Chapter 8 – The Complete Response of RL and RC Circuits
Exercises
Ex 8.3-1 Before the switch closes:
After the switch closes:
Therefore R t =
2
= 8 Ω so τ = 8 ( 0.05 ) = 0.4 s .
0.25
Finally, v (t ) = voc + (v (0) − voc ) e
−
t
τ
= 2 + e −2.5 t V for t > 0
157
Ex 8.3-2 Before the switch closes:
After the switch closes:
2
6
= 8 Ω so τ = = 0.75 s .
0.25
8
t
−
1 1 −1.33t
Finally, i (t ) = isc + (i (0) − isc ) e τ = +
e
A for t > 0
4 12
Therefore R t =
Ex. 8.3-3
At steady-state before t = 0:
i =
10 12 10+40 16+ 40||10
= 01
. A
158
After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be
so I sc = 0.3 A, R th = 40Ω, τ =
Finally:
Ex. 8.3-4
i(t) = (0.1 − 0.3)e
−2t
L
R th
=
20
1
=
40
2
+ 0.3= 0.3 − 0.2e
−2t
A
At steady-state for t < 0
After t = 0, replace the circuit connected to the capacitor by its Thèvenin equivalent
so Voc = 12V, R th = 200Ω, τ = R th C = (200)(20 ⋅ 10
Finally:
v(t) = (12 − 12)e
−
t
4
−6
) = 4 ms
+ 12 = 12 V
159
Ex. 8.3-5 Before t = 0, i(t) = 0 so I o = 0
After t = 0, replace the circuit connected to the inductor by its Norton equivalent
I sc = 0.2 A, R th = 45Ω, τ =
L
25
5
=
=
R th
45
9
So
i (t) = 0.2 (1 − e −1.8 t ) A
Finally: v(t) = 40 i(t)+25
Ex. 8.3-6
d
−1.8t
−1.8t
−1.8t
i(t) = 8(1 − e
) + 5(1.8)e
= 8+e
V
dt
t < 0:
⇒ i (0 ) = 0.5 A
t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get
Continued
I sc = 93.75mA, R th = 640Ω, t =
L
.1
1
=
=
R th
640
6400
160
So
i (t) = 406.25e −6400 t + 93.75 mA
Finally
v(t) = 400 i(t) + 0.1
d
−6400t
−6400t
−6400t
i(t) = 400 (.40625e
V
+ .09375 ) + 0.1 (−6400 ) (0.40625e
) = 37.5 − 97.5e
dt
Ex. 8.4-1
(
)(
)
τ = 210
⋅ −3 110
⋅ −6 = 2 ⋅ 10
v c (t) = 5+ (1.5−5 ) e
−
v c (1) = 5 − 3.5e
t
2
−1
2
−3
where t is in ms
= 2.88V
So v c (t) will be equal to v T at t=1 ms if v
T
= 2.88 V
Ex. 8.4-2
i L (0) = 1mA, I sc = 10mA
L
R th = 500Ω, τ =
500
16
(K
)K ⇒ i 1t6 = 10 − 9e
*
L
v R (t) = 300 i L t = 3 − 2.7e
−
500
t
L
−500
t
L mA
V
We require that v R = 1.5V at t = 10ms = 0.01 s
That is
15
. = 3 − 2.7e
−
e
−
5
L
=
−
500
( 0.01)
L
1.5− 3
= 0.555
−2.7
5
= ln (0.555) = −.588
L
5
L =
= 8.5 H
0.588
161
Ex. 8.6-1
0 < t < t1
v(t) = v(∞ ) + Ae
v(t) = 1+Ae
− t/RC
− t/(1)(.1)
where v(∞ ) = (1A)(1Ω )= 1V
= 1+Ae
−10t
Now v(0 − ) = v(0 + ) = 0 = 1 + A ⇒ A = − 1 ∴ v(t) =1− e −10 t V
t > t1
−
t 1 = .5s
v(t) = v(t 1 )e
t − .5
(1)(.1)
= v(.5)e −10(t −.5)
Now v(.5) = 1 − e −10(.5) = .993V
∴ v(t) =.993e −10(t −.5) V
Ex. 8 6-2
t < 0 no sources ∴ v(0 − ) = v(0 + ) = 0
0 < t < t1
v (t) = v ( ∞ ) + A e −
− t
t/ RC
= v(∞ ) + A e
2 × 1 0 5 (1 0 − 7 )
where for t = ∞ (steady-state)
∴capacitor becomes an open ⇒ v(∞) = 10V
v(t) = 10 + Ae −50t
Now v(0) = 0 = 10+A ⇒ A = −10 ∴ v(t) = 10(1− e −50t ) V
t > t 1 , t 1 = .1s
v(t) = v(.1)e −50(t −.1)
where v(.1) = 10(1− e −50(.1) ) = 9.93 V
∴ v(t) = 9.93e −50(t −.1) V
162
Ex. 8.6–3
for t < 0
i=0
0 < t < .2
(K di
)K dt + 10i = 50
*
KCL: − 5 + v/2 + i = 0
also: v = 0.2
di
dt
∴ i(t) = 5+ Ae −10 t
i(0) = 0 = 5+A ⇒ A = − 5
so have i (t) = 5 (1− e −10t ) A
t > .2
i (.2) = 4.32 A ∴ i(t) = 4.32e −10(t −.2) A
Ex. 8.7-1
v s = 10sin 20t V
 dv 
KVL a: −10sin20t +10 .01 + v = 0 ⇒
 dt 
dv
+10v = 100 sin 20t
dt
Natural response: s + 10 = 0 ⇒ s = − 10 ∴ v n (t) = Ae −10 t
Forced response: try v f (t) = B1 cos 20t + B2 sin 20 t
plugging v f (t) into the differential equation and equating like terms
yields: B1 = − 40 & B2 = 20
Complete response: v(t) = v n (t) + v f (t)
v(t) = Ae −10t − 40 cos 20t + 20 sin 20t
Now v(0 − ) = v(0 + ) = 0 = A − 40 ∴ A = 40
∴ v(t) = 40e −10 t − 40 cos 20t + 20 sin 20 t V
Ex. 8.7-2
i s = 10e −5t
KCL at top node: −10e −5t + i + v /10 = 0
Now v = .1
di
di
⇒
+100i = 1000e −5t
dt
dt
Natural response: s + 100 = 0 ⇒ s = − 100 ∴ i n (t) = Ae −100t
Forced response: try i f (t) = Be −5t & plug into D.E.
⇒−5Be −5t + 100 Be −5t = 1000e −5t
⇒ B = 10.53
Complete response: i(t) = Ae −100t + 10.53e −5t
Now i(0 − ) = i (0 + ) = 0 = A + 10.53 ⇒ A = −10.53
∴ i(t) = 10.53 (e −5t − e −100 t ) A
163
Ex. 8.7-3
A current i L = v s / 1 flows in the inductor with the switch closed. When the switch opens, i L
cannot change instantaneously. Thus, the energy stored in the inductor dissipated in the spark.
Add a resistor (say 1 kΩ across the switch terminals.)
Problems
Section 8.3: The Response of a First Order Circuit to a Constant Input
P8.3-1
We know that
16 1
6
i L t = I o − I SC e
1 6
where I o = i L t o and τ =
−
t − t o τ +I
SC
L
. In this problem t o = 0 and I o = i L (0)=3mA.
R TH
The Norton equivalent of
is
So R th = 1333Ω and Isc = 5mA.
L= 5 H so τ =
Finally
L
5
=
= 3.75 ms
R th 1333
iL ( t ) = −2e
−
t
3.75
+5 mA t > 0
where t has units of ms.
164
P8.3-2
We know that
16 1
6
v c t = V0 − Voc e
1 6
−
t − t 0 τ +V
oc
where v 0 = v C t 0 and τ = R th C,
In this problem, t 0 = 0 and v 0 = v c (0) = 8V.
The Thèvenin equivalent of
is
so R th = 1333 and Voc = 4 V.
4
9
Next, C = 0.5µF so τ = 0.5⋅10 −6 1333 = 0.67ms
Finally
16
v c t = 4e
−
t
.67
where t has units of ms.
+4 V
P 8.3-3 Before the switch closes:
165
After the switch closes:
Therefore R t =
−6
= 3 Ω so τ = 3 ( 0.05 ) = 0.15 s .
−2
Finally, v (t ) = voc + (v (0) − voc ) e
P 8.3-4
−
t
τ
= −6 + 18 e −6.67 t V for t > 0
Before the switch closes:
After the switch closes:
166
−6
6
= 3 Ω so τ = = 2 s .
−2
3
t
−
10 −0.5 t
Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 +
e
A for t > 0
3
Therefore R t =
P8.3-5
Before the switch opens, v o (t ) = 0 V
⇒ v o ( 0 ) = 0 V . After the switch opens the part of the circuit connected
to the capacitor can be replaced by it's Thevenin equivalent circuit to get:
Therefore
τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s .
Next, v C (t ) = voc + (v (0) − voc ) e
−
Finally, v 0 (t ) = vC (t ) = 10 − 10 e
t
τ
= 10 − 10 e −12.5 t V for t > 0
−12.5 t
P8.3-6
Before the switch opens, v o (t ) = 0 V
V for t > 0
⇒ v o ( 0 ) = 0 V . After the switch opens the part of the circuit connected
to the capacitor can be replaced by it's Norton equivalent circuit to get:
Therefore
τ=
5
= 0.25 ms .
20 × 103
Next, i (t ) = isc + (i (0) − isc ) e
Finally, vo (t ) = 5
−
t
τ
= 0.5 ×10−3 (1 − e −4000 t ) A for t > 0
d
i L (t ) = 10 e −4000 t V
dt
for t > 0
167
P8.3-7
Since the input to this circuit is constant, the capacitor will act like an open circuit when the
circuit is at steady-state:
t<0
P8.3-8
t>0
Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit
is at steady-state:
t<0
t>0
i L = 4mA
i L = 2ma
P8.3-9
Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit
is at steady-state:
t<0
t>0
i L = 2mA
i L = 2mA
P8.3-10
t<0
Since the input to this circuit is constant, the capacitor will act like an open circuit when the circuit
is at steady-state:
t>0
168
P8.3-11
4 9=0
at t = 0 − (steady-state)
v L 0−
4 9 = 6A = i 40 9
∴ i L 0−
+
L
for t > 0
16
16
i L t = i L 0 e − (R L) t = 6e −20 t A
P8.3-12
Assume the capacitor is charged at t = 5τ, i.e.,
Vc
Vc
= 1 − e −5 = .993
max
Similarly, assume the capacitor is discharged when
Vc
Vc
= e
−5
−3
= 6.74 × 10 .
max
Now determine C from discharging condition
vc (t ) = vc (to ) e
−
(t −to )
C R bulb
⇒ 6.74 × 10
−3
= e
− 0.5
CR bulb
⇒ C = 10−5F = 10µ F
Now determine a condition for R from charging circuit at the instant
vc = 0 ⇒
6V
−6
<100 × 10 A ⇒ R >60 kΩ
R
then for the charging ckt.
.993 = 1− e −5/ RC
5
= 100.8 kΩ
−4.96 = − 5 RC ⇒ R =
4.96 (10 −5 )
1 6
~ 100kΩ > 60kΩ.
and see that R −
169
P8.3-13
First, use source transformations to obtain the equivalent circuit
for t < 0:
for t > 0:
i L = 2A
So I 0 = 2A, I sc = 0, R th
and i L ( t ) = 2e −24t
1
L
1
= 3Ω +9Ω =12Ω, τ =
= 2 =
R th
12
24
t>0
Finally v ( t ) = 9 i L ( t ) = 18 e−24t
t>0
P 8.3-14 Before the switch opens, v C (t ) = 0 V
Therefore R t =
⇒ v C ( 0 ) = 0 V . After the switch opens:
10
= 20 kΩ so τ = ( 20 ×103 )( 4 ×10−6 ) = 0.08 s .
−3
0.5 ×10
Next, v C (t ) = voc + (v (0) − voc ) e
−
Finally, v 0 (t ) = −vC = −10 + 10 e
t
τ
= 10 − 10 e −12.5 t V for t > 0
−12.5 t
V for t > 0
170
Section 8-4: Sequential Switching
P 8.4-1 Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get:
Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc = 5 V and Rt = 4 Ω so
τ = 4 × 0.5 = 2 s . Therefore
v (t ) = voc + (v (0) − voc ) e
At t =1.5 s, v (1.5) = 5 + 5e
Therefore
−0.5 (1.5)
−
t
τ
= 5 + 5e −0.5 t V for 0 < t < 1.5 s
= 7.36 V . For 1.5s < t, voc = 10 V and Rt = 8 Ω so τ = 8 × 0.5 = 4 s .
v (t ) = voc + (v (1.5) − voc ) e
−
t −1.5
τ
= 10 − 2.34 e
−0.25 (t −1.5 )
V for 1.5 s < t
Finally
 5 + 5 e −0.5 t V
for 0 < t < 1.5 s
v (t ) = 
−0.25 (t −1.5 )
for 1.5 s < t
V
10 − 2.34 e
P 8.4-2 Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:
Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 A and Rt = 6 Ω so
τ=
12
= 2 s . Therefore
6
i (t ) = isc + (i (0) − isc ) e
At t =1.5 s, i (1.5) = 2 + e
−0.5 (1.5)
−
t
τ
= 2 + e −0.5 t A for 0 < t < 1.5 s
= 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ =
12
= 1.5 s . Therefore
8
171
i (t ) = isc + (i (1.5) − isc ) e
i (t ) = isc + (i (1.5) − isc ) e
−
t −1.5
τ
−
t −1.5
τ
= 3 − 0.64 e
−0.667 (t −1.5 )
V for 1.5 s < t
= 3 − 0.64 e
−0.667 (t −1.5 )
V for 1.5 s < t
Finally
2 + e −0.5 t A
for 0 < t < 1.5 s

i (t ) = 
−0.667 (t −1.5 )
for 1.5 s < t
A
3 − 0.64 e
P8.4-3
t = 0−
(steady-state)
KVL : − 52+18i+ (12 8 )i=0
−
⇒
i(0 ) = 104 39 A
 6 
+
∴ iL = i 
 =2A = i L (0 )
 6+ 2 
0 < t < 51ms
i L (t) = i L (o) e − (R
i L (t) = 2e
−6t
∴ i(t) = i L ( t )
L) t
R = 6 12 + 2 = 6Ω
A
6 = 2 3 e
6+12 −6 t
A
t > 51ms
i L (t) = i L (51m s) e − ( R
L ) (t − .051)
i L (51m s) = 2e − 6(.051) = 1.473
i L ( t ) = 1.473 e − 14(t − .051) A
P8.4-4
t = 0−
Assume V1 = voltage across 10µF capacitor = 3V
0 < t < 10mS
With R negligibly small, we may assume a static steady-state situation is obtained in the circuit nearly
instantaneously (t = 0+ ). Thus with both capacitors in parallel, the common voltage is obtained by
considering charge conservation.
at t = 0 − , q 100µF = CV = (100µF) (3V) = 300 µC
q 400µF = CV = (400µF) (0) = 0
q ToT = q 100 + q 400 = 300µC
+
at t = 0 , q100 + q 400 = 300µ C
Now using q = CV ⇒ (100µ F) (V) + (400µ F) (V) = 300µ C ⇒ V = 0.6 V
172
10ms < t < ls
Combine 100µF & 400µF in parallel to obtain
with V(10ms) = 0.6V
v(t) = V(10ms) e
− (t −.01) RC
= 0.6e
− ( t −.01) (103 ) (5 x10−4 )
v(t) = 0.6 e−2(t −.01) V
P8.4-5
Voc = VT = (40 ×
20
 40 
) − 5i1 = 20 − 5 =   = 15 V
20+20
 40 
for R T , kill source with V = 0
Note i1 = 12 A
v
1
R T = 0 = 1(10Ω) − 5( A ) = 7.5Ω
1
2
R eq = 7.5Ω
Forced response
i = 2A
τ = LR =
15×10 −3
= 2ms
7.5
natural: i = Be − t τ = Be −500t
total : i = Be −500t + 2 now i(0) = 0 ⇒ B = − 2
i(t) = 2(1− e −500t )A
time to 99%:
for e −500t = .01 or 500t = 4.605 ⇒ t = 9.2ms
P8.4-6
(K) ⇒ v (t) = 5e
*K
τ =RC=105 ×10 −6 =.1s
v c (0) =5 V
−t/τ
c
Now 5/2 =v c (t 1 )=5e − t1 / τ
e − t1 / τ =.5
i(t 1 ) =
v(t 1 )
100 kΩ
⇒ t 1 =.0693s
=
5/ 2
= 25µ A
105
173
P8.4-7
1
t = 0 − steady-state
6
v(0 − ) = v(0 + ) = (2A) (5Ω) = 10V
0 < t < 100ms
v(t) = v(0)e − t
RC
v(t) = 10e − t (5) (.01) = 10e −20t V
−
t > 100ms
v(t)=v(100ms) e
(t −1)
(4) (.01)
v(100ms)=10e −20(.1) =1.35 V
∴ v(t) = 1.35e−25(t −.1) V
P8.4-8
t = 0 − (steady-state)
1
7= 7A
8
1+4+2+1
−
+
v c (0 ) = v c (0 ) = 2i 2 = 7 / 4 V
i2 =
0 < t < .35 Closing the rightmost switch shorts out 1Ω in parallel with 7A source and isolates 7A
source leaving
−vx
v
+ 5i c + i c + c = 0 (1)
2
4
vx
vc + v x
− 5ic +
KCL at b :
= 0 (2)
4
1
KCL at a :
also: i c = .2
dVc
dt
(3)
Plugging (3) into (1) & (2) & then eliminating v c yields ⇒
dv c
7
+
vc = 0
dt
10
dv c
= −.245e
dt
So from (1) we have v x (t) = 24i c + 2v c = −2.38 e −.7t
So v c (t) = v c (0) e − 0.7 t = 7 4 e − 0.7 t , i c = .2
t > .35
−.7 t
v c (.35) = 7 4 e − ( 0.7 ) (.35) = 1.37V
now = v x = − v c
KCL:
vc
4 + 5i c +
also : i c = 0.2
From (1) & (2) ⇒ dvc dt + 5 8 vc = 0 ⇒ v c (t ) = v c (.35) e
dVc
− ( t −.35)5 8
vc
2 + ic = 0
dt
(1)
(2)
= −1.37 e−.625(t −.35)
174
Section 8-5: Stability of First Order Circuits
P8.5-1
This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the
inductor is positive.
R th =
R1
IT
R 1 +R 2
VT = R 2 i(t) − Ri(t)
i(t) =
(K
)K ⇒ R
*
th
=
VT
IT
(R 2 − R)R 1
R 1 +R 2
Then R th > 0 requires R 2 > R. In this case R 2 = 400Ω so 400 > R is required to guarantee stability.
P8.5-2
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
R th =
v(t) = RI T
VT = I T ( R 1 +R 2 ) − ARI T
() ⇒ R
*
th
VT
IT
= R 1 + R 2 − AR 2
The circuit will be stable when R th > 0, that is,
R +R 2
>A
R th > 0 ⇒ 1
R2
When R 1 = 4 kΩ and R 2 = 1kΩ, then A < 5 is required to guarantee stability.
175
P8.5-3
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
R th =
VT
IT
VT = − R 1i(t) = R 2 (i(t) + Bi(t) + I T )
−R2
i(t) =
IT
R 1 +R 2 +R 2 B
VT = − R 1
R th =
−R2
IT
R 1 +R 2 +R 2 B
R 1R 2
R 1 +R 2 +R 2 B
The circuit is stable when R th > 0, that is
R +R 2
R 1 +R 2 + R 2 B > 0 ⇒ B > − 1
R2
when R 1 = 6k Ω and R 2 = 3kΩ, B > − 3 is required to guarantee stability.
P8.5-4
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
R th =
VT
IT
(K
)K
*
R 1R 2
IT
R R
R 1 +R 2
R th = 1 2 (1 − A )
R 1 +R 2
VT = v (t) − Av(t)
v(t) =
The circuit will be stable when R th > 0, that is, when A<1.
176
Section 8-6: The Unit Step Response
P8.6-1
10u(t) − 4 =
%&10(0) − 4 =−4
'10(1) − 4 = 6
6u( − t) + 4u(t) =
t < 0:
t>0
t > 0:
t < 0:
P8.6-2
t<0
%&6(1) + 4(0) = 6
'6(0) + 4(1) = 4
t < 0
t > 0
t > 0:
177