Download UNIT 3 QUIZ SOLUTIONS (100 points) 1. (10 points) Given the AC

UNIT 3 QUIZ SOLUTIONS
(100 points)
1. (10 points) Given the AC current wave i = 3.0 sin
t
a.) Find the angular velocity, frequency and period of the wave.
SOLUTION:
The angular velocity is the coefficient of t;
So ω = rad/sec
The frequency, f =
=
= 0.5 Hz
The period is the reciprocal of the frequency; T=
= 2 ms
b.) Sketch the graph of current versus time for one cycle of the wave from t=0.
SOLUTION:
2. (10 points)
a.) In the following figure, find the frequency, period, and angular velocity of each ac
wave.
SOLUTION:
The period is the time for one cycle, so for IP, T=10 ms, and for VP, T= 10 ms
The frequency of IP , f = =
= 100 Hz,
and the frequency of VP , f =
=
= 100 Hz
The angular velocity of IP and VP is ω = 2πf = 2π (100) = 200 π radians
b.) Find the phase angle difference and time difference between the two ac waves.
SOLUTION:
The phase angle difference between the two waves is because the waves are
shifted ¼ of a cycle with respect to each other
Since t = =
=
s = 2. 5 ms (and we can check that on the graph).
3. (10 points)
Given the AC wave, V=180 sin
), find the peak value (amplitude), angular
velocity, phase angle, frequency and period of the wave.
SOLUTION:
Peak value = 180 volts
Angular velocity = ω =
Phase angle =
rad
Frequency, f =
Period, T= =
=
= 65 Hz
15 ms
4. (20 points)
The tip of a one-link robot is located at
= rad at time t=0 sec,
as shown in the figure. It takes 2 seconds for the robot to move from
rad to
+ 2 rad. If l = 10 in, plot the x- and y-components as a function of time. Also find the
amplitude, period, phase angle, and timeshift.
SOLUTION:
Amplitude = l = 10 in
The period is equal to the time required to travel
The frequency is determined as f = = = 0.5 Hz
Also, ω = 2πf = 2π (0.5) = π rad/s
rad, thus T=2 sec
The phase shift is equal to the initial position (in radians)
Therefore, the time shift is found by solving τ =
The equations of the tip are:
x(t) = A cos (ωt +
= 10 cos (πt+ ) in
y(t) = A sin (ωt +

= 10 sin (πt+ ) in
=

= sec (to the left)

5. (10 points)
Use the trigonometry identity sin(A+B)=sin(A)cos(B)+cos(A)sin(B) to re-write this sine
function representing voltage, M sin (2 +
SOLUTION:
M sin (2 +
= M sin(2t) cos + M cos (2t) sin
6. (15 points)
a. An AC voltage wave takes 3ms to go from 0 to peak value. What are the
frequency, period, and angular velocity of the wave?
SOLUTION:
83 Hz
12 ms
520 rad/s
b. If the peak value of the voltage wave is 120V and the phase angle = 0 rad, write
the equation of the wave.
Solution:
V=120sin520t
7. (25 points)
A series RL circuit is subjected to a sinusoidal voltage of frequency 120
The current i(t) = 10 cos (120 ) A is flowing through the
circuit. The voltage across the resistor and inductor are given by :
vR (t) = 10 cos (120 )
vL (t) = 12 cos (120
) volts, where t is in seconds.
a. Write down the amplitude, frequency (in Hz), period (in seconds), phase shift (in
degrees), and time shift (in seconds) of the voltage, vL (t).
SOLUTION:
Amplitude = 12 V
Frequency = f
F= =
= 60 Hz = f
The period is then T=
The phase shift,

Therefore, the time shift is τ =
 τ =
=
b. Plot one cycle of the voltage, vL (t), and indicate the earliest time (after t=0) when
the voltage is a maximum.
SOLUTION:
c. The total voltage across the circuit is given by v(t) = vR (t) + vL (t).
Write v(t) in the form, v(t) = M cos (120
+ ) (i.e) find M and )
SOLUTION:
v(t) = vR (t) + vL (t) = M cos (120
vL (t) = 12 cos (120
+ )
) = 12 cos (120
)
- 12 sin (120
= -12 sin (120
M cos (120
+ ) = M cos (120
)
) – M sin (120
vR (t) + vL (t) = 10 cos (120 ) -12 sin (120
=
M cos (120
) – M sin (120
Comparing coefficients on cos (120
cos (120




√


: 10 =
) 
)
 
= 50.2 o
v(t) = 15.62 cos (120
+
) volts
) = 12
)
)