1. Electric Circuits Assessment Problems
Chapter 7
Circuit Elements

2. Ans: (a)RT=3+(6//30)=8 IT= 120 8 =15A 𝑖=15 30 30+6 =-12.5(方向相反) # (b)
7.1 The switch in the circuit shown has been closed for a long time and is opened at t = 0.
a) Calculate the initial value of 𝑖.
b) Calculate the initial energy stored in the inductor.
c) What is the time constant of the circuit for t > 0?
d) What is the numerical expression for 𝑖(t) for t > 0?
e) What percentage of the initial energy stored has been dissipated in the 2 Ω resistor 5 ms after the switch has been opened?
Ans:
(a)RT=3+(6//30)=8
IT= =15A
𝑖= =-12.5(方向相反) #
(b)
W= m (-12.5)2 = 625mJ #
(c) τ= 8𝑚 2 = 4ms #

3. (d)
𝑖(t)= 𝑖(0)e- 𝑡 τ =-12.5e- 𝑡 4m =-12.5 e-250t #
(e)
𝑖(5m) = e-250(5m) =-3.58A
W= m (-3.58)2 = 51.3mJ
WR= 625m-51.3m = 573.7mJ
WR WL 100% = 573.7𝑚 625𝑚 100% =91.8% #

4. 7.2 At t = 0, the switch in the circuit shown moves instantaneously from position a to position b.
a) Calculate vo for t ≥ 0+.
b) What percentage of the initial energy stored in the inductor is eventually dissipated in the 4 Ω, resistor?
Ans: (a) 𝑖(0-)= = 4A RT=(6+10)//4=3.2 Ω τ= =0.1s 𝑖(t)=4e- 𝑡 0.1 =4e-10t 𝑖 1 = 4e-10t ( ) =0.8e-10t ; Vo = 0.8e-10t (10) = -8e-10t (極性相反) #
𝑖 1
𝑖 2

5. (b) 𝑖 2 = 4e-10t ( 16 16+4 ) =3. 2e-10t PR=(3. 2e-10t)2 (4) =40
(b) 𝑖 2 = 4e-10t ( ) =3.2e-10t PR=(3.2e-10t)2 (4) =40.96e-20t WR= ∫0 PR dt = ∫ e-20t dt = 2.048J WL = = 2.56J WR WL 100%= % = 80% # ∞ ∞

6. a) the initial value of v(t), b) the time constant for t > 0,
7.3 The switch in the circuit shown has been closed for a long time and is opened at t = 0. Find
a) the initial value of v(t),
b) the time constant for t > 0,
c) the numerical expression for v(t) after the switch has been opened,
d) the initial energy stored in the capacitor, and
e) the length of time required to dissipate 75% of the initially stored energy.
Ans:
(a)
𝑖 1= = 4A
𝑣 (t1) = 4(50) = 200V#
(b)
τ=RC= 0.4μ (50k) =20ms #
𝑖 1

7. (c) 𝑣 (t)= 𝑣 (t1)e- 𝑡 τ =200e- 𝑡 20𝑚 =200e-50t ;t≥0 # (d)W= 1 2 CV2= μ 2002 =8mJ # (e) 8m =6mJ (電阻消耗) WL=8m-6m=2mJ 2m= μ (200e-50t) 2 2m=8me-100t 1 4 =e-100t ln 1 4 =lne-100t t=13.86ms #

8. 7.4 The switch in the circuit shown has been closed for a long time before being opened at t = 0.
a) Find 𝑣o(t) for t > 0.
b) What percentage of the initial energy stored in the circuit has been dissipated
after the switch has been open for 60 ms?
Ans:
(a)
𝑖 T = 𝑘 =0.2mA
V20k=0.2m (20k) = 4V
V40k=0.2m (40k) = 8V
τ1=RC=20k(5μ) =100ms
τ2=RC=40k(1μ) =40ms
Vo1(t) =4e- 𝑡 100m =4e-10t
Vo2(t) =8e- 𝑡 40m =8e-25t
Vo(t)= Vo1(t) + Vo2(t) = 4e-10t + 8e-25t V; t≥0 #

9. (b) Wc1 = 1 2 5μ 42 =40μJ Wc2 = 1 2 1μ 82 =32μJ WT=40 μ +32 μ = 72 μJ
WC1(60ms) = μ (4e-10(60m))2 = μJ
WC2(60ms) = μ (8e-25(60m))2 =1.59 μJ
WCT = μ μ = μJ
Wdiss = 72 μ μ = μJ
Wdiss 𝑊𝑇 100%= % =81.05% #

10. 7. 5 Assume that the switch in the circuit shown in Fig. 7
7.5 Assume that the switch in the circuit shown in Fig has been in position b for a long time, and at t = 0 it moves to position a. Find (a) 𝑖(0+); (b) 𝑣(0+); (c) T,t ≥ 0; (d) 𝑖(t), t ≥ 0; and (e) 𝑣(t), t ≥ 0+.
Ans:
(a)
𝑖(0+)= 24 2 =12A #
(b)
𝑣(0+)=20(10)=-200(極性相反) #
(c)
T= 𝐿 𝑅 = 200𝑚 10 =20ms #
(d)
𝑖(t)=(12-(-8)) e-50t +(-8) =20e-50t-8 A#
(e)
𝑣(t)= L 𝑑𝑖 𝑑𝑡 = 200m (20e-50t-8)dt=-200e-50t V#

11. Ans: (a) 𝑉𝐴−𝑉𝑜 8 + 𝑉𝐴 160 + 𝑉𝐴+75 40 =0 (b) t≥0# 8kΩ 40kΩ + VA 75V
7.6
(a)Find the expression for the voltage across the 160 k Ω resistor in the circuit shown in Fig Let this voltage be denoted VA, and assume that the reference polarity for the voltage is positive at the upper terminal of the 160 k Ω resistor.
(b) Specify the interval of time for which the expression obtained in (a) is valid.
Ans:
(a)
Vo=90e-100t-60
𝑉𝐴−𝑉𝑜 8 + 𝑉𝐴 𝑉𝐴 =0
25VA-20Vo+300=0
25VA-20(90e-100t-60)+300=0
VA=72e-100t - 60 V #
(b)
t≥0#
8kΩ
40kΩ + VA
75V
160kΩ
0.25μF -