Alternating Current and Frequency

This problem set deals with alternating current and frequency.

1. An electric generator that produces 600 cycles in 10 sec has a frequency of 60 Hz.

2. The common unit of frequency for one cycle/s is 1 Hz.

3. If a wave has a frequency of 10 cycles/s (10 Hz), what is its period?

T = 1/f = 1/(10 cycles/sec) = 0.1 sec/cycle

4. Frequency is a measure of how often events repeat themselves. The graph below is a sine wave. If point A is the starting point, at what later points does the wave repeat itself? That is, when does the cycle start over again?

The cycle starts over at points E and H.

5. In the graph above, how many complete cycles are shown for the wave?

Two cycles (A to E and E to H)

6. Given the oscilloscope below with the indicated settings, find the
1. time/division and volts/division
2. period of one cycle of this waveform
3. peak-to-peak voltage of the signal.
1. time/div is 2 ms/div and volts/div is 10 volts/div
2. T = (8 div)(2 ms/div) = 16 ms
3. Vpp = (7 div)(10 volts/div) = 70 volts

7. Name the type of waveform shown below. Find its period, frequency and peak-to-peak voltage.
• square wave
• T = (5 div)(0.2 sec/div) = 1 sec
• f = 1/T = 1/(1sec/cycle) = 1 Hz
• Vpp = (5 div)(0.1 V/div) = 0.5 V

8. Find the peak-to-peak voltage, period and frequency of the sine waveform.
• Vpp = (6 div)(2 V/div) = 12 V
• T = (8 div)(50 msec/div)(1 sec/106 msec) = 0.0004 sec (400 msec)
• f = 1/T = 1/(0.0004 sec/cycle) = 2500 Hz

9. Find the peak-to-peak voltage, period and frequency of the sawtooth waveform shown below. Note that a "×10" oscilloscope probe was used to obtain the signal. This means that the voltage must be multiplied by a factor of ten to get the proper answer.
• Vpp = (8 div)(5 V/div)(×10) = 400 V
• T = (8 div)(20 msec/div) = 160 msec (0.16 s)
• f = 1/T = 1/(0.16 sec/cycle) = 6.25 Hz

10. Find the peak-to-peak voltage and frequency of the sine wave shown below. Note that a "x10" oscilloscope probe was used to obtain the signal. This means that the voltage must be multiplied by a factor of ten to get the proper answer.
• Vpp = (8 div)(2 V/div)(×10) = 160 V
• f = 1/T= 1/[(10 div)(0.002 sec/div)] = 50 Hz

11. The electron gun in a TV picture tube scans the width of the screen. To form an image, it produces 15.75 ×104 lines on the screen during a 5 minute period. Find the frequency of the electron gun.

 Knowns Unknowns Equations/Remarks n = 157,500 lines t = 5 min = 300 sec f = ? f = n/t f = n/t = (157500 lines)/(300 sec)f = 525 lines/sec Divide the number of cycles (lines in this case) by the amount of time it takes.

12. The oscilloscope of an engine analyzer shows that the waveform for an electronic ignition device on a 6-cylinder engine has a frequency of 6000 pulses/min at an engine speed of 2000 rpm. Find the
1. number of times the device pulses during a 5 minute test
2. period in seconds between successive pulses.

 Knowns Unknowns Equations/Remarks f = 6000 pulses/min t = 5 min 1 min = 60 sec a. n = ? b. T = ? f = n/t T = 1/f a. f = n/t     n = ft = (6000 pulses/min)(5 min)     n = 30,000 pulses b. T = 1/f = 1/[(6000 pulses/min)(1 min/60 sec)]     T = 0.01 sec Rearrange the frequency equation to solve for the number of cycles (pulses). Substitute in the known values and calculate the answer.Convert the minutes to seconds to find the period

13. A train of identical flat cars travels at constant speed past an observer. The observer sees a new flat car go by every 4 sec. What is the frequency at which the cars pass the observer?

f = 1/T = 1/(4 sec/cycle) = 0.25 Hz

14. What is the period of the following waves:
1. 50 Hz
2. 50 kHz (50,000 Hz)
3. 50 MHz (50,000,000 Hz)
1. T = 1/f = 1/(50 cycles/sec) = 0.02 sec
2. T = 1/f = 1/(50,000 cycles/sec) = 2 ×10-5 sec (20 msec)
3. T = 1/f = 1/(50,000,000 cycles/sec) = 2 ×10-8 sec