A four-pole alternator has a uniform magnetic flux density of 0,5 T. A square coil with 200 turns and sides of 18cm is rotated at 20 r/s inside the field. Determine the frequency of the value generated.

Answer 1

The frequency of the value generated by the alternator can be determined using the formula:

f = (P * N) / 120

Where:
- f is the frequency in hertz (Hz)
- P is the number of poles
- N is the rotational speed of the coil in revolutions per minute (RPM)

Given:
- A four-pole alternator has a uniform magnetic flux density of 0.5 T.
- A square coil has 200 turns.
- The sides of the square coil are 18 cm.
- The coil is rotated at 20 r/s.

First, we need to convert the rotational speed from revolutions per second (r/s) to revolutions per minute (RPM).

N = (20 r/s) * (60 s/1 min) = 1200 RPM

Next, we need to calculate the number of poles (P). Since it is a four-pole alternator, P = 4.

Now, we can substitute P = 4 and N = 1200 RPM into the formula to calculate the frequency:

f = (4 * 1200) / 120

f = 40 Hz

Therefore, the frequency of the value generated by the alternator is 40 Hz.