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Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The cuida ni 10 Q www 0.5 H ell C= 10 mF

Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The cuida ni 10 Q www 0.5 H ell C= 10 mF

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 questionnn 19

2A (4) le 4.8 am Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s d) 2 rad/s e) 5 rad/s Q13) Resonant frequency wo is a) 4 rad/s b) 5 rad/s c) 4 rad/s d) 16 rad/s e) 25 rad/s Q14) The voltage v (t=1s) can be found as a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V e) 12.61 V a) R=0.22, C=0.2 F d) R=192, C=1F m 1.92 (2 34 149 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as a) 5W b) 24 W c) 96 W d) 192 W c) 48W Q11) The Va for the the circuit shown in Fig. 7 as seen from the terminal a-b can be found as: a)-/220 V d)-110 V )-165 V b)-/330 V c)-j55 V Please refer to the circuit of Fig. 8 for the following questions (Q12, Q13 and Q14). Assuming it(0)-8 A, and V.(0)-40 V. te th L m L T 1.61, 80 30 34 L m 1520 A Z Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L d) 5/8 L e) 4/7 L b) 4/9 L c) 7/4 L Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: SH 6 MA 24 V + L 3k0z www 492 -NN 202 ΣΕΩΣ 1022 t=0 3kQ Fig. 8 2mF HF 210 ww 330 455 + Fig. 10 SkΩ Σ 4 m² Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ d) 8 mJ e) 16 mJ b) 128 mJ c) 256 mJ Q17) If v(t)=15 cos(1000t+66°) V and i(t)-2cos(1000t+450°) A, then v(t) leads i(t) by a) 156° b) 0-24° c) 204° d) 24° e) 66° Q18) Assuming that the input impedance is given as Zin= 1+j1 22 and o-1 rad/s, then the input admittance can be represented as the parallel combination of: b) R=0.2 02, L=0.2 H e) R=202, L=2H c) R=10, L=1H 100 0.5H C= 10 mF Fig. 12 a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The switch in Fig. 12 has been in position A for long time. At t=0, the switch moves to position B. Please refer to the circuit of Fig. 12 for the following questions (Q20, Q21 and Q22) Fig. 11 Q20) v(0) can be found as: d) 15 V e) 24 V a)30 V b) 12.5 V c) 9V Q21) v(co) can be found as: a) 24 V b)30 V c) 12.5 V d) 9V Q22) The voltage v(t) at t = 1 s is: b)24.9 V c) 30 V e) 15 V a) 20.9 V d) 39.1 V e) 27.97 V B 04 f 12.5m. 9.51, 4kQ www P0.5 mF b 410 30 V
Transcribed Image Text:2A (4) le 4.8 am Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s d) 2 rad/s e) 5 rad/s Q13) Resonant frequency wo is a) 4 rad/s b) 5 rad/s c) 4 rad/s d) 16 rad/s e) 25 rad/s Q14) The voltage v (t=1s) can be found as a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V e) 12.61 V a) R=0.22, C=0.2 F d) R=192, C=1F m 1.92 (2 34 149 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as a) 5W b) 24 W c) 96 W d) 192 W c) 48W Q11) The Va for the the circuit shown in Fig. 7 as seen from the terminal a-b can be found as: a)-/220 V d)-110 V )-165 V b)-/330 V c)-j55 V Please refer to the circuit of Fig. 8 for the following questions (Q12, Q13 and Q14). Assuming it(0)-8 A, and V.(0)-40 V. te th L m L T 1.61, 80 30 34 L m 1520 A Z Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L d) 5/8 L e) 4/7 L b) 4/9 L c) 7/4 L Q19) For the circuit shown in Fig. 11, the value of C needed to make the response underdamped with unity damping factor (a = 1) is: SH 6 MA 24 V + L 3k0z www 492 -NN 202 ΣΕΩΣ 1022 t=0 3kQ Fig. 8 2mF HF 210 ww 330 455 + Fig. 10 SkΩ Σ 4 m² Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ d) 8 mJ e) 16 mJ b) 128 mJ c) 256 mJ Q17) If v(t)=15 cos(1000t+66°) V and i(t)-2cos(1000t+450°) A, then v(t) leads i(t) by a) 156° b) 0-24° c) 204° d) 24° e) 66° Q18) Assuming that the input impedance is given as Zin= 1+j1 22 and o-1 rad/s, then the input admittance can be represented as the parallel combination of: b) R=0.2 02, L=0.2 H e) R=202, L=2H c) R=10, L=1H 100 0.5H C= 10 mF Fig. 12 a) 40 mF b) 15 mF c) 26 mF d) 2.5 mF e) 7.5 mF The switch in Fig. 12 has been in position A for long time. At t=0, the switch moves to position B. Please refer to the circuit of Fig. 12 for the following questions (Q20, Q21 and Q22) Fig. 11 Q20) v(0) can be found as: d) 15 V e) 24 V a)30 V b) 12.5 V c) 9V Q21) v(co) can be found as: a) 24 V b)30 V c) 12.5 V d) 9V Q22) The voltage v(t) at t = 1 s is: b)24.9 V c) 30 V e) 15 V a) 20.9 V d) 39.1 V e) 27.97 V B 04 f 12.5m. 9.51, 4kQ www P0.5 mF b 410 30 V
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
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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