Vs= 30
v(0) = 24 = 30 + A1 + A2
−6= A1 + A2
Cdv(0)
i(0) = dt =0
dv(0)
= − 47.833A1 − 0.16A2 = 0
dt
A1 = 0.021 A2 = − 6.021
ANSW ER:
�(�) = �� + [�. ����−��.���� − �. ��� �−�.���� ] V
8 .5 9 The make before break switch in Fig. 8.105 has been in position 1 for At it is moved
instantaneously to position 2. Determine v(t).
v(0) = 0
40
i(0) = 4+16 = 2�
@ t>0
R
16
α =
=
=2
2L
2(4)
;
ωo =
α = ωo , CRITICALLY DAMPED
1
LC
1
=
4(1/16)
= 2
i(t) =A1e−2t + −A2 te−2t
i(0) = 2A
��
= A1 e−2t +A2 te−2t −2A2 te−2t
��
��(0)
��
1
= −2A1 +A2 = − L [Ri (0) -v(0)]
1
2
-2A1 +A2 =− 4 (32 − 0)
A2 =− 4
i(t) =2e−2t − 4 te−2t
1
v=
C
1
0
−idt + v(0) =− 31
ANSW ER:
t
0
� =− ����−�� �
−2t
e
dt + 64
t
0
te−2t dt = 16e−2t | +
64 −2t
e ( − 2t − 1) |
4