to Solution Problem Set IX
Chapter 10, Problem 2. (a) If -10 cos t + 4 sin t = A cos( t + ) , where A > 0 and - 180 o < 180 o , find A and . (b) If 200 cos(5t + 130 o ) = F cos 5t + G sin 5t , find F and G. (c) Find three values of t, 0 t 1 s, at which i (t ) = 5 cos 10t - 3 sin 10t = 0 . (d) In what time interval between t = 0 and t = 10 ms is 10 cos 100t 12 sin 100t ?
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Chapter 10, Solution 2. (a)
-10 cos t + 4sin t + ACos ( wt + ), A > 0, - 180 < 180 A = 116 = 10.770, A cos = -10, A sin = -4 tan = 0.4, 3d quad = 21.80 = 201.8, too large = 201.8 - 360 = -158.20
(b)
200 cos (5t + 130) = Fcos 5t + G sin 5t F = 200cos130 = -128.6 G = -200sin130 = -153.2
(c)
i(t ) = 5cos10t - 3sin10t = 0, 0 t 1 s sin10t 5 = , 10t = 1.0304, cos10t 3 t = 0.10304 s; also, 10t = 1.0304 + , t = 0.4172 s; 10t = 1.0304 + 2, t = 0.7314 s
0 < t < 10ms, 10 cos100t 12sin100t ; let 10cos100t =12sin100t 10 tan100t = , 100t = 0.6947 t = 2.211 ms 0 < t < 2.211 ms 12
(d)
Chapter 10, Problem 7. Determine which waveform in each pair is lagging: (a) 6 cos( 2 60t - 9 o ) and - 6 cos( 2 60t + 9 o ) . (b) cos( t - 100 o ) and - cos(t - 100 o ) . (c) - sin t and sin t. (d) 7000 cos(t - ) and 9 cos(t - 3.14 o ) .
Chapter 10, Solution 7. (a) 6 cos (2p60t 9o) -6 cos (2p60t + 9o) 6-9o 6189o -6 cos (2p60t + 9o) lags 6 cos (2p60t 9o) by 360 9 189 = 162o.
P P
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6189o
P P
6-9o
P P
Solution to Problem Set IX
2
(b)
cos (t - 100o) -cos (t - 100o)
180o
1 -100o -1 -100o = 1 80o -cos (t - 100o) lags cos (t - 100o) by 180o.
1-100o (c) -sin t sin t -1-90o = 190o 1 -90o -sin t lags sin t by 180o. 190o
1 -90o
(d)
7000 cos (t p) 9 cos (t 3.14o)
7000 -p = 7000 -180o 9 -3.14o 7000 cos (t p) lags 9 cos (t 3.14o) by 180 3.14 = 176.9o.
7000 -180o 9 -3.14o
Chapter 10, Problem 10.
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Solution to Problem Set IX
Household electrical voltages are typically quoted as either 110 V, 115 V, or 120 V. However, these values do not represent the peak ac voltage. Rather, they represent what is known as the root mean square of the voltage, defined as 1 T 2 Vm cos 2 ( t )dt T 0 where T = the period of the waveform, Vm is the peak voltage, and = the waveform frequency ( f = 60 Hz in North America). (a) Perform the indicated integration, and show that for a sinusoidal voltage, V V rms = m 2 (b) Compute the peak voltages corresponding to the rms voltages of 110, 115, and 120 V. Vrms =
3
Chapter 10, Solution 10. V2 = m T V2 = m T V2 = m 2T V2 = m 2T
(a)
Vrms
0
T
0
cos t dt
2
1
2
T
2t cos 2 dt T
1
2
T
0
4t dt 1 + cos T V2 dt + m 2T
1
2
T
0
T
0
4t cos dt T
1 2
1
2
V2 V2 4 = m T + m cos u 0 8 2T Vm = 2
*
(b)
Vm = 110 2 = 155.6 V , 115 2 = 162.6 V , 120 2 = 169.7 V
Chapter 10, Problem 15. A sinusoidal voltage source v s = 100 cos 10 5 t V, a 500 resistor, and an 8 mH are inductor in series. Determine those
1 T , at which zero power is being: (a) delivered to the resistor; (b) delivered to the inductor; (c) 2 generated by the source.
instants of time, 0 t
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Solution to Problem Set IX
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Chapter 10, Solution 15. (a)
i=
800 5 cos 105 t - = 0.10600 cos (10 t - 57.99) A 500 500 + 800 57.99 pR = 0 when i = 0 105 t - = , t = 25.83s 180 2 100
2 2
(b)
vL = Li = 8 10-3 0.10600 (-105 ) sin (105 t - 57.99) vL = -84.80sin (105 t - 57.99) pL = vL i = -8.989sin (105 t - 57.99) cos (105 t - 57.99) = -4.494 sin (2 165 t - 115.989) pL = 0 when 2 105 t - 115.989 = 0, 180, t = 10.121 or 25.83s
(c)
ps = vs iL = 10.600 cos105 t cos (105 t - 57.99) ps = 0 when 105 t = , t = 15.708s and also t = 25.83s 2
Chapter 10, Problem 22.
o o Convert the following to rectangular form: (a) 7 - 90 ; (b) 3 + j + 7 - 17 o ; (c) 14e j15 ; (d) 10 . Convert the following to polar form: (e) -2 (1 + j9); ( f) 3.
o
Chapter 10, Solution 22. (a) 7 -90o = -j 7 (b) 3 + j + 7 -17o = 3 + j + 6.694 j 2.047 = 9.694 j 1.047 (c) 14ej15 = 14 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3. 623 (d) 1 0o = 1 (e) 2 (1 + j 9) = -2 j 18 = 18.11 - 96.34o (f) 3 = 3 0o
o
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Solution to Problem Set IX
5
Chapter 10, Problem 23. Perform the indicated operation(s), and express the answer as a single complex number in rectangular form: (a) (1 + j )(2 - j 7) 3 + 15 - 23 o ; (b) 12 j (17180 o ) ; (c) 5 - 16 . 33 - 9 o
Chapter 10, Solution 23.
(a) 3 + 15 -23o = 3 + 13.81 j 5.861 = 16.81 j 5.861 (b) (j 12)(17 180o) = (12 90o)(17 180o) = 204 270o = j 204 (c) 5 16(9 j 5)/ (33 -9o) = 5 (164 -29.05o)/ (33 -9o) = 5 4.992 -20.05o = 5 4.689 j 1.712 = 0.3109 + j 1.712
Chapter 10, Problem 26. Convert these complex numbers to rectangular form: (a) 5 - 110 o ; (b) 6e j160 ; (c) (3 + j 6)( 250 o ) . Convert to polar form: (d) -100 - j 40 ; (e) 250 o + 3 - 120 o .
o
Chapter 10, Solution 26.
(a) (b) (c) (d) (e)
5 - 110 = -1.7101 - j 4.698
6e j160 = -5.638 + j 2.052
(3 + j 6) (250) = -5.336 + j12.310 -100 - j 40 = 107.70 - 158.20
250 + 3 - 120 = 1.0873 - 101.37
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Solution to Problem Set IX
6
Chapter 10, Problem 31. Express each of the following currents as a phasor: (a) 12 sin(400t + 110 o ) A; (b) -7 sin 800t - 3 cos 800t A; (c) 4 cos(200t - 30 o ) - 5 cos(200t + 20 o ) A. If = 600 rad/s, find the instantaneous value of each of these voltages at t = 5 ms: (d) 7030 o V; (e) -60 + j40 V.
Chapter 10, Solution 31.
(a) (b)
12 sin (400t + 110) A 1220A
-7 sin 800t - 3cos 800t j 7 - 3 = -3 + j 7 = 7.616113.20 A 4 cos (200t - 30) - 5cos (200t + 20) 4 - 30 - 520 = 3.910 - 108.40 A
(c)
(d)
= 600, t = 5ms : 7030 V 70 cos (600 5 10-3rad + 30) = -64.95 V
(e)
= 600, t = 5ms : 60 + j 40 V = 72.11146.3 72.11cos (3rad + 146.31) = 53.75 V
Copyright 2002 The McGraw-Hill Companies Inc.