Chapter 7 Energy Storage Elements
Exercises
Ex. 7.3-1
2 2<t <4
d

and
i C ( t ) = 1 v s ( t ) = −1 4 < t < 8
dt
 0 otherwise

2 t − 2 2 < t < 4

so i (t ) = i C ( t ) + i R ( t ) =  7 − t
4<t <8
 0
otherwise

2 t − 4 2 < t < 4

i R (t ) = 1 v s (t ) =  8 − t
4<t <8
 0
otherwise

Ex. 7.3-2
1 t
1 t
v(t ) = ∫ i s (τ ) dτ + v(t 0 ) = ∫ i s (τ ) dτ − 12
1 0
C t0
3
t
v(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
0
t
v(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
4
In particular, v(4) = 36 V.
In particular, v(10) = 0 V.
t
v(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
10
Ex. 7.4-1
Cv 2
1
2
= ( 2×10 −4 ) (100 ) = 1 J
2
2
+
−
vc ( 0 ) = vc ( 0 ) = 100 V
W =
7-1
Ex. 7.4-2
a)
W ( t ) = W ( 0 ) + ∫ 0 vi dt
t
First, W ( 0 ) = 0 since v ( 0 ) = 0
Next, v( t ) = v( 0 ) +
∴ W (t ) =
1 t
4 t
i
dt
10
2 dt = 2×104 t
=
∫
∫
0
0
C
∫ ( 2×10 ) t ( 2 )dt
t
4
0
= 2 × 104 t 2
W (1s ) = 2 ×104 J = 20 kJ
b)
W (100s ) = 2 × 104 (100 )
2
= 2 × 108 J = 200 MJ
Ex. 7.4-3
We have v (0+ ) = v (0− ) = 3 V
vc ( t ) =
t
1 t
i (t ) dt + vc (0) = 5 ∫ 0 3 e5t dt + 3 = 3 ( e5t −1)+ 3 = 3 e5t V, 0<t <1
∫
0
C
a)
v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1
b)
 W (t ) t =0.2 s = 6.65 J
2
W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒ 
2
2
W ( t ) t =0.8 s = 2.68 kJ
Ex. 7.5-1
7-2
Ex. 7.5-2
v1 = v2 ⇒
dv1 dv2
i
i
C
=
⇒ 1 = 2 ⇒ i1 = 1 i2
dt
dt
C1 C2
C2
C

KCL: i = i1 + i2 =  1 + 1 i2
 C2

⇒
i2 =
C2
i
C1 +C2
Ex. 7.5-3
(a) to (b) :
(c) to (d) :
1
1
= mF ,
1 1 1 9
+ +
1 1 1
3 3 3
1
1
1
1
= + +
2
2 10
Ceq
9
(b) to (c) : 1+
⇒ Ceq =
1 10
=
mF ,
9
9
10
mF
19
7-3
Ex. 7.6-1
 2 2<t <4
d

and
v L ( t ) = 1 i s ( t ) = −1 4 < t < 8
dt
 0 otherwise

2 t − 2 2 < t < 4

so v(t ) = v L ( t ) + v R ( t ) =  7 − t
4<t <8
 0
otherwise

2 t − 4 2 < t < 4

4<t <8
v R (t ) = 1 is (t ) =  8 − t
 0
otherwise

Ex. 7.6-2
i (t ) =
1 t
1 t
v s (τ ) dτ + i (t 0 ) = ∫ v s (τ ) dτ − 12
∫
1 0
L t0
3
t
i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
0
t
i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
4
In particular, i(4) = 36 A.
In particular, i(10) = 0 A.
t
i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
10
Ex. 7.7-1
v = L
di
1 d
=  
4 t e − t ) = (1−t ) e− t V
(
dt
 4  dt
P = vi =  (1−t ) e− t  ( 4 t e− t ) = 4 t (1−t ) e −2t W
2
1
11
W = Li 2 =   ( 4 t e− t ) = 2 t 2 e − 2t J
2
24
7-4
v (t ) = L
 0
 2t

and i ( t ) = 
−2( t − 2 )
 0

di 1 di
=
dt 2 dt
p (t ) = v (t ) i (t )
Ex. 7.7-2
t <0
 0

0<t <1
 1
⇒ v( t ) = 
1<t < 2
 −1
 0
t >2
t <0
0<t <1
1<t < 2
t >2
t <0
 0
2t
0<t <1

= 
2( t − 2 ) 1<t < 2
 0
t >2

W ( t ) = W ( t0 ) + ∫ tt p( t ) dt
0
i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 0
0 < t < 1: W ( t ) =
t
∫ 2 t dt = t
1<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t
2
0
t
1
2
− 4t + 4
t >2 : W (t ) = W ( 2) = 0
Ex. 7.8-1
7-5
Ex. 7.8-2
Ex. 7.8-3
i1 =
1 t
∫ v dt + i1 ( t0 ) ,
L1 t 0
i2 =
1 t
∫ v dt + i 2 ( t0 )
L 2 t0
but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0
1
1 t
1  t
1 t
t
∫ t v dt + ∫ t v dt =  +
∫ v dt
 ∫ t v dt =
L1 0
LP t 0
0
 L1 L2  0
1 t
1
∫ t v dt
i
L2
L 0
L1
∴1 = 1
=
=
1 t
1 1
i
L1 + L2
+
∫ t v dt
LP 0
L1 L2
i = i1 + i2 =
7-6
Problems
Section 7-3: Capacitors
P7.3-1
v (t ) = v (0) +
1 t
i (τ ) dτ
C ∫0
and q = Cv
In our case, the current is constant so
∴ Cv ( t ) = Cv ( 0 ) + i t
t
∫ i (τ ) dτ .
0
−6
−6
q −Cv( 0 ) 150×10 −(15×10 )( 5 )
∴ t=
=
= 3 ms
i
25×10−3
P7.3-2
i (t ) = C
d
1d
1
v (t ) =
12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A
dt
8 dt
8
P7.3-3
( 3×10 ) cos ( 500t + 45 ) = C dtd
−3
so
°
12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° )
= C ( 6000 ) cos ( 500t + 45° )
3×10−3 1
1
= ×10−6 = µ F
C=
3
6×10
2
2
7-7
P7.3-4
v (t ) =
0 < t < 2 ×10−9
1 t
1
i (τ ) dτ + v ( 0 ) =
∫
0
C
2 × 10−12
is ( t ) = 0 ⇒ v ( t ) =
2 × 10−9 < t < 3 × 10−9
1
2 × 10−12
t
∫ i (τ ) dτ − 10
−3
0
t
∫ 0 dτ − 10
0
−3
= −10−3
is ( t ) = 4 ×10−6 A
t
1
4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t
−12 ∫2ns (
2 × 10
In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 × 106 ) ( 3 × 10−9 ) = 10−3
⇒ v (t ) =
3 × 10−9 < t < 5 ×10−9
is ( t ) = −2 × 10−6 A
t
1
−2 × 10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t
−12 ∫3ns (
2 × 10
In particular, v ( 5 × 10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V
⇒ v (t ) =
5 ×10−9 < t
is ( t ) = 0 ⇒ v ( t ) =
1
2 × 10−12
∫
t
5ns
0 dτ − 10−3 = −10−3 V
P7.3-5
(b)
(a)
 0 0 < t <1

d
 4 1< t < 2
i (t ) = C v(t ) = 
dt
 −4 2 < t < 3
 0
3<t
t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ
0
C 0
t
v ( t ) = ∫ 0 dτ + 0 = 0 V
For 0 < t < 1, i(t) = 0 A so
0
For 1 < t < 2, i(t) = (4t − 4) A so
(
t
v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ
1
( )
t
) 1=2 t
2
− 4t + 2 V
v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so
(
t
v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ
2
( )
t
) 1 +2= ( −2 t
2
)
+ 12 t − 14 V
v(3) = −2 32 + 12 ( 3) − 14 = 4 V
t
For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V
0
7-8
P7.3-6
(a)
(b)
 0 0<t <2
d

i (t ) = C v(t ) = 0.1 2 < t < 6
dt
0
6<t

t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ
0
C 0
t
For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V
0
For 2 < t < 6, i(t) = 0.2 t − 0.4 V so
(
t
v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = 0.2τ 2 − 0.8τ
1
( )
t
) 2 =0.2 t
2
− 0.8 t + 0.8 V
v(6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 V . For 6 < t, i(t) = 0.8 A so
t
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V
6
P7.3-7
v (t ) = v ( 0) +
t
1 t
i (τ ) dτ = 25 + 2.5 × 104 ∫ 0 ( 6×10−3 ) e −6τ dτ
∫
0
C
t
= 25 + 150∫ 0 e −6τ dτ
t
 1

= 2 5 + 150  − e −6τ  = 50 − 25e−6t V
 6
0
P7.3-8
v
1
= (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) µ A
3
200×10
40
dv
iC = C
= 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t µ A
dt
i = iR + i C = 200 e−2t + 25 − 50 e −2t
iR =
(
)
= 25 + 150e −2t
µA
7-9
Section 7-4: Energy Storage in a Capacitor
P7.4-1
Given
0
t<2


i ( t ) = 0.2 ( t − 2 ) 2 < t < 6
 0.8
t >6

The capacitor voltage is given by
v (t ) =
For t < 2
t
1 t
0
2
i
τ
d
τ
+
v
=
(
)
(
)
∫0 i (τ ) dτ + v ( 0 )
0.5 ∫ 0
t
v ( t ) = 2 ∫ 0 dτ + 0 = 0
0
In particular, v ( 2 ) = 0. For 2 < t < 6
v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V
t
t
2
2
In particular, v ( 6 ) = 3.2 V. For 6 < t
t
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V
t
6
Now the power and energy are calculated as
0


2
p ( t ) = v ( t ) i ( t ) = 0.04 ( t − 2 )
 1.28 ( t − 4 )

t<2
2<t <6
6<t
and
W (t ) = ∫
t
0
0

t<2

4
2<t<6
p (τ ) dτ =  0.01( t − 2 )

2
6<t
0.8 ( t − 4 ) − 0.64
7-10
7-11
These plots were produced using three MATLAB scripts:
capvol.m
function v = CapVol(t)
if t<2
v = 0;
elseif t<6
v = 0.2*t*t - .8*t +.8;
else
v = 1.6*t - 6.4;
end
capcur.m
function i = CapCur(t)
if t<2
i=0;
elseif t<6
i=.2*t - .4;
else
i =.8;
end
c7s4p1.m
t=0:1:8;
for k=1:1:length(t)
i(k)=CapCur(k-1);
v(k)=CapVol(k-1);
p(k)=i(k)*v(k);
w(k)=0.5*v(k)*v(k);
end
plot(t,i,t,v,t,p)
text(5,3.6,'v(t), V')
text(6,1.2,'i(t), A')
text(6.9,3.4,'p(t), W')
title('Capacitor Current, Voltage and Power')
xlabel('time, s')
% plot(t,w)
% title('Energy Stored in the Capacitor, J')
% xlabel('time, s')
7-12
P7.4-2

ic ( 0 ) = 0.2 A
dv
= (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒ 
−19
dt
 ic (10ms ) = 8.5×10 A
1
W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0
2
−3
v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J
ic = C
P7.4–3
dvc
so read off slope of vc (t ) to get i (t )
dt
p (t ) = vc (t ) i (t ) so multiply vc (t ) & i(t ) curves to get p (t )
i (t ) = C
P7.4-4
1 t
1 t
π

i dτ = vc ( 0 ) + ∫0 50 cos10t +  dτ =
∫
0
2
6
C

π
5
Now since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t )
2
6
vc ( t ) = vc ( 0 ) +
( 2×10 ) ( 2.5) = 6.25 µ J
1
= C v2 =
c max
2
2
−6
∴ Wmax
5 π 5
π


 vc ( 0 ) − 2 sin 6  + 2 s in 10t + 6 
π
5

= sin 10t +  V
2
6

2
First non-negative t for max energy occurs when: 10t +
π
6
=
π
2
⇒t =
π
30
= 0.1047 s
7-13
P7.4-5
Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 µ C
∆q
60×10−6
=
= 6 sec to charge
i
10×10−6
1
1
2
stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 µ J
2
2
∆t =
Section 7-5: Series and Parallel capacitors
P7.5-1
2µ F 4 µ F = 6µ F
6µ F in series with 3µ F =
i (t ) = 2 µ F
6µ F⋅3µ F
= 2µ F
6µ F+3µ F
d
(6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA
dt
P7.5-2
4 µ F in series with 4 µ F =
4 µ F×4 µ F
= 2 µF
4 µ F+4 µ F
2 µF 2 µF = 4 µF
4 µ F in series with 4 µ F = 2 µ F
i (t ) =(2×10−6 )
d
(5+ 3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA
dt
P7.5-3
C in series with C =
C C
C ⋅C C
=
C +C 2
C
5
= C
2
2
C⋅ 5 C
5
5
2
C in series with C =
= C
5
2
7
C+ C
2
5  d
5 
(25×10−3 ) cos 250t =  C  (14sin 250t ) =  C (14)(250) cos 250t
 7  dt
7 
so 25×10−3 = 2500 C ⇒ C = 10×10−6 = 10 µF
7-14
Section 7-6: Inductors
P7.6-1
di
= 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V
dt
8 ×1 0 6 V
= 4 ×1 0 6 V
∴ v m ax = 8 ×1 0 6 V th u s h a ve a fie ld o f
m
m
2
6
w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m
∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d .
F in d m a x . v o ltag e a c ro ss c o il:
v (t ) = L
P7.6-2
v=L
di
+ R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V
dt
P7.6-3
(a)
(b)
 0 0 < t <1
 4 1< t < 2
d

v(t ) = L i (t ) = 
dt
 −4 2 < t < 3
3<t
 0
i (t ) =
t
1 t
v
τ
d
τ
+
i
=
0
(
)
(
)
∫0 v (τ ) dτ
L ∫0
For 0 < t < 1, v(t) = 0 V so
t
i ( t ) = ∫ 0 dτ + 0 = 0 A
0
For 1 < t < 2, v(t) = (4t − 4) V so
t
t
i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 A
0
1
i (2) = 4 ( 22 ) − 4 ( 2 ) + 2 = 2 A
For 2 < t < 3, v(t) = −4t + 12 V so
t
t
i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) A
2
2
i (3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 A
t
For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A
3
7-15
P7.6-4
v (t ) = (250 × 10 −3 )
d
(120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° )
dt
= 15 cos(500t − 30° )
P7.6-5
iL (t ) =
for
1
5 ×10−3
0< t < 1 µ s
iL (t ) =
1
5×10−3
t
∫ v (τ ) dτ
0
s
− 2 ×10−6
vs (t ) = 4 mV
 4×10−3 
−3
−6
d
τ
t − 2×10−6 = 0.8− 2×10−6 A
×
−
×
=
4
10
2
10

−3 
∫0
×
5
10


t
 4×10−3

6
iL (1µs) = 
1×10−6 )  − 2×10−6 = − ×10−6 A =−1.2 A
−3 (
5
 5×10

for 1µ s <t < 3 µ s vs (t ) = −1 mV
t
1
6
1×10−3
6
−3
−6
d
τ
−
×
−
×
=−
1
10
10
(t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A
)
−3 ∫1µ s (
−3
5×10
5
5×10
5
 1×10−3

iL ( 3µ s ) =  −
+ 3×10−6  − 1×10−6 = −1.6 µA
−3
 5×10

iL (t ) =
for 3µ s < t vs (t ) = 0 so iL (t ) remains − 1.6 µA
7-16
P7.6-6
v(t ) =
( 2 ×10 ) i
3
s
for 0<t <1 µ s
(t ) +
( 4 ×10 ) dtd
−3
is (t ) (in general)
 1×10−3 
d
is (t ) = (1) 
t = 103 t ⇒ is (t ) = 1×103
−6 
dt
 1×10 
v(t ) = (2×103 )(1×103 ) t + 4×10−3 (1×103 ) = ( 2×106 t + 4 ) V
d
is (t ) = 0
dt
v(t ) = (2×103 )(1×10−3 ) + ( 4×10−3 )×0 = 2 V
for 1µ s <t < 3µ s is (t ) = 1 mA ⇒
 1×10−3 
t ⇒
for 3µ s< t < 5µ s is (t ) = 4×10−3 −
−6 
 1×10 
d
1×10−3
=−103
is (t ) = −
−6
dt
1×10
v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t
when 5µ s <t< 7µ s
is (t ) = −1×10−3 and
v(t ) = ( 2×103 )(10−3 ) = − 2 V
d
is (t ) = 0
dt
 1×10−3 
d
when 7µ s< t < 8µ s is (t ) = 
t − 8×10−3 ⇒ is (t ) = 1×103
−6 
dt
 1×10 
v(t ) = ( 2 × 103 )(103 t − 8 × 10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t
when 8µ s < t , then is (t ) = 0 ⇒
d
is (t ) = 0
dt
v(t )= 0
P7.6-8
(a)
(b)
 0 0<t <2
d

v(t ) = L i (t ) = 0.1 2 < t < 6
dt
0
6<t

t
1 t
i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ
0
L 0
t
For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A
0
For 2 < t < 6, v(t) = 0.2 t − 0.4 V so
t
t
i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A
2
2
( )
i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A .
For 6 < t, v(t) = 0.8 V so
7-17
t
i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A
6
Section 7-7: Energy Storage in an Inductor
P7.7-1
t<0
0
d

v( t ) =100×10
i ( t ) = 0.4 0≤ t ≤1
dt
0
t>1

t <0
0

p( t ) =v( t ) i( t ) = 1.6t 0≤t ≤1
0
t >1

−3
W (t ) = ∫
t
0
 0

p (τ ) dτ = 0.8t 2
 0.8

t <0
0<t <1
t >1
7-18
P7.7-2
 d

p (t ) = v (t ) i (t ) = 5 (4sin 2t )  (4sin 2t )
 dt

= 5 (8cos 2t ) (4sin 2t )
= 80 [2 cos 2t sin 2t ]
= 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W
t
t
80
W(t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = −
[cos 4τ |t0 ] = 20 (1 − cos 4t )
0
0
4
P7.7-3
t
1
6 cos 100τ dτ + 0
25×10−3 ∫ 0
6
[sin 100τ | 0t ] = 2.4sin100 t
=
−3
(25×10 )(100)
p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t )
i(t ) =
= 7.2 [ 2(cos100 t )(sin100 t ) ]
= 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t
t
t
W (t ) = ∫0 p (τ ) dτ = 7.2 ∫0 sin 200τ dτ
7.2
 cos 200τ |t0 
200
= 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJ
=−
Section 7-8: Series and Parallel Inductors
P7.8-1
6×3
= 2 H and 2 H + 2 H = 4 H
6+3
1 t
6
sin100τ | t0  = 0.015sin100 t A = 15sin100 t mA
i (t ) = ∫ 0 6 cos100τ dτ =
4
4×100 
6H
3H =
P7.8-2
(8×10 )×(8×10 )
−3
4 mH + 4 mH = 8 mH , 8mH 8mH =
−3
8×10−3 +8×10−3
= 4 mH
and 4 mH + 4 mH = 8 mH
d
v(t ) = (8×10−3 )
(5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V
dt
7-19
P7.8-3
L⋅ L
L
L
5
L
=
and L + L +
=
L+ L
2
2
2
5  d
5 
25cos 250 t =  L 
14×10−3 ) sin 250 t =  L (14×10−3 )(250) cos 250 t
(
 2  dt
2 
25
so L =
= 2.86 H
5
−3
(14×10 ) (250)
2
L
L =
(
)
Section 7-9: Initial Conditions of Switched Circuits
P7.9-1
Then
i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 12 V
Next
7-20
P7.9-2
Then
i L ( 0+ ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 V
Next
P7.9-3
Then
i L ( 0+ ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 V
Next
7-21
P7.9-4
at t = 0−
KVL: − vc (0− ) + 32 − 15 = 0 ⇒ vc (0− ) = vc (0+ ) =17 V
at t = 0+
Apply KCL to supernode shown above:
−15 − 9 15
ic ( 0+ ) +
−
+ 0.003 = 0 ⇒ ic ( 0+ ) = 6 mA
4000 5000
Then
ic ( 0+ ) 6 ×10−3
dvc
V
=
=
= 3000
−6
2 ×10
s
dt t =0+
C
7-22
Section 7-10: Operational amplifier Circuits and Linear Differential Equations
P7.10-1
P7.10-2
7-23
P7.10-3
P7.10-4
7-24
Verification Problems
VP7-1
We need to check the values of the inductor current at the ends of the intervals.
? − 1 + 0.065 = 0.025 ( Yes!)
at t = 1 0.025 =
25
at t = 3
−
3
3
+ 0.065 =?
− 0.115
25
50
−0.055 = −0.055
( Yes!)
9
− 0.115 =? 0.065
50
at t = 9
0.065 = 0.065 ( Yes!)
The given equations for the inductor current describe a current that is continuous, as
must be the case since the given inductor voltage is bounded.
VP7-2
We need to check the values of the inductor current at the ends of the intervals.
1
? − 1 + 0.03 ( Yes!)
+ 0.025 =
at t = 1 −
200
100
4
? 4 − 0.03 ( No!)
+ 0.03 =
100
100
The equation for the inductor current indicates that this current changes
instantaneously at t = 4s. This equation cannot be correct.
at t = 4
−
Design Problems
DP7-1
i (t )
d
= 0.5 F .
v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C =
d
dt
v (t )
dt
v (t )
d
= 0.5 H .
b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L =
d
dt
i (t )
dt
v (t )
c) v(t) is proportional to i(t) so the element is a resistor. R =
= 2 Ω.
i (t )
a)
7-25
DP7-2
1.131cos ( 2t + 45° ) = 1.131  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) 
= 0.8 cos 2 t − 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the
resistor. The noticing that
∫
t
−∞
t
v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
−∞
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign.
Then
4 cos 2 t
4 cos 2 t
R=
=
= 5 Ω and
i1 (t )
0.8 cos 2 t
−0.8 sin 2 t
i2 (t )
=
= 0.1 F
C=
d
−
t
8
sin
2
4 cos 2 t
dt
1.131cos ( 2t − 45° ) = 1.131  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) 
= 0.8 cos 2 t + 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the
resistor. Then noticing that
t
t
−∞
−∞
∫ v (τ ) dτ = ∫
4 cos 2 t dτ = 2 sin 2t
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the plus sign. Then
4 cos 2 t
4 cos 2 t
=
= 5 Ω and
i1 (t )
0.8 cos 2 t
R=
L=
∫
t
−∞
4 cos 2 t dτ
i2 (t )
=
2 sin 2 t
= 2.5 H
0.8 sin 2 t
7-26
DP7-3
a)
11.31cos ( 2t + 45° ) = 11.31  cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) 
= 8 cos 2 t − 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
t
t
−∞
−∞
∫ i (τ ) dτ = ∫
4 cos 2 t dτ = 2 sin 2t
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the minus sign. Then
v (t )
8 cos 2 t
v2 (t )
−8 sin 2 t
R= 1
=
= 2 Ω and L =
=
=1H
d
4 cos 2 t 4 cos 2 t
4 cos 2 t −8 sin 2 t
dt
b)
11.31cos ( 2t + 45° ) = 11.31  cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) 
= 8 cos 2 t + 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
∫
t
−∞
t
i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
−∞
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign. Then
v (t )
8 cos 2 t
R= 1
=
= 2 Ω and C =
4 cos 2 t 4 cos 2 t
∫
t
−∞
4 cos 2 t dτ
v2 (t )
=
2 sin 2 t
= 0.25 F
8 sin 2 t
7-27
DP7-4
iL ( 0− ) = 0
at t=0−
By voltage division: vC ( 0− ) =
We require vC ( 0− ) = 3 V so
VB
4
VB = 12 V
at t=0+
Now we will check
First:
and
dvC
dt
t = 0+
iL ( 0+ ) = iL ( 0− ) = 0
vC ( 0+ ) = vC ( 0− ) = 3 V
Apply KCL at node a:
VB − vC ( 0+ )
+
+
iL ( 0 ) + iC ( 0 ) =
3
0 + iC ( 0+ ) =
12 − 3
⇒ iC ( 0+ ) = 3 A
3
Finally
iC ( 0
dvC
=
dt t =0+
C
as required.
+
)=
3
V
= 24
0.125
s
DP7-5
1
1
L i L2 = C v C2 where iL and vC are the steady state inductor current and capacitor
2
2
v
voltage. At steady state, i L = C . Then
R
We require
2
v 
L  C  = C vC2
R
⇒
C=
L
R2
⇒
R =
L
=
C
10−2
=
10−6
104 = 10 2 Ω
so R = 100 Ω .
7-28