The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit of Fig. 1 for the following questions (Q1, and Q2) Q1) The time constant t can be found as: a) 6.67 s b) 0.3 s c) 10 s Q2) The current i(t) at t-Ims is: a) 2.02 A b) 6 A c) 4.02 A a) 1.23 cos(10t +30°) V d) 2.25 cos(10t-53.6%) V d) 0.1 s e) 0.15 s d) 5.98 A e) 4 A Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5) Fig. 1 Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value of vi(t) is: 2cos101 V b) 1.23 cos(10t-30%) V e) 1.79 cos(10t-26.57°) V Q5) The value of the inductance of the j2 02 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H 2002 -4.0 www HE pa Q4) By applying KCL to the node v/(t), the value of the voltage labeled vi(t) is (V): a) 2.86 cos(10t +77.9°) b) 2.86 cos(10t-77.9°) d) 4.1 cos(10t-62.3°) c) 4.1 cos(10t +62.3°) f) 3.92 cos(10t -77.9°) e) 3.92 cos(10t +77.9°) Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a) 22+j6 2 b) 18 +j6 02 c) 22-j6 2 Q7) The current in(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) 552 5-ods 10rV c) 2.25 cos(10t +53.6%) V 1.79 cos(10t+26.57°) V d) 18-j62 e)-18 +j6 2 b) 12.5cos(500t+0.107°) e) 12.5 cos(500t + 0.205°) 20 (2 10.02 150 1=0 Q9) The complex power absorbed by voltage source is (VA) b)-751.3-j457. c)-823.5+j294.1 a) -823.5-j294.1 d)-751.3+j457.7 e) 751.3-j457.7 1.5 H m tacos5001 V Fig. 2 Fig. 3 Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9) Q8) The current through the-j10 2 can be found as (Arms) a) 8.75/19.65* b) 8.752-19.65* c) 10.25290* d) 10.25Z-90* e) 202-53.26* f) 20253.26* c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 100/2 100/0° V ms 1052 0.21 Fig. 4 46 Fig. 5 200 m 2002 0.3 mH www 102
The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit of Fig. 1 for the following questions (Q1, and Q2) Q1) The time constant t can be found as: a) 6.67 s b) 0.3 s c) 10 s Q2) The current i(t) at t-Ims is: a) 2.02 A b) 6 A c) 4.02 A a) 1.23 cos(10t +30°) V d) 2.25 cos(10t-53.6%) V d) 0.1 s e) 0.15 s d) 5.98 A e) 4 A Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5) Fig. 1 Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value of vi(t) is: 2cos101 V b) 1.23 cos(10t-30%) V e) 1.79 cos(10t-26.57°) V Q5) The value of the inductance of the j2 02 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H 2002 -4.0 www HE pa Q4) By applying KCL to the node v/(t), the value of the voltage labeled vi(t) is (V): a) 2.86 cos(10t +77.9°) b) 2.86 cos(10t-77.9°) d) 4.1 cos(10t-62.3°) c) 4.1 cos(10t +62.3°) f) 3.92 cos(10t -77.9°) e) 3.92 cos(10t +77.9°) Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a) 22+j6 2 b) 18 +j6 02 c) 22-j6 2 Q7) The current in(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) 552 5-ods 10rV c) 2.25 cos(10t +53.6%) V 1.79 cos(10t+26.57°) V d) 18-j62 e)-18 +j6 2 b) 12.5cos(500t+0.107°) e) 12.5 cos(500t + 0.205°) 20 (2 10.02 150 1=0 Q9) The complex power absorbed by voltage source is (VA) b)-751.3-j457. c)-823.5+j294.1 a) -823.5-j294.1 d)-751.3+j457.7 e) 751.3-j457.7 1.5 H m tacos5001 V Fig. 2 Fig. 3 Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9) Q8) The current through the-j10 2 can be found as (Arms) a) 8.75/19.65* b) 8.752-19.65* c) 10.25290* d) 10.25Z-90* e) 202-53.26* f) 20253.26* c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 100/2 100/0° V ms 1052 0.21 Fig. 4 46 Fig. 5 200 m 2002 0.3 mH www 102
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Question
circuits, pleaseeeee solve questionnn 9
Transcribed Image Text:The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit of
Fig. 1 for the following questions (Q1, and Q2)
Q1) The time constant t can be found as:
a) 6.67 s b) 0.3 s
c) 10 s
Q2) The current i(t) at t- Im s is:
a) 2.02 A b) 6 A
c) 4.02 A
a) 1.23 cos(10t +30°) V
d) 2.25 cos(10t-53.6%) V
d) 0.1 s e) 0.15 s
Fig. 1
Refer to the circuit of Fig. 2 for the following 3 questions (Q3, Q4 and Q5)
Q3) By using superposition technique, the contribution of the 2cos10t voltage source to the value of
vi(t) is:
a) 12.5cos(500t - 0.107°)
d) 12.5 cos(500t + 89.9°)
d) 5.98 A e) 4 A
Q5) The value of the inductance of the j2 02 impedance is:
a) 0.2 H b) 10 H
c) 20 H d)1.6 H
e) 16 H
2cos101 V
Q7) The current in(t) of Fig. 4 can be found as (mA):
2002 -4.0
www
HE
b) 1.23 cos(10t-30%) V
e) 1.79 cos(10t-26.57°) V
Q4) By applying KCL to the node v/(t), the value of the voltage labeled vi(t) is (V):
a) 2.86 cos(10t +77.9°)
b) 2.86 cos(10t-77.9°)
c) 4.1 cos(10t +62.3°)
d) 4.1 cos(10t-62.3°)
e) 3.92 cos(10t +77.9°)
f) 3.92 cos(10t -77.9%)
Q6) Referring to the circuit of Fig. 3, Zen can be determined as:
a) 22+j6 2 b) 18+j62 c) 22-j62 d) 18-j62 e) -18+j6
pa
a)-823.5-j294.1
d)-751.3+j457.7
552
5 cos 10rV
b) 12.5cos(500t+0.107°)
e) 12.5 cos(500t + 0.205°)
10.02
20:2
c) 2.25 cos(10t +53.6%) V
1.79 cos(10t+26.57°) V
1502
1=0
Q9) The complex power absorbed by voltage source is (VA)
b)-751.3-j457. c)-823.5+j294.1
e) 751.3-j457.7
Fig. 2
Fig. 3
Refer to the circuit of Fig. 5 for the following 2 questions (Q8, and Q9)
Q8) The current through the-j10 2 can be found as (Arms)
a) 8.75/19.65*
b) 8.752-19.65*
c) 10.25290*
d) 10.25Z-90°
e) 202-53.26*
f) 20253.26
1.5 H
m
tacos5001 V
c) 12.5cos(500t - 89.9°)
f) 12.5 cos(500t - 0.205°)
100/2
100/0° V ms
1052
0.21
Fig. 4
46
Fig. 5
0.3 mH
720 (2
m
2002
www
10
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