익힘문제
7.2>
i = C
dv dt
100 ×10−12 × 0 = 0, t < 1 ms 2−0 10−12 × = 100 × 10 −3 = 200 nA , 1 ≤ t ≤ 2 ms × 1 10 100 ×10−12 × 0 = 0, t > 2 ms 익힘문제
7.3>
v (50 µ s) = 1.5 cos (10
5
w= 익힘문제
1 2
CV 2
× 50 ×10−6 ) V = 0.4255 V
= 0.5 × 1000 ×10−6 × 0.42552 = 90.53 µ
J
7.4>
2 × 10−3 0.2 × 1× 10−3 = 0.4 V, di 4−2 v L = L = 0.2 × = 0.2 V, dt 3 −1 0.2 × 0 − 4 = −0.2667 V, 6 −3
t < 1 ms 1 ≤ t ≤ 3 ms t > 3 ms
Thus, (a) (b) (c)
v L = 0.4 V @ t = t = 0 v L = 0.2 V @ t = t = 2 ms v L = − 0.2667 V @ t = t = 6 ms
익힘문제 7.6>
(a)
v=L
di dt
= 25 × 10−3 (10e−100t − 1000t e−100 t )
so v (12 ms) = − 15.06 mV
(b)
i (0.1) =
1 L
0.1
∫ 0
6e
−12 t
0.1
−1 e−12t + 10 dt + 10 = 0.025 12 0 6
= 23.98 A
(c)
(d)
p(t ) = Li
di
= 0.025 8(1 − e−40t ) ×10−3 8 ×10−3 (40) e−40 t
dt so p(50 ms) = 7.489 µ W w L (t ) =
1
(0.025) 8 ×10−3 (1 − e −40t )
2
2 so w L (40 ms) = 509.6 nJ
익힘문제 7.7>
Viewing the circuit right to left, we see that C eq
= { [ (1// 5 // 2) + 12] // 0.4 // 0.8} + (7 // 5) = { [ 0.5882 + 12] // 0.2667} + 2.917 = 3.178 µ F
익힘문제 7.8>
•
define a clockwise current i(t) KVL yields
−vS + 0.002 vC =
1
di dt
+ vC = 0
t
[1]
i dt ′ = 4 cos10 t C ∫ 5
−∞
V
[2]
− − Thus, i (t ) = −4(80 ×10 ) (10 ) sin10 t = −32 × 10 sin10 t di = −32 × 10−3 (105 ) cos105 t = −3200 cos105 t and dt 9
5
5
Substituting this into Eq. [1], v s
익힘문제
= −0.002 × 3200 cos105 t + 4 cos105 t = − 2.4 cos 105 t V
7.10>
3
5
•
define iC flowing downward through the capacitor 3 10 8 × 10− e −
6
t
=
where iC = 0.2 × 10
v
10 −6 dv
+ iC
[1]
dt
Thus, Eq. [1] becomes 3 10 8 ×10− e −
6
t
= 0.1v + 0.2 ×10−6
dv
[1]
dt
6
Substituting v = −80 × 10−3 e−10 t , 6
(0.1) ( −80 ×10−3 e−10 t ) + (0.2 ×10−6 ) ( −80 ×10 −3 ) ( −10 6) e−10
6
t
= (−8 × 10−3 + 16 × 10−3 ) e−10 t = 8 × 10−3 e−10 t 6
6
which verifies that v = −80e−10 t mV is indeed a solution to Eq. [1]. 6
(a)
Invoking duality, v1 = 0.1v = −8e−10 t mV
(b)
by KVL, v2
(c)
by duality, i = −80e −10 t mA
6
= 8 ×10−3 e−10 t − v1 = 16e−10 t mV 6
6
6
익힘문제
8.3>
i (t ) = I o e − t /τ
(a)
(b)
(c) (d)
i (2τ ) i (τ )
and
=
i (0.5τ ) i (0) i (t ) i (0)
e −2
i (0) = I o
= 0.3679
e −1
= e−0.5 = 0.6065
= e−t / = 0.2, so
if i (0)
t
τ
−
i (t )
=
τ
= −l n 0.2 = 1.609
i(0) l n 2
Then i (t ) = i(0) − i(0) l n 2 = i(0) [1 − l n 2 ] Thus, 1 − l n 2 = e
− t / τ
and so
t τ
= i(0) e−t /
τ
= −l n (1− l n 2) = 1.181
익힘문제
8.4>
Before the switch is thrown, the 800-Ω resistor is connected only by one of its terminals and therefore may be ignored. (i = 0) With no current flow permitted through the capacitor (it is assumed any transients have long since died out), v = 50 V. After the switch is thrown, the only remaining circuit is a simple source-free RC circuit. We know v(0) = 50 since the capacitor voltage cannot change in zero time. With τ RC 1.6 ms , =
=
v (t) = v(0) e
−t /τ
so
−2 ×10−3 v(2ms) = 50 exp −3 1.6 ×10 = 14.33 V
익힘문제 8.5> − − For t < 0: i2 (0 ) = 0, i1 (0 ) = 2 ×
2
2+8 − and i L (0 ) = 2 − 0.4 = 1.6 A
= 0.4 A,
For t > 0, 100% of the 2-A source contributes to i2. The 8− Ω resistor is shorted out so i1 = 0. Thus, i2 (t ) = 2 − iL (t ) where i L (t ) = iL (0 + ) e− t /τ i L (0+ ) = iL (0 − ) so and 익힘문제
= 1.6 A
and τ =
L R
=
0.4 2
= 0.2 s
−0.15 = 1.6 exp = 755.6 mA 0.2 i2 ( 0.15 ) = 1.244 A i L ( 0.15 )
8.6>
− For t < 0, vC (0 ) = 120 ×
1250 1250 + 250
v (0− ) = 120 ×
= 100 V
× 400 = 38.4 V 600 + 2000 // 500 400 + 100 2000 // 500
−
At t = 0+, the capacitor voltage cannot differ from its value at t = 0 . Thus, vC (0+) = 100 V and with the source removed,
2000 // 500 × 400 = 25.6 V 850 + 2000 // 500 100 + 400
v (0+ ) = vC (0 + ) × τ
[1]
= R eq ⋅ C = [ 1250 //(850 + 2000 // 500) ] × 4 ×10−6 = 2.5 ms
vC (1.3 ms) = 100e−1.3/2.5
= 59.45 V
Replacing vC (0+) in Eq. [1] with vC (1.3 ms) yields v (1.3 ms) =
59.45 100
× 25.6 = 15.22 V
익힘문제
8.8>
(a)
3 − 0 + 0.8 = 3.8
(b)
[4] (0) = 0
(c)
2 sin 0.8 π = 1.176
익힘문제 8.9>
60
(a)
− − t = 0 , so only 60 V is across the RL circuit. Thus v L (0 )
(b)
At t = 0+, the source voltage changes to 60 − 40 = 20 V. The inductor current cannot + change, so i L (0 )
= 6 A.
= 0 and
For t
> 0,
= 0.
i L (t )
where i L (t )
=
=2A
The direction of i L has not changed. iL (∞) + iL (0 + ) − iL ( ∞) e − t / τ and
10
=6A
= 40 V
At t = ∞ , the source voltage is 20 V but all transients have died out. Thus, A and v L (∞)
(d)
=
The current through the resistor is 6 A, so the voltage dropped
+ across the inductor is 20 − 10 (6) = − 40 V. Thus, v L (0 )
(c)
i L (0− )
τ
= L R =
50 × 10−3 10
= 5 ms .
−3/ 5 Thus, i L (3 ms) = 2 + (6 − 2)e = 4.195 A
We then find that v L (3 ms)
=
20 − 4.195 (10)
= 21.95 V
i L (∞)
=
20 10
=
2
익힘문제 8.10>
(a)
vS (0− ) = 0 so iL (0 − ) = 0
(b)
i L (0+ ) = iL (0 − ) so iL (0+ ) = 0 vS (t ) = 200iL (t ) + vL ( t)
or 20e −100 t = 200i L i L (t ) = i f
+4
diL
[1]
dt
+ in
in (t ) = Ae− t /τ where τ =
L R
=
4 200
= 20 ms
−100t −50 t If we assume i f (t ) = B e then i L (t ) = Ae
so
di L dt
+ B e−100t
= −50 Ae−50t − 100 B e−100t
Substituting back in Eq. [1], 200i L
+4
di L dt
= 200 A e−50t + 200 B e−100t − 200 Ae−50 t − 400 B e−100 t = −200 B e−100 t
Thus, referring to Eq. [1], 20 = − 200 B so B = −50 t Thus, i L (t ) = Ae + since i L (0 )
−20 200
= −0.1
− 0.1 e−100t
= A − 0.1 = 0,
(c)
i L (8 ms) = 22.10 mA
(d)
i L (15 ms) = 24.92 mA
A = 0.1 and
i L (t )
=
0.1(e
50 t
−
−
e
100 t
−
) amperes
익힘문제 8.11> −
(a)
i R (0 ) = 0
(b)
i L (0 ) = 0 so i L (0+) = 0
−
Thus, all of the source current is shunted through the 60− Ω resistor; hence, i R (0+) = 10 mA (c)
i R (∞) = 10 ×
(d)
τ
=
L
=
R eq
40 40 + 60 0.1
40 + 60
= 4 mA
= 1 ms
i R (t ) = iR (∞) + iR (0 + ) − i( ∞) e− t /τ 3
= 4 + [ 10 − 4] e−10 t mA so i R (1.5 ms) = 5.339 mA 익힘문제
(a)
8.12>
−
At t = 0 , only the current source is on, so vC (0− ) = 1× [ 25 //(20 + 80) ]
= 20 V
(b)
vC (0 + ) = vC (0 − ), so vC (0 + ) = 20 V
(c)
At t = ∞ , both sources are on, so vC (∞) = 1× [ 25 //(20 + 80) ] + 10 ×
(100) 125
= 20 + 8 = 28 V (d)
vC (t ) = vC (∞) + vC (0 + ) − vC ( ∞) e− t /τ where τ = R eq C R eq = 25//100 = 20 k Ω , so τ = 100 ms
−80/100 Thus, vC (80 ms) = 28 + [ 20 − 28] e
= 24.41 V
