、
、
一一
一
一
一
二
《
言
ECE 2202 – CIRCUIT ANALYSIS II
HOMEWORK #2
1) The current iC(t) through the capacitor in Figure 1 is given by the plot shown in Figure 2. It is
given that vC(2[ms]) = 6[V]. Find vC(8[ms]).
iC(t), [A]
3
+
vC(t)
150[pF]
iC(t)
Figure 1
romF the figure 2
ic (t)
In
=
丫
-3
,
FUA)
t
时S
]
,
we
O ≤ t
<
:
RImS]
,
Ʃ ttidt
+
Active
:
UlHo
,
C
150 IPFJ
=
/
insg t + VcGIm)=
Vc (Bims
_
Ʃ
7)
)
位
] ]kg +IVJ
6
]EIMAJ [ ]
16 67 [V]+ 6[ v]
ms
=
=
Vc( 8ins)
12
t,[ms]
Active
3 Ims]
t- 4 IuA33 [ ms] ≤
'
9
Figure 2
that :
know
Capacitor circuit
U (t)
6
3
-
=
.
)[ms]
=
- Ʃ ms 信4 im ) +Vc ( Ems )
t
>dt
后[
)
3
8Ims]
| 4
PF > 證 3 insg
“
(
-
=
TI
·上 IMA ] Ims - 10 6]
72 2 [U]
] ]
-
10 H τV]
.
2.1
=
61
.
55 IV]
+
-
6 []
10 67 IV]
.
t=0
+
vS(t)
1[mF]
2.2[k]
4.7[k]
600[H]
islt ]
→
因 亞亞素
R
2
is (t)
=
B
. / 132≥ 1 . 5[k]
=+ +÷
C
=
/
.
,
卡
3
Vs(t) 6 [ ] t t > 0
=
,
Vslt dt
)
(4 103t + 1 . ×102 t +÷00t IA] t70
×
2
,
)
Pdelbyvoltage source =Vs(t islt ]
)
=② 4×102
t
4 + 7. 2×
.
when t 5 G]
[
=
Pdelbyvoltafe
sorce =
86 5IW]
.
,
2.2
心
一
vS (t ) = 6  V 2  t 2 ; for t  0.
 s 
文
2) In the circuit shown below, the switched closed at t = 0. No energy was stored in the
capacitor, and no energy was stored in the inductor, at t = 0. The expression for vS(t) is given
below.
Find the power delivered by the voltage source at t = 5[s].
t5 >
102t 3+ 50
E
t=0
4.7[k]
Switch SWB
t=0
7[V]
4iX
3.3[k]
L1 =
3[mH]
-
t
=
iX
0
:
纮可以
” d
a)
.
ix ( 0)
b>
.
iQ
=
-
[kn]
=
-
1 489 [mA]
.
i可
()
.ia10
i , 10
)
"
"
divi
=
-
i 1 l0
fromt =δ
:
]
=
-
[u)
=]
K
=
=
- 4 ix
() i< ) ( tia]
ilo )= IlO
5ix 10) =Ʃ l 0+ )
= 5 ixlo)
-
-
icloytialoy
2.3
=
2 x 10)
=
-
2
,
55 ImA]
ImAJ
3 182
2 . 12
0
)
-
,
lot)
点
ix
州
2.2[k]
iQ (δ) = OIA ]
鴓藏
系”
“ 产箭
節韓
4我…
心
二
比
t = 2[s]
Switch SWA
+
一
3) In the circuit shown below, the two switches were open for a long time before t = 0, until all
voltages and currents stopped changing. Then, both switches closed at t = 0. Then, switch SWA
opened again 2[s] later.
a) Find iQ(0-).
b) Find iX(0-).
c) Find iQ(0+).
d) Find iX(0+).
EmA]
t=0
R1 = 300[]
R2 = 10[]
t = 0.5[s]
SW1
SW2
iX
+
vS1 =
300[V]
-
R3 = 20[]
R4 =
30[]
L=
0.2[H]
iL
&^
:
m—x
/
iS = -2iX
+
一
4) In the circuit shown below, switches SW1 and SW2 have been closed for a long time before
t = 0, allowing all voltages and currents to stop changing.
At t = 0, switch SW1 opens and remains open.
At t = 0.5[s], switch SW2 opens and remains open.
a) Find iL(0-).
b) Find iX(0-).
c) Find iL(0+).
d) Find iX(0+).
e) Find the energy stored in the inductor just before t = 0, wSTO.BY.L(0-).
f) Find the energy stored in the inductor just after t = 0, wSTO.BY.L(0+).
vS2 =
100[V]
-
R3 = 2OIN
…恆
-
0
的
300I]
R
,
1
IA ]
) i<lo-
)
=i - i
a
,
t b
=
+
:
1
:
2
=
-
)
…
<
.
2
=
84 IIJ ]
f) Wsv BYL 1O ) =ƩL [ i 16]]2 = 48 I [ J ] 2.4
:
+
.
3
=i 2 =30IA
直
WsTO BY 1 (σ) =Ʃ Li
o
]
e) :
]
@
jiiizio
iao
: 60ix +R 4 li - i 2 )+13323f/0V:O
-
]b . ixl0
2 q[ A
③
i3= 2OIA ]
i 2 = 30 [A]
=
= ora )
② R2 .iz -60ix +R 4 ( 22 -23 ) = O
Pi = ,
i
+
)
c) :
+
2 < 10 ) 2cl 0 ]
d) : ix0+)
=
=
-
=
-
29 IA ]
iil0 ) 29 [A ]
+
=
t = 10[ms]
R2 = 680[]
+ vQ
SWA
-
t=0
+
。
5) In the circuit shown below, the energy stored in inductor L1 was 0
zero when switch SWA
closed and switch SWB opened
at t = 0. Then, 10[ms] later switch SWA opened.
→
a) Find vQ(0-).
b) Find vQ(0+).
c) Find the energy stored in the inductor just before t = 0, wSTO.BY.L1(0-).
d) Find the energy stored in the inductor just after t = 0, wSTO.BY.L1(0+).
vS1 =
5[V]
-
SWB
R1 =
220[]
t=0
R3 =
750[]
L1 =
7[mH]
—唯亞“
—
…
∵
t
=
3
3
(
∴
=
b) :
is1
-5)
) iL( σ =
)
=
:WSTO .BY , 2 , 0
(
)
)
-]
,(
=Ʃ 210
L
]0
WSπ BY L ( 0) =Ʃ L [Z : 10+ ) ] 三
,
.
,
l∵ kf 6 G
=
1
_rs
± 750
纮磊道迎
M≥
①∴ -5[V ] +R , ( i -τ2 ) =
m
l+ :
∵ 2(0 +
> lalt)
,…
iS1 =
33[mS]vQ
.
2.5
② : B, lizi 1 ] + R2i2 + R3 iz
③ : Valoy
OQ (0 + )
=
=
t iz
·
186 2 [mV ]
.
=
750is )
Numerical Solutions:
1. 61.6[V]
2. 86.55[W]
3. a) 0, b) -1.489[mA], c) -3.182[mA], d) -2.55[mA]
4. a) -29[A], b) 30[A], c) -29[A], d) 29[A], e) 84.1[J], f) 84.1[J]
5. a) 0, b) 186.2[mV], c) 0, d) 0
2.6