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Chapter 7, Problem 1.
In the circuit shown in Fig. 7.
v ( ) t = 56 e −^200 t V, t > 0
i ( ) t = 8 e −^200 t mA, t > 0
(a) Find the values of R and C.
(b) Calculate the time constant τ.
(c) Determine the time required for the voltage to decay half its initial value at
t = 0.
Figure 7.
For Prob. 7.
Chapter 7, Solution 1.
(a) τ=RC = 1/
For the resistor, V=iR=
56 8Re 10 7 k 8
t t e x R
− − − = ⎯⎯→ = = Ω
3
C F
R X X
(b) τ =1/200= 5 ms (c) If value of the voltage at = 0 is 56.
− = ⎯⎯→ =
t t x e e
200 ln2 ln2 3.466 ms 200
t o to
Chapter 7, Problem 2.
Find the time constant for the RC circuit in Fig. 7.82.
Figure 7.
For Prob. 7.2.
Chapter 7, Solution 2.
τ=R (^) th C
where R (^) th is the Thevenin equivalent at the capacitor terminals.
R (^) th= 120 || 80 + 12 = 60 Ω
τ = × × =
60 0. 5 10 30 ms
Chapter 7, Problem 3.
Determine the time constant for the circuit in Fig. 7.83.
Figure 7.
For Prob. 7.3.
Chapter 7, Solution 3.
R = 10 +20//(20+30) =10 + 40x50/(40 + 50)=32.22 kΩ
3 12
τ RC 32.22 X 10 X 100 X 10 3.222 μS
− = = =
Figure 7.
For Prob. 7.5.
Chapter 7, Solution 5.
Let v be the voltage across the capacitor.
For t <0,
4 (0 ) (24) 16 V 2 4
v
− = =
For t >0, we have a source-free RC circuit as shown below.
i
4 Ω
v 1/3 F
τ = RC = + = s
/ / 3 ( ) (0) 16
t t v t v e e
− τ − = =
(^1 1) / 3 / 3 ( ) ( ) 16 1.778 A 3 3
dv t t i t C e e dt
− − = − = − − =
Figure 7.
For Prob. 7.6.
Chapter 7, Solution 6.
v(t) 4 e V
v(t) v e , RC 40 x 10 x 2 x 10
( 24 ) 4 V
v v( 0 )
- 5 t
t/ 6 3 o
o
−
− τ −
= τ= = =
Figure 7.
For Prob. 7.8.
Chapter 7, Solution 8.
(a) 4
τ=RC =
dt
dv
- i=C
- 0.2e = C( 10 )(-4)e ⎯⎯→ C =
-4t -4t 5 mF
4 C
R 50 Ω
(b) τ = = = 4
RC 0. 25 s
(c) = = ( 5 × 10 )( 100 )= 2
CV
w (^) C ( 0 ) 02 -^3 250 mJ
(d) (^ )
τ = × = −^0 2 - 2t 0
2 R 0 CV^1 e 2
CV
w
- 5 1 e^0 e^0
or e 0 2
8t
= ln( 2 ) = 8
t 0 86. 6 ms
Figure 7.
For Prob. 7.9.
Chapter 7, Solution 9.
For t < 0, the switch is closed so that
4 (0) (6) 4 V 2 4
v o = =
For t >0, we have a source-free RC circuit. 3 3
τ RC 3 10 x x 4 10 x 12 s
− = = =
/ / ( ) (0) 4 V
t t vo t vo e e
− τ − = =
Chapter 7, Problem 11.
For the circuit in Fig. 7.91, find i (^) 0 for t > 0.
Figure 7.
For Prob. 7.11.
Chapter 7, Solution 11.
For t<0, we have the circuit shown below.
3 Ω 4H
24 V 8 Ω
4H
i (^) o
8 A 3 Ω 8 Ω
3//4= 4x3/7=1.
i o
− = =
A
For t >0, we have a source-free RL circuit. 4 1/ 3 4 8
L
R
/ 3 ( ) (0) 1.4118 A
t t io t io e e
− τ − = =
Chapter 7, Problem 12.
The switch in the circuit of Fig. 7.92 has been closed for a long time. At t = 0 the switch
is opened. Calculate i ( t ) for t > 0.
Figure 7.
For Prob. 7.12.
Chapter 7, Solution 12.
When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω
resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).
4 A
i( 0 )= =
−
Since the current through an inductor cannot change abruptly,
i( 0 )= i( 0 )=i( 0 )= 4 A
− +
When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
R
L
τ= = =
Hence,
= =
-2t
(b)
12 V
−
(a)
i(
- )
