Ejercicios De Faosores

ejercicios de fasoresChapter 9, Problem 1.
Given the sinusoidal voltage v(t) = 50 cos (30t + 10 o ) V, find: (a) the amplitude V m ,(b)
the period T, (c) the frequency f, and (d) v(t) at t = 10 ms.
Chapter 9, Solution 1.
(a) Vm = 50 V.

2π
= 0.2094s = 209.4ms
ω 30
(c ) Frequency f = ω/(2π) = 30/(2π) = 4.775 Hz.
(d) At t=1ms, v(0.01) = 50cos(30x0.01rad + 10˚)
= 50cos(1.72˚ + 10˚) =44.48 V and ωt = 0.3 rad.
(b) Period T =

2π

=

Chapter 9, Problem 2.

A current source in a linear circuit has
i s = 8 cos (500 π t - 25 o ) A
(a) What is the amplitude of the current?
(b) What is the angular frequency?
(c) Find the frequency of the current.
(d) Calculate i s at t = 2ms.
Chapter 9, Solution 2.

(a)

amplitude = 8 A

(b)

ω = 500π = 1570.8 rad/s

(c)

f=(d)

Is = 8∠-25° A
Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°)
= 8 cos(π − 25°) = 8 cos(155°)
= -7.25 A

ω
= 250 Hz
2π

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limiteddistribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.

Chapter 9, Problem 3.

Express the following functions in cosine form:
(a) 4 sin ( ω t - 30 o )
(b) -2 sin 6t
(c) -10sin( ω t + 20 o )
Chapter 9, Solution 3.
(a)
4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4cos(ωt – 120°)

(b)

-2 sin(6t) = 2 cos(6t + 90°)

(c)

-10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)

Chapter 9, Problem 4.

(a) Express v = 8 cos(7t = 15 o ) in sine form.
(b) Convert i = -10 sin(3t - 85 o ) to cosine form.
Chapter 9, Solution 4.
(a)
v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

(b)

i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°)= 10 cos(3t + 5°)

Chapter 9, Problem 5.

Given v 1 = 20 sin( ω t + 60 o ) and v 2 = 60 cos( ω t - 10 o ) determine the phase angle
between the two sinusoids and which one lags the other.
Chapter 9, Solution 5.

v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°)
v2 = 60 cos(ωt − 10°)
This indicates that the phase angle between the two signals is 20° and that v1 lags
v2.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student usingthis Manual,
you are using it without permission.

Chapter 9, Problem 6.

For the following pairs of sinusoids, determine which one leads and by how much.
(a) v(t) = 10 cos(4t - 60 o ) and i(t) = 4 sin (4t + 50 o )
(b) v 1 (t) = 4 cos(377t + 10 o ) and v 2 (t) = -20 cos 377t
(c) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t -11.8 o )
Chapter 9, Solution 6.
(a)
v(t) = 10 cos(4t – 60°)i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°)
Thus, i(t) leads v(t) by 20°.

(b)

v1(t) = 4 cos(377t + 10°)
v2(t) = -20 cos(377t) = 20 cos(377t + 180°)
Thus, v2(t) leads v1(t) by 170°.

(c)

x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°)
X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04°
x(t) = 13.928 cos(2t – 21.04°)
y(t) = 15 cos(2t – 11.8°)
phasedifference = -11.8° + 21.04° = 9.24°
Thus, y(t) leads x(t) by 9.24°.

Chapter 9, Problem 7.

If f( φ ) = cos φ + j sin φ , show that f( φ ) = e jφ .
Chapter 9, Solution 7.
If f(φ) = cosφ + j sinφ,
df
= -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ )
dφ
df
= j dφ
f
Integrating both sides
ln f = jφ + ln A

f = Aejφ = cosφ + j sinφ
f(0) = A = 1
i.e. f(φ) = ejφ = cosφ + j sinφ...