Chapter 9 Homework Solution
Chapter 9(1):
P9.2-1, 5, 9
P9.3-2, 4
P9.4-1, 3
P9.5-2, 3, 4
P9.6-1, 2, 5
Chapter 9(2):
P9.7-1, 2
P9.8-2, 4, 9, 15, 17
P9.9-1
P 9.2-1 Find the differential equation for the
circuit shown in Figure P 9.2-1 using the direct
method.
Figure P 9.2-1
P 9.2-5 The input to the circuit shown in Figure P 9.2-5 is the voltage of the voltage source, vs. The
output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows
how the output of this circuit is related to the input, for t > 0.
Hint: Use the direct method.
Figure P 9.2-5
Solution:
After the switch closes, use KCL to get
i=
(t )
v (t )
d
+ C v (t )
R2
dt
Use KVL to get
v s =R1 i ( t ) + L
d
i (t ) + v (t )
dt
Substitute to get
R1
d
L d
d2
v s = v ( t ) + R1C v ( t ) +
v ( t ) + CL 2 v ( t ) + v ( t )
R2
dt
R 2 dt
dt

R1 + R 2
d2
L d
= CL 2 v ( t ) +  R1C +
v (t ) +
v (t )


dt
R 2  dt
R 2CL

Finally,
 R1
vs
R1 + R 2
d2
1 d
=
v t +
+
v (t ) +
v (t )

2 ( ) 
 L R 2C  dt
CL dt
R
CL
2


P 9.2-9 The input to the circuit shown in
Figure P 9.2-9 is the voltage of the voltage source,
vs. The output is the capacitor voltage v(t).
Represent the circuit by a second-order differential
equation that shows how the output of this circuit
is related to the input, for t > 0.
Hint: Use the direct method.
Figure P 9.2-9
Solution:
After the switch closes
i (t ) = C
d
v (t )
dt
KCL and KVL give

1  d
d

vs = R2  i (t ) +  L i (t ) + v (t )   + L i (t ) + v (t )


R1  dt
dt


Substituting gives
 R2 
 R2 
d2
d
v s = 1 +
LC
v ( t ) + R 2C v ( t ) +  1 +

 v ( t )
2



R
dt
dt
R
1
1




 R2 
 R2 
d2
d
= 1 +
LC 2 v ( t ) + R 2C v ( t ) + 1 +

 v (t )


R1 
dt
dt
R1 


Finally
R1 R 2
1
d2
d
v (t ) +
v (t ) +
v (t )
=
LC
LC ( R1 + R 2 ) dt
L ( R1 + R 2 ) dt
R1v s
P 9.3-2 Find the characteristic equation and its
roots for the circuit of Figure P 9.3-2.
Answer: s2 + 400s + 3 × 104 = 0
roots: s = – 300, – 100
Figure P 9.3-2
Solution:
KVL: 40 (is − iL ) = 100 ×10−3
i=
L
i=
c
diL
+ vc
dt
1
−3  dvc
 ×10 
3
 dt
di 40
di 100
d 2iL
40
×10−3 s − ×10−3 L −
×10−6
dt 3
dt
3
3
dt 2
di
d 2iL
di
+400 L +30000i L =
400 s
2
dt
dt
dt
2
−100,
s + 400 s + 30000 =
0 ⇒ ( s + 100)( s + 300) =
0 ⇒ s1 =
iL =
P 9.3-3 Find the characteristic equation and
its roots for the circuit shown in Figure P 9.3-3.
Figure P 9.3-3
Solution:
v − vs
dv
0
+ i L + 10 ×10−6
=
1
dt
di
KVL: v = 2i L +10−3 L
dt
KCL:
diL
di
d 2iL
− vs + iL + 10 ×10−6 ⋅ 2 L + 10 ×10−6 ×10−3
dt
dt
dt
2
di
d i
vs= 3iL + .00102 L + 1×10−8 2L
dt
dt
2
d iL
di
+ 102000 L + 3 ×10−8 iL = 1×108 vs
dt
dt
2
s + 102000 s + 3 ×108 = 0, ∴ s1 = 3031, s2 = − 98969
0=
2iL + 10−3
−300
s2 =
P 9.3-4 German automaker Volkswagen, in
its bid to make more efficient cars, has come
up with an auto whose engine saves energy by
shutting itself off at stoplights. The “stop–
start” system springs from a campaign to
develop cars in all its world markets that use
less fuel and pollute less than vehicles now on
the road. The stop–start transmission control
Figure P 9.3-4
has a mechanism that senses when the car does
not need fuel: coasting downhill and idling at an intersection. The engine shuts off, but a small starter
flywheel keeps turning so that power can be quickly restored when the driver touches the accelerator.
A model of the stop–start circuit is shown in Figure P 9.3-4. Determine the characteristic equation
and the natural frequencies for the circuit.
Answer: s2 + 20s + 400 = 0
s = – 10 ± j17.3
Solution:
Assume zero initial conditions
1 di1
1 di2
10 − 7
−
=
2 dt
2 dt
1 di1
1 di2
loop 2 : −
7
+
+ 200 ∫ i2 dt =
2 dt
2 dt
1 
1


− s
10 + 2 s 

2



determinant : 

1
200  
1
 s+

 −2s
s  
2

loop 1 : 10i1 +
s 2 + 20 s + 400 =0,
∴ s =− 10 ± j 17.3
P 9.4-1 Determine v(t) for the circuit of
Figure P 9.4-1 when L = 1 H and vs = 0 for t ≥
0. The initial conditions are v(0) = 6 V and
dv/dt(0) = – 3000 V/s.
Answer: v(t) = – 2e–100t + 8e–400t V
Figure P 9.4-1
Solution:
v ( 0 ) = 6,
dv ( 0 )
= −3000
dt
Using operators, the node equation is: Csv +
v
( v −vs )
L


+ = 0 or  LCs 2 + =
s + 1 v
R
sL

R

So the characteristic equation is: s 2 +
v (t )
So=
v ( 0 )=
Ae −100t + Be −400t
6=
v
s
1
1
=
s+
0
RC
LC
⇒ s1,2 =− 250 ± 2502 − 40, 000 =− 100, − 400
A+ B
dv ( 0 )
 A = −2
=
− 3000 =
− 100 A − 400 B 
dt
 B = 8
∴ v (t ) =
− 2e −100t + 8e −400t
t>0
P 9.4-2 An RLC circuit is shown in
Figure P 9.4-2, where v(0) = 2 V. The switch
has been open for a long time before closing at
t = 0. Determine and plot v(t).
Figure P 9.4-2
Solution:
v ( 0)
=
2, i ( 0 )
=
0
Characteristic equation s 2 +
v=
(t )
1
1
s+
= 0 ⇒ s 2 + 4s + 3 = 0 ⇒ s =
RC
LC
Ae − t + Be −3t
Use eq. 9.5 − 12 ⇒ s1 A + s2 B =−
−1A − 3B = −
also have v ( 0 )=
2=
2
−0 = − 8
1
4
A+ B
From (1) & ( 2 ) get A = −1, B = 3
∴ v (t ) =
−e − t + 3 e −3t V
v ( 0) i ( 0)
−
RC
C
(1)
( 2)
− 1, − 3
P 9.4-3 Determine i1(t) and i2(t) for the circuit of
Figure P 9.4-3 when i1(0) = i2(0) = 11 A.
Figure P 9.4-3
Solution:
di1
di
−3 2 =
0
dt
dt
di
di
KVL : − 3 1 + 3 2 + 2i2 =
0
dt
dt
KVL : i1 + 5
(1)
( 2)
in operator form
(1 + 5s ) i1 + ( −3s ) i2 = 0 

0 
( −3s ) i1 + ( 3s + 2 ) i2 =
Thus i=
1 (t )
Ae
i 2 ( t ) = Ce
-t
-t
6
6
thus ∆ = (1 + 5s )( 3s + 2 ) − 9 s 2 =6 s 2 + 13s + 2 =
0 ⇒ s =− 1 6, − 2
+ Be -2t
+ De-2t
Now i1 ( 0 ) =
11 =
A + B; i2 ( 0 ) =
11 =
C+ D
from (1) & ( 2 ) get
di1 ( 0 )
di2 ( 0 )
33
A
143
C
=−
= − − 2B ;
=−
= − − 20
dt
dt
2
6
6
6
− 1, 0 =
which yields A =
3, B =
8, C =
12
i1 (t ) =3e − t /6 + 8e −2t A
&
i2 (t ) =
− e − t /6 + 12e −2t A
P 9.5-2 Find vc(t) for t > 0 for the circuit of
Figure P 9.5-2. Assume steady-state
conditions exist at t = 0–.
Answer: vc(t) = – 8te–2t V
Figure P 9.5-2
Solution:
( )
dvc
t
=
KCL at v c : ∫−∞ vc dt + vc + 1
0
4 dt
d 2 vc
dv
⇒
+ 4 c + 4vc =
0
dt
dt
t >0
s 2 + 4 s + 4 =0, s =
−2, − 2
t = 0- (Steady-State)
⇒ vc ( t ) =
A1 e −2t + A2 t e −2t
vc ( 0− )= 0= vc ( 0+ ) & iL ( 0− )=
20 V
=
10 Ω
2 A=
Since vc ( 0+ ) =
0 then ic ( 0+ ) =
−iL ( 0+ ) =
−2 A
∴
So vc ( 0+ =
)
dvc ( 0+ )
0=
A1
= − 8 = A2
dt
∴ vc ( t ) =
− 8te −2t V
dvc ( 0+ )
dt
ic ( 0+ )
=
=
−8 V
S
1
4
iL ( 0+ )
P 9.5-3 Police often use stun guns to
incapacitate potentially dangerous felons. The
hand-held device provides a series of highvoltage, low-current pulses. The power of the
pulses is far below lethal levels, but it is
enough to cause muscles to contract and put
Figure P 9.5-3
the person out of action. The device provides a
pulse of up to 50,000 V, and a current of 1 mA
flows through an arc. A model of the circuit for one period is shown in Figure P 9.5-3. Find v(t) for 0 < t <
1 ms. The resistor R represents the spark gap. Select C so that the response is critically damped.
Solution:
Assume steady − state at t =0− ∴ vc ( 0− ) =104 V & iL ( 0− ) =0
t >0
diL
+ 106 iL =0
(1)
dt
 d 2i
di 
dv
: iL =
−C c
=
−C .01 2L + 106 L 
dt
dt
dt 

KVL a : − vc + .01
Also
∴ 0.01C
d 2iL
di
+ 106 C L + iL =
0
2
dt
dt
Characteristic eq. ⇒ 0.01C s + 10 s + 1 =
2
6
0 ⇒ s=
−106 C ±
(10 C )
6
⇒ C = 0.04 pF ∴ s =−5 ×107 , −5 ×107
7
7t
A1e −5×10 t + A2te −5×10
diL +
0 ) = 100 vc ( 0+ ) − 106 iL ( 0+ )  = 106 A
(
s
dt
di ( 0 )
7
=
So i L ( 0 ) =
0=
106 =
106 te −5×10 t A
A1 and L
A2 ∴ iL ( t ) =
dt
Now from (1) ⇒
6
=
=
Now
v ( t ) 10
iL ( t ) 1012 te −5×10 t V
7
2
2 (.01C )
for critically damped: 1012 C2 − .04C = 0
=
So iL ( t )
( 2)
− 4 (.01C )
P 9.5-4 Reconsider Problem P 9.4-1 when L = 640
mH and the other parameters and conditions remain the
same.
Answer: v(t) = (6 – 1500t)e – 250t V
Figure P 9.4-1
Solution:
s2 +
1
1
1
1
s+
=
0 with
=
500 and
=
62.5 ×103 yields s =
−250, −250
RC
LC
RC
LC
=
v (t )
Ae −250t + Bte −250t
v ( 0 )= 6=
A
dv ( 0 )
=−3000 =− 250 A + B ⇒ B =− 1500
dt
∴ v ( t ) = 6e −250t − 1500te −250t
P 9.6-1 A communication system from a
space station uses short pulses to control a
robot operating in space. The transmitter
circuit is modeled in Figure P 9.6-1. Find the
output voltage vc(t) for t > 0. Assume steadystate conditions at t = 0–.
Answer:
vc(t) = e– 400t [3 cos 300t + 4 sin 300t] V
Figure P 9.6-1
Solution:
t>0
KCL at vc :
also :
Solving for i
L
vc
250
+ iL + 5 ×10−6
vc = 0.8
dvc
=0
dt
diL
dt
(1)
( 2)
in (1) & plugging into ( 2 )
d 2 vc
dv
+800 c +2.5×105 vc =
0 ⇒ s 2 +800 s +250,000 =
0, s =
−400± j 300
2
dt
dt
=
∴ v (t )
c
t = 0−
e
−400t 

 A1 cos300t + A 2 sin 300t 
(Steady − State)
−6 V −6
=
=
A iL ( 0+ )
500
500 Ω
vc ( 0−=
) 250 −6 500 + 6= 3 V= vc ( 0+ )
−
i=
L (0 )
(
Now from (1) :
So vc ( 0 =
)
+
dvc ( 0+ )
dt
∴=
vc ( t )
)
dvc ( 0+ )
3=
dt
A1
= − 2 ×105 iL ( 0+ ) − 800vc ( 0+ ) = 0
=
− 400 A1 + 300 A2 ⇒ A2 =
0 =
4
e −400t [3cos 300t + 4sin 300t ] V
P 9.6-2 The switch of the circuit shown in Figure P 9.6-2 is opened at t = 0. Determine and plot v(t) when
C = 1/4 F. Assume steady state at t = 0–.
Answer: v(t) = – 4e–2t sin 2t V
Figure P 9.6-2
Solution:
t = 0−
i ( 0) = 2 A
v ( 0) = 0
t = 0−
KCL at node a:
t
v
dv 1
+ C + ∫ vdt + i ( 0 ) =
0 (1)
dt L 0
1
in operator form have v + Csv +
with s 2 + 4 s + 8 =0
1
v + i=
( 0 ) 0 or
Ls
1 
 2 1
s + s+ =
v 0
C
LC 

⇒ s =−2 ± j 2
v ( t ) e −2t [ B1 cos 2t + B2 sin 2t ]
=
v ( 0 )= 0= B1
dv ( 0 ) 1
=  −i ( 0 ) − v ( 0 )  =−4 [ 2] =−8 =2 B2 or B2 =−4
dt
C
−2 t
So v ( t ) = −4e sin 2 t V
From (1) ,
P 9.6-5 The photovoltaic cells of the
proposed space station shown in
Figure P 9.6-5a provide the voltage v(t) of the
circuit shown in Figure P 9.6-5b. The space
station passes behind the shadow of earth (at
t = 0) with v(0) = 2 V and i(0) = 1/10 A.
Determine and sketch v(t) for t > 0.
Figure P 9.6-5
Solution:
=
v ( 0 ) 2=
V and i ( 0 ) 1 A
10
1
1
s+
= 0 or s 2 + 2 s + 5 = 0 thus the roots are s =−1 ± j 2
RC
LC
−t 
So have v(t ) =
e
B cos 2t + B2 sin 2t  now v(0+ ) =
2=
B1
 1

Char. eq. ⇒ s 2 +
Need
So
dv ( 0+ )
dt
dv ( 0+ )
dt
v ( 0+ )
1
1 V
+
+
ic ( 0 ) . KCL yields ic ( 0 ) =
=
−
− i ( 0+ ) =
−
5
2 s
C
 1
10  −  =
=
− B1 + 2 B2 ⇒ B2 =
−3
2
 2
3
Finally, we have v ( t ) =
2e − t cos 2t − e − t sin 2t V
2
t >0
P 9.7-1 Determine the forced response for
the inductor current if when (a) is = 1 A, (b) is =
0.5t A, and (c) is = 2e–250t A for the circuit of
Figure P 9.7-1.
Figure P 9.7-1
Solution:
v
dv
+ iL + C
R
dt
di
KVL : v = L L
d 2iL
L diL dt
=
is
+ iL + LC
R dt
dt 2
KCL : is =
(a)
is =
l u (t ) ∴ assume i f =
A
d 2iL
1 diL
1
+
+
iL =
is
2
dt
RC dt LC
1
= 1 ⇒ A = 1×10−5 = i f
to get: 0 + 0 + A
−3
(.01) (1×10 )
if =
A in
Let iL =
(b)
0.5t u (t ) ∴ assume i f =
is =
At + B
0+ A
65
1
0.5 t
+ ( At + B )
=
(.01)(.001)
(100 ) (.001)
100000 B 0 and =
100000 At 0.5t
⇒ 650 A +=
A=
=
B
5 ×10−6
3.25 ×10−8
if =
5 ×10−6 t − 3.25 ×10−8 A
(c)
is
2=
e −250t Assumming i f Ae −250t does not work
because i f cannot have the same form as is ∴ we choose i f =
Bte −250t
Be −250t −250 Bte −250t Bte −250t
2 e −250t
+
+
=
RC
RC
LC
150 B = 2
B = 0.0133
if
= 0.0133 te −250t A
P 9.7-2 Determine the forced response for the
capacitor voltage, vf, for the circuit of Figure P 9.7-2
when
(a) vs = 2 V, (b) vs = 0.2t V, and (c) vs = 1e–30t V.
Figure P 9.7-2
Solution:
Represent the circuit by the differential equation:
(a)
vs =
2 ∴ assume v f =
A
Then 0 + 0 + 12000
=
A 2 so
=
A
(b)
d 2 v R dv 1
+
+
v=
vs
dt L dt LC
1 =
6000
vf
vs =
0.2t ∴ assume v f =
At + B
70 A + 12000 At + 12000 B = 0.2t ⇒ 70 A + 12000 B = 0 and 12000 At = 0.2t
1
70 A
,=
B
60000
12000
t
∴=
vf
+ 350 V
60000
=
A
(c)
vs =
e −30t
⇒
=
B 350
∴ assume v f =
Ae −30t
900 A − 2100 Ae −30t + 12000 Ae −30t =
vf
e −30t
=
V
10800
e −30t
⇒ 10800 Ae −30t =
e −30t
⇒
A=
1
10800
P 9.8-2
Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-2.
d2
d
Hint: Show that 1 = 2 i (t ) + 5 i (t ) + 5i (t ) for t > 0
dt
dt
Figure P 9.8-2
Answer:
i(t) = 0.2 + 0.246 e–3.62t – 0.646 e–1.38t A
for t > 0.
Solution:
First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open
circuit at steady state, and the inductor will act like a short circuit.
i=
( 0)
−1
= 0.2 A
1+ 4
and
4
v ( 0 ) = ( −1) =
−0.8 V
1+ 4
For t > 0
Apply KCL at node a:
v − Vs
d
+ C v + i =0
R1
dt
Apply KVL to the right mesh:
R2 i + L
d
d
i − v= 0 ⇒ v= R 2 i + L i L
dt
dt
After some algebra:
L + R1 R 2C d
R1 + R 2
Vs
d2
i+
i+
=
i
2
dt
R1 L C dt
R1 L C
R1 L C
The forced response will be a constant, if = B so 1=
⇒
d2
d
i + 5 i +=
5i 1
2
dt
dt
d2
d
B + 5 B + 5 B ⇒ B= 0.2 A .
2
dt
dt
To find the natural response, consider the characteristic equation:
0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 )
The natural response is
=
in A1 e−3.62 t + A2 e−1.38 t
so
i ( t ) = A1 e−3.62 t + A2 e−1.38 t + 0.2
Then
d


v (t ) =
−10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8
 4 i (t ) + 4 i (t )  =
dt


At t=0+
−0.2= i ( 0 + )= A1 + A2 + 0.2
−0.8 =v ( 0 + ) =−10.48 A1 − 1.52 A2 + 0.8
so A1 = 0.246 and A2 = -0.646. Finally
i (t ) =
0.2 + 0.246 e−3.62 t − 0.646 e−1.38 t A
P 9.8-4 Find v(t) for t > 0 for the circuit shown in
Figure P 9.8-4 when v(0) = 1 V and iL(0) = 0.
Figure P 9.8-4
Answer:
v =25e −3t −
1
 429e −4t − 21cos t + 33 sin t  V
17
Solution:
di


=0
KCL at top node :  0.5 L − 5cos t  + iL + 1 dv
12 dt
dt


diL
1 dv + v
=
KVL at right loop : 0.5
( 2)
12 dt
dt
t > 0
d
d
d 2iL diL 1 d 2 v
+
+
=
−5sin t (3)
dt 2
dt 12 dt 2
d 2iL
d 2 v dv
⇒ 0.5 2 = 1
+
( 4)
12 dt 2 dt
dt
dt
of (1) ⇒ 0.5
dt
of ( 2 )
Solving for
d 2iL
di
in ( 4 ) and L in ( 2 ) & plugging into ( 3)
2
dt
dt
d 2v
dv
+ 7 + 12v =
− 30sin t
2
dt
dt
so v(t ) = A1e −3t + A2 e −4t + v f
0 ⇒ s =
⇒ s 2 + 7 s + 12 =
− 3, − 4
Try v f = B1 cos t + B2 sin t & plug
into D.E., equating like terms
yields B1 = 21 , B2 = − 33
17
17
t = 0+
5 −1
dv(0+ )
2
ic ( 0+ ) = =
2A ∴
=
=
24 V
1
s
11
dt
12

So v(0+ ) = 1 = A1 + A2 + 21
A1 = 25
17


429
dv(0+ )
 A2 = −
=24 =− 3 A1 − 4 A2 − 33
17
17
dt

∴ v(t ) = 25e −3t − 1 ( 429e −4t − 21cos t + 33sin t ) V
17
(1)
P 9.8-9 In Figure P 9.8-9, determine the
inductor current i(t) when is = 5u(t) A. Assume
that i(0) = 0, vc(0) = 0.
Answer: i(t) = 5 + e–2t [ – 5 cos 5t – 2 sin 5t] A
Figure P 9.8-9
Solution:
dv
v
+ i + =i
dt
2 s
2
d i
 L  di
LC
+ i +   =5 u ( t )
dt
 2  dt
KCL : C
1 d 2i
 4  di
+ i +   =5 u ( t )
29 dt
 29  dt
Characteristic eqn: s 2 + 4 s + 29 = 0 ⇒
d 2i
di
+4 +i =
145 u ( t )
dt
dt
roots : s = − 2 ± j5
145 =
∴ =
in e −2t [ A cos 5t + B sin 5t ] and i=
f
29
−2t
So i (t ) =
5 + e [ A cos 5t + B sin 5t ]
5
Now i (0) ==+
0 A 5 ⇒ A=
−5
di (0)
=
0=
−2 A + 5 B ⇒ B =
−2
dt
P 9.8-15 The circuit shown in Figure P 9.8-15 is at steady state before the switch closes. Determine the
capacitor voltage, v(t), for t > 0.
Figure P 9.8-15
Solution:
First, we find the initial conditions;
For t < 0, the switch is open and the circuit is at steady
state. At steady state, the capacitor acts like an open
circuit and the inductor acts like a short circuit.
v ( 0 − ) =0 V and i ( 0 − ) =0 A
also
i ( 0)
v ( 0)
d
v ( 0) = −
=
0
dt
0.005 50 × 0.005
Next, represent the circuit after the switch closes by a differential equation.
After the switch closes, use KCL to get
i=
(t )
v (t )
d
+ C v (t )
R2
dt
Use KVL to get
v s =R1 i ( t ) + L
d
i (t ) + v (t )
dt
Substitute to get
R1
d
L d
d2
v s = v ( t ) + R1C v ( t ) +
v ( t ) + CL 2 v ( t ) + v ( t )
R2
dt
R 2 dt
dt
= CL

R1 + R 2
d2
L d
v
t
+
R
C
+
v (t ) +
v (t )
(
)


1
2

dt
R 2  dt
R2

Finally,
 R1
vs
R1 + R 2
d2
1 d
=
v
t
+
+
v (t )
(
)

 v ( t ) +
2

CL dt
R 2CL
 L R 2C  dt
Compare to
d2
dt
to get
i t + 2α
2 ( )
d
i ( t ) + ω 02 i ( t ) =
f (t )
dt
R1
R1 + R 2
1
2a =
+
, ω 0 2 = and
L R 2C
R 2CL
vs
f (t ) =
CL
With the given element values, we have α = 14.5 and ω 0 2 = 200 . Consequently, the roots of the
characteristic equation are s 1 = −11.3 and s 2 = −17.7 so the circuit is overdamped. The natural response
is
=
v n ( t ) A1 e −11.3 t + A 2 e −17.7 t
Next, determine the forced response.
The steady state response after the switch opens will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
=
vf
1
=
v s 10 V
2
So
v n (t ) =
10 + A1 e−11.3 t + A 2 e−17.7 t
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0 = v ( 0 ) =10 + A1 + A 2
and
d
0=
v ( 0) =
−11.3 A1 − 17.7 A 2
dt
Solving these equations gives
A1 =
−27.6 and A 2 =
17.6
Finally,
10 − 27.6 e−11.3 t + 17.6 e−17.7 t
v (t ) =
P 9.8-17 The circuit shown in
Figure P 9.8-17 is at steady state before the
switch opens. Determine the inductor current,
i2(t), for t > 0.
Figure P 9.8-17
Solution:
First, we find the initial conditions;
For t < 0, the switch is closed and the
circuit is at steady state. At steady state, the
inductors act like short circuits.
i1 ( 0 − )=
and
20
= 1.333 A
15
i 2 ( 0 − ) =0 A
Next, represent the circuit by a differential equation.
After the switch opens, KVL gives
d
d
L1 =
i1 ( t ) R 2 i 2 ( t ) + L 2 i 2 ( t )
dt
dt
KVL and KCL give
L1
d
i1 ( t ) + R1 ( i1 ( t ) + i 2 ( t ) ) =
0
dt
Use the operator method to get
L=
R2 i 2 + L2s i 2
1 s i1
L1s i1 + R1 ( i1 + i 2 ) =
0
L1s 2i1 + R1s i1 + R1s i 2 =
0
s ( R 2i 2 + L 2 s i 2 ) +
R1
L1
(R i
2 2
+ L 2 s i 2 ) + R1s i 2 =
0


L2
R1 R 2
L 2 s 2 i 2 +  R 2 + R1
+ R1  s i 2 +
i2 =
0


L1
L1


 R 2 R1 R1 
R1 R 2
s 2i 2 + 
+
+  s i2 +
i2 =
0
 L 2 L 2 L1 
L
L
1
2


so
 R 2 R1 R1  d
R1 R 2
d2
i
t
+
+
+  i 2 (t ) +
i 2 (t ) =
0
(
)

2 2

dt
L1 L 2
 L 2 L 2 L1  dt
Compare to
d2
dt
i t + 2α
2 ( )
d
i ( t ) + ω 02 i ( t ) =
f (t )
dt
to get
2a =
R2
L2
+
R1
L2
+
R1
L1
, ω 02 =
R1 R 2
L1L 2
and
f (t ) = 0
With the given element values, we have α = 33.9 and ω 0 2 = 281.25 . Consequently, the roots of the
characteristic equation are s 1,2 =−α ± α 2 − ω 0 2 =−4.4, − 63.4 so the circuit is overdamped. The natural
response is
=
i n ( t ) A1 e−4.4 t + A 2 e−63.4 t
Next, determine the forced response.
The steady state response after the switch opens will
be used as the forced response. At steady state the
inductors act like short circuits.
if = 0 A
So
i 2 ( t ) = i n ( t ) + i f ( t ) = A1 e−4.4 t + A 2 e−63.4 t
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
=
0 i 2 ( 0=
) A1 + A 2
L2
d
i 2 ( 0 ) + R 2 i 2 ( 0 ) + R1 i 1 ( 0 ) + R1 i 2 ( 0 ) ⇒
dt
d
i 2 ( 0) =
−20
dt
and
d
−20 = i ( 0 ) =
−4.4 A1 − 63.4 A 2
dt
Solving these equations gives A1 = −0.339 and A2 = 0.339 so
−0.339 e−4.4 t + 0.339 e−63.4 t for t ≥ 0
i 2 (t ) =
P 9.9-1 Find v(t) for t > 0 using the state variable method of
Section 9.9 when C = 1/5 F in the circuit of Figure P 9.9-1.
Sketch the response for v(t) for 0 < t < 10 s.
Answer: v(t) = – 25e–t + e– 5t + 24 V
Figure P 9.9-1
Solution:
=
t 0− circuit is source free ∴ iL=
(0) 0 & v(0)
=
0
t>0
1
KCL at top node: i L +   dv
4
(1)
=
 5  dt
di
KVL at right loop: ( v − 1) L
0
− 6i L =
dt
d 2v
dv
Solving for i in (1) & plugging into (2) ⇒
+ 6 + 5v =
120
1
dt
dt 2
The characteristic equation is: s 2 + 6 s + 5 =
0,
The roots of the characteristic equation are s =−1, −5
∴ Tthe natural response is: v=
A1 e − t + A2 e −5t
n (t )
Try vf = B & plug into D.E. ⇒ B =24 = vf
dv(0)
=−
20 5iL (0) =
20 V
s
dt
So v(0) =0 = A1 + A2 + 24 
A1 =
−25, A2 =
1


dv(0)
−25 e − t + e −5t + 24 V
=
20 =
− A1 − 5 A2  ∴ v(t ) =
dt

From (1)