2A (4) 4.8 m 1.92 0 Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s mm 1.61, -1452 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as a) 5W c) 96 W d) 192 W b) 24 W e) 48W d) 16 rad/s Q13) Resonant frequency , is a) 4 rad/s b) 5 rad/s c) 4 rad/s Q14) The voltage v (t=1s) can be a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V found as we Q11) The Vah 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 e)-/165 V b)-/330 V c)-j55 V 80 d) 2 rad/s Q20) v(0) can be found as: a)30 V b) 12.5 V c) 9V 021 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. L me D L m 34 e) 5 rad/s e) 25 rad/s 15/02 A e) 12.61 V Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L c) 7/4 L d) 5/8 L e) 4/7 L b) 4/9 L b) R=0.202, L=0.2 H e) R=222, L=2 H 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 6 mA 3H 10Q 492 ww 3kQ SANAY 20 ww 8023 0.5 H BG Fig. 8 2mF 2102 www 3kQ 5KQ Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ b) 128 mJ c) 256 mJ d) 8 mJ e) 16 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)-24° c) 204° d) 24⁰ e) 66° Q18) Assuming that the input impedance is given as Zin = 1+j1 2 and 0-1 rad/s, then the input admittance can be represented as the parallel combination of: a) R=0.22, C=0.2 F d) R=102, C=1F c) R=10, L=1H 4 mF Fig. 10 C 12.5 mF www. 9.51, 10 mF EQ Fig. 11 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) d) 15 V e) 24 V A B 4k0
2A (4) 4.8 m 1.92 0 Q12) Neper frequency a is a) 4 rad/s b) 16 rad/s c) 25 rad/s mm 1.61, -1452 Fig. 6 Fig.7 Q10) The average power supplied by the dependent source of Fig. 6 can be determined as a) 5W c) 96 W d) 192 W b) 24 W e) 48W d) 16 rad/s Q13) Resonant frequency , is a) 4 rad/s b) 5 rad/s c) 4 rad/s Q14) The voltage v (t=1s) can be a) 10.61 V b) 14.61 V c) 16.61 V d) 8.61 V found as we Q11) The Vah 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 e)-/165 V b)-/330 V c)-j55 V 80 d) 2 rad/s Q20) v(0) can be found as: a)30 V b) 12.5 V c) 9V 021 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. L me D L m 34 e) 5 rad/s e) 25 rad/s 15/02 A e) 12.61 V Fig. 9 Q15) For the circuit shown in Fig. 9, the equivalent inductance Leq is: a) 1/2 L c) 7/4 L d) 5/8 L e) 4/7 L b) 4/9 L b) R=0.202, L=0.2 H e) R=222, L=2 H 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 6 mA 3H 10Q 492 ww 3kQ SANAY 20 ww 8023 0.5 H BG Fig. 8 2mF 2102 www 3kQ 5KQ Q16) For the circuit shown in Fig. 10, the energy stored in the 4 mF capacitor under de conditions is: a) 32 mJ b) 128 mJ c) 256 mJ d) 8 mJ e) 16 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)-24° c) 204° d) 24⁰ e) 66° Q18) Assuming that the input impedance is given as Zin = 1+j1 2 and 0-1 rad/s, then the input admittance can be represented as the parallel combination of: a) R=0.22, C=0.2 F d) R=102, C=1F c) R=10, L=1H 4 mF Fig. 10 C 12.5 mF www. 9.51, 10 mF EQ Fig. 11 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) d) 15 V e) 24 V A B 4k0
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Question
circuits, pleaseeeee solve questionnn11
Transcribed Image Text:2A (4)
le
4.8 2
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 o 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=0.202, C=0.2 F
d) R=102, C=1F
m
1.92 02
34
L
m
L
Fig. 6
Fig.7
Q10) The average power supplied by the dependent source of Fig. 6 can be determined as
b) 24 W
d) 192 W
a) 5W
c) 96 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
c)-j55 V
b)-/330 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.
T
1.61, 80
30
34
L
m
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:
1520 A
e) 24 V
e) 15 V
SH
6 MA
24 V +
14
3k0z
www
402
-NN
20
+49
ΣΕΩΣ
ΔΩΣ
t=0
3kQ2
Fig. 8
2mF
HF
2102
ww
J30
400
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)-24°
c) 204°
d) 24°
e) 66°
Q18) Assuming that the input impedance is given as Zin= 1+j1 02 and co-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
Fig. 10
+
100 0.5H C= 10 mF
SKΩ Σ
4 m² =
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:
a)30 V b) 12.5 V c) 9V
d) 15 V
Q21) v(co) can be found as:
a) 24 V b)30 V c) 12.5 V d) 9 V
Q22) The voltage v(t) at t = 1 s is:
a) 20.9 V
b)24.9 V
c) 30 V
d) 39.1 V e) 27.97 V
Fig. 12
12.5mFv
04
B
4kQ
co www
9.51,
P0.5 mF
b
410
30 V
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
34
-
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
31
2002
0.3mH
200
ww
1002
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