circuits, pleaseeeee solve questionnn7 The switch in Fig. 1 has been closed for long time. It opens at t=0. Please refer to the circuit ofFig. 1 for the following questions (Q1, and Q2)Q1) The time constant t can be found as:a) 6.67 s b) 0.3 sc) 10 sd) 0.1 sQ2) The current i(t) at t= 1m s is:a) 2.02 A b) 6 Ac) 4.02 Aa) 1.23 cos(10t +30°) Vd) 2.25 cos(10t-53.6%) Ve) 0.15 sd) 5.98 A e) 4 A2cos101 Vb) 1.23 cos(10t-30°) Ve) 1.79 cos(10t -26.57°) V20 400 40wwwHERefer 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 ofvi(t) is:Q5) The value of the inductance of the j2 2 impedance is:a) 0.2 H b) 10 Hc) 20 H d)1.6 He) 16 HQ7) 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°)paQ6) 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 22Q4) 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°)59235 cos 10rvb) 12.5cos(500t+ 0.107°)e) 12.5 cos(500t+ 0.205°)20:2-wwc) 2.25 cos(10t +53.6°) Vf) 1.79 cos(10t+26.57°) V10.0wwT-2 -15.02341=01.5 HmQ9) The complex power absorbed by voltage source is (VA)b)-751.3-j457. c)-823.5+j294.1a) -823.5-j294.1d) -751.3+j457.7e) 751.3-j457.7Fig. 1102 cos5001 VFig. 2Fig. 3Refer 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) 202-53.26f) 20253.26c) 12.5cos(500t - 89.9°)f) 12.5 cos(500t - 0.205°)10Ω (1) 6A100/0° V ms10042www0.2 iFig. 4–ΠΟΥFig. 53120020.3mH200ww100

Answered: Image /qna-images/answer/4fcfc15a-3797-41b0-876d-c1f427699862.jpg

a) 22 +j6Ω b)18+j6Ω c) 22-j6 (2 d) -J6 12 Q7) The current i(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) 2cos101 V 20 400 40 www j20 b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 5-ods 10rV 20 (2 PWY Z 10-02 78502 Fig. 3 10.02 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10042 cos5001 y 0.2 i Fig. 4 İL 0.3 mH

a) 22 +j6Ω b)18+j6Ω c) 22-j6 (2 d) -J6 12 Q7) The current i(t) of Fig. 4 can be found as (mA): a) 12.5cos(500t - 0.107°) d) 12.5 cos(500t + 89.9°) 2cos101 V 20 400 40 www j20 b) 12.5cos(500t+ 0.107°) e) 12.5 cos(500t+ 0.205°) 5-ods 10rV 20 (2 PWY Z 10-02 78502 Fig. 3 10.02 c) 12.5cos(500t - 89.9°) f) 12.5 cos(500t - 0.205°) 10042 cos5001 y 0.2 i Fig. 4 İL 0.3 mH

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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