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This is solution for problem related to Basic Engineering Circuit and Network Analysis course. It was suggested by Prof. Kanaka Balasubramanium at Agra University. It includes: Answers, Image, Marked, Answers, Selected, Problems, Underdamped, Bode, Plot
Typology: Exercises
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1.11 (a) (b) (c) (d) 1.14 (a) (b) 1.18 (a)
(b)
1.21 P 2 = 48 Wabsorbed
5 V, + top
5 V, + bottom
5 V, + bottom
5 V, + top
I = 2 A 1.24 (a) absorbed absorbed supplied (b) absorbed absorbed supplied supplied 1.28 absorbed absorbed absorbed supplied PD.S. = 8 W
2.53 RAB = 2 k
R (^) AB = 12 k
IL = 2.4 mA
Vo = 36 V
Vx = 10 V
Vbe = 9 V
Vda = - 9 V
Ix = - 3 mA
I 2 = 3 mA
I 1 = 6 mA
Gx = 200 S
Rx = 5 k 2.
**2.
2.113** Io = 0.67 A
Io = 2 A
g = 4
Io = - 4 mA
Vo = 3 V
Vo = - 12 V
Io = 0.33 mA
I 1 = - 3 mA
Imin = 17.51 mA Pmin = 175.1 mW
Imax = 18.22 mA Pmax = 182.2 mW
3.9 Vo = -3.6 V
Io = 0.6 mA
Io = 1 mA 3. 3. VB = -0.33 V
Io = - 1 mA
3.49 Vo = 15 V
Vo = 1.33 V
Vo = 8 V
Vo = 5 V
Vo = 2 V **3.
3.95** Io = 2.88 mA
Vo = -3.33 V
Vo = 6 V
Vo = 7.57 V
Io = 1.64 mA
Io = 2.67 mA
Vo = 6.67 V
Io = 2.88 mA
Vo = - 5 V
Io = 606 A
Vo Vin^ =^ 7.
0.5 1.
0.75 V
–2.25 V
v o (V)
t (s)
–0.
–1.
–2.
1
(^0 1 )
**4.17 (a) (b) (c)
4.38** vo = v 1 c 1 +
R 4 d^ a^ -^
R 1 b
Vo = -11.43 V
Vo = -11.2 V
Vo = 0 V
io v 1 = -^
Vo = 2 V
Vo = 4V 1 - V 2
I 1 = 1 mA I 2 = 0 A I 3 = 1 mA
5.44 Io = 1.6 mA
Vo = 65.79 V
Io = 5 mA
Vo = 4.8 V
Io = 1.6 mA
Io = 0.909 mA
Vo = 14.4 V
Io = - 2 mA
Io = -3.2 mA
Io = 1.14 mA **5.
5.98** RL = 99.0
Io = 1.6 mA
Io = -0.5 mA
Vo = 2 V
P 2 mA = 12 mW
Vo = - 5 V
Vo = -0.316 V
RTh = 400
Io = 0.909 mA
Io = 2.57 mA
i 1 (q) = 0 A
i 2 A 0 +B = 0 A
vRA 0 +B = - 8 V
vo(t) = b 8 -^ 8e
0 t t 7
vo(t) = 16 - 2.8e-1.5t^ V
i(t) = - 5 + 2.5e-4t^ A
vo(t) = 12 - 3e-6t^ V
io(t) = 2.4 - 2.4e-2.5^ *^105 t^ mA
io(t) = 3 + 0.33e-1.67t^ mA
vo(t) = 4.36e-2.73t^ V
io(t) = 2 + 0.5e-3.75t^ mA
io (A) 1
0 –0. –0. –0.
0.125 0.25 0.375 0.5 0.675^ t^ (s)
io(t) = 0.375 - 1.125e-8t^ A
io(t) = 0.67e-5t^ A 7.
7.75 (a) (b) (c) underdamped!
7.95 i(t) = e-3t[ 3 cos t + 5.4 sin t] A
i(t) = (4.04e-T^2 t^ - 1.54e T^1 t) A T T^1 =^ 0.08^ s 2 =^ 12.42^ s
0
5
10
15
20
–100 0 100 200 300 400 500 600 t (ms)
v o
( t )(V)
vo(t) = 12e-20t^ + 720te-20t^ V
v(t) = e-4t[ 10 cos 2t + 40 sin 2t] V
v 1 (t) = e-t^ CA 1 cos 2t + A 2 sin 2tD
s1,2 = -^2 ;^22 4 -^20 = - 1 ; j2 rads
s 2 + 2s + 5 = 0
8.5 (a)
(b)
**8.
8.** I (^) m
Re
j 0.
–0.
j 0.
–0.
vo(t) = 7.08 cos A 104 t + 36.8°B V
L = 10 mH
v(t) = 5.16 cos A377t + 45.1°B V
Z = 2.83/16.9ø^
Z = 1.00 + j3.77
i(t) = 6 sin (377t + 45 °) = 6 cos (377t - 45 °) A
i = vR i(t) = 5 cos (377t + 180 °) A
f = 63.7 Hz T = 15.7 ms (^) **8.
8.**
**8.
8.103 I** o = 17.7/- 137 ° A
V o = 37.6/-60.7° V
V o = (^4) / 90 ° V
V oc = - 4 + j4V Z Th = 2.24/ 117 °
V o = 5.55/86.8° V
I o = 2 /-36.9° A
V o = 4.71/-98.1° V
V o = 2.53/71.6° V
V o = 5.55/86.8° V
V o = 4.56/37.9° V
V o = 3.58/153.4° V
Z = -0.508 + j0.586
I o = 2.83/ 45 ° A
V o = 45 /-23.1° V
9.18 supplied supplied 9. 9.
9.
9. PL = 5.00 W
Z L = 4 + j1
Z L = 2.8 + j0.4
Z L = 2 + j2
p(t) = 40 cos ( 2 t + 96.9°) + 32.0 W
v(t) = 20 cos (t + 66.9°) V **9.
9.57** lagging 9. lagging 9.67 lagging
9.
9.79 @ I man @ = b (^238) 14.4^ mA A drywet^ skinskin
S (^) V S = 46.3/43.8° kVA
pfS = 0.
pfS = 0.
V S = 303 /11.1° Vrms
pf = 0.
pf = 0.
Irms = 3.27 A
Vrms = 1.87 V
10.47 wAt = 1 msB = 94.1 J
Z in = 4.6 + j3.2
V o = 2.5/36.9° V
V o = 12.2/15.0° V
V o = 1.24/- 120 ° V
I o = 3.25/66.0° A
V o = 2.17/5.19° V
V o = 10.15/10.8° V
V o V S^ =^ 0.140/24.8°^ 10.
V o = 1.80/- 140 ° V
I 2 = - I 1 n = 1.90/ 138 ° A
V 2 = n V 1 = 10.72/3.4° V
V ca = (^171) /- 165 ° V rms
V bc = (^171) /- 45 ° V rms
V ab = (^10023) / 75 ° V rms = (^171) / 75 ° V rms
V cn = (^100) / 165 ° V rms
V bn = (^100) /- 75 ° V rms
V an = (^100) / 45 ° V rms
Real
V an V cn
V ca
V bn (^) V bc
V ab Im
= 0.
L (^10) (0.1(^2 )()( 100100 )) = 200 = 46 dB
= 0.
100 = -^40 dB
= 1001
M 2 * 10 -^3 = - 54 dB
H (j) = 100 (j) (j + 1 )(j + 10 )(j + 50 )
0
0.01 0.1 1 10 100 1k (rad/s)
| H | (dB)
+20 dB/dec (^) –20 dB/dec –40 dB/dec
-600.1 1 (rad/s)
| G | (dB)
10 100
0
20 dB/dec –40 dB/dec
0
20
40
60
80
(rad/s)
| G
| (dB)
1 10 100 1k
–20 dB/dec
–20 dB/dec
–40 dB/dec
12.57 Filter is highpass 12.63 (a)
(b) 12.69 (a) (b) 12.75 Arbitrarily select C = 1 nF,yields R = 6.87 k
0.952 mS gm 1.053 mS
RG = 100 k
Cnew = 12.5 F
Lnew = 50 H
L = 10 nH
BW = 300 rads
@ V omax @ = 84.9 V
max = 7.06 krads
o = 7.07 krads
G (j) = 288 (j + 100 )^2 j(j + 900 )C(j)^2 + j4 + 400 D
H (j) = 5.33 * 104 (j + 1 )(j + 120 ) (j + 10 )(j + 80 )^2
13.12 (a) (b) 13.16 (a) (b)
13.21 (a)
(b) 13.27 (a) (b) f(t) = 161 Ct - 4te-4t^ - e-4t^ Du(t)
f(t) = Cte-t^ - 3e-t^ + 4e-2t^ Du(t)
f(t) = C(t) - e-2t^ cos (t + 90 °)Du(t)
f(t) = c - 12 e-2t^ + 1.58e-t^ cos (t - 18.4°) d u(t)
f(t) = 10e-2t^ cos (t)u(t)
f(t) = 10e-t^ cos (t - 90 °)u(t)
f(t) = 12 C 1 - 2e-t^ + 3e-2t^ Du(t)
f(t) = 16 C 1 + 3e-2t^ - 4e-3t^ Du(t)
F (s) = e
F (s) = e-(s^ +^ a)^ e (^) (s + acos) 2 + 2 + (s + a) sin (s + a)^2 + ^2 f
LCf(t)D = e-(s^ +^ a)^ **13.
13.41** Initial values (a) (b) (c) Final values **(a) (b) (c)
13.48** vo(t) = 2.67e-1.67tu(t) V
i(t) = 2e-4.5tu(t) A
tlim S q f(t)^ =^0
t^ lim S q f(t)^ =^0
t^ lim S q f(t)^ =^4
lim t S 0 f(t) = 2
lim t S 0 f(t) = 2
lim t S 0 f(t) = 2
y(t) = Ce-t^ - e-2t^ Du(t)
f(t) = 2e-^8 (t^ -^1 )^ - e-^4 (t^ -^1 )u(t - 1 )
14.25 vo(t) = C1.5A 1 - e-4tB Du(t) V
vo(t) = C 4 - 5e-t^ + 2e-2t^ Du(t) V
vo(t) = C 2 A 1 - e-tB Du(t) V
vo(t) = C 222 e-t^ cos (t - 45 °)Du(t) V
vo(t) = C 422 e-t^ cos (t - 45 °)Du(t) V
Z (s) = (^) 6s 2 +6s 16s+^8 + 11^ **14.
14.**
14.
14.45 V V o i
= (^) 14ss^ + +^1
a 4 - 43 e-^4 (t^ -^1 )^ b u(t - 1 ) V
vo(t) = a 4 - 43 e-4t^ b u(t) -
vo(t) = 12e-2tu(t) V
vo(t) = 1.15Ce-0.42t^ - e-1.58t^ Du(t) V
vo(t) = 9.6e-tu(t) V
2
V 2 = 0
2
2
I 2 = 0
2
2
2
I 2 = 0
1
16.42 V o = 2.32/ 157 ° V
Y in = 2 + j 1 + j2 S
z 22 = y 11 z 12 = ¢ y = 5
z 21 =