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