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Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(t) is (V): a) 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°) b) 2.86 cos(10t-77.9°) e) 3.92 cos(10t +77.9°)

Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(t) is (V): a) 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°) b) 2.86 cos(10t-77.9°) e) 3.92 cos(10t +77.9°)

Introductory Circuit Analysis (13th Edition)
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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circuits, pleaseeeee solve questionnn4
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 d) 0.1 s Q2) The current i(t) at t= 1m 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 e) 0.15 s d) 5.98 A e) 4 A 2cos101 V b) 1.23 cos(10t-30°) V e) 1.79 cos(10t -26.57°) V 20 400 40 www HE 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: Q5) The value of the inductance of the j2 2 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H 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°) pa Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a)22+j6Ω b)18+j6Ω c) 22-j6 Ω d) 18-j62 e)-18+j6 22 Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(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°) 5923 5 cos 10rv b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 20:2 -ww c) 2.25 cos(10t +53.6°) V f) 1.79 cos(10t+26.57°) V 10.0 ww 7 1.5 H m 2 -15.02 - 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 Fig. 1 102 cos5001 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 (A rms) a) 8.75/19.65* b) 8.75-19.65* c) 10.25/90* d) 10.25Z-90° e) 202-53.26 f) 20253.26 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10Ω (1) 6A 100/0° V ms 10042 www 0.2 i Fig. 4 –ΠΟΥ Fig. 5 11 2002 0.3mH 200 ww 1002
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 d) 0.1 s Q2) The current i(t) at t= 1m 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 e) 0.15 s d) 5.98 A e) 4 A 2cos101 V b) 1.23 cos(10t-30°) V e) 1.79 cos(10t -26.57°) V 20 400 40 www HE 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: Q5) The value of the inductance of the j2 2 impedance is: a) 0.2 H b) 10 H c) 20 H d)1.6 H e) 16 H 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°) pa Q6) Referring to the circuit of Fig. 3, Zin can be determined as: a)22+j6Ω b)18+j6Ω c) 22-j6 Ω d) 18-j62 e)-18+j6 22 Q4) By applying KCL to the node v/(t), the value of the voltage labeled v/(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°) 5923 5 cos 10rv b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 20:2 -ww c) 2.25 cos(10t +53.6°) V f) 1.79 cos(10t+26.57°) V 10.0 ww 7 1.5 H m 2 -15.02 - 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 Fig. 1 102 cos5001 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 (A rms) a) 8.75/19.65* b) 8.75-19.65* c) 10.25/90* d) 10.25Z-90° e) 202-53.26 f) 20253.26 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10Ω (1) 6A 100/0° V ms 10042 www 0.2 i Fig. 4 –ΠΟΥ Fig. 5 11 2002 0.3mH 200 ww 1002
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