Question

at rated speed, the fiold excitation in adjusted to give rated armature-winding current. The short circuit is then removed, and, with the same field current and speed, the open-circuit voltage between stator terminals is found to be 1,744. The effective (a-c) resistance per phase of the armature winding is 1.2ohms. Caleulate the per cent regulation of the alternator at a power factor of 0.8 lagging. 84. A 25-kva alternator has a total loss of 2,000 watts when it delivera rated kilovoltamperes to a load at a power factor of 0.76. Caleulate its per cent efficiency.

Answer 1

The frequency of the output voltage is 4314.69 Hz and the strength of the magnetic field is 0.001 T.

Step-by-step explanation:

To calculate the frequency of the output voltage, we need to know the number of poles of the generator. The number of poles can be calculated using the formula:

Number of poles = Number of cycles x Number of phases / 2 x RPM

Substituting the given values: Number of cycles = 200 turns, Number of phases = 1, RPM = 3600, we get:

Number of poles = (200 x 1) / (2 x 3600) = 1/36 = 0.0278

Since a generator has a synchronous speed of 120f/P, where f is the frequency, and P is the number of poles, we can calculate the frequency using this formula:

f = 120 / P

Substituting the value of P, we get:

f = 120 / 0.0278 = 4314.69 Hz

To calculate the strength of the magnetic field, we can use Faraday's law of electromagnetic induction, which states that the induced emf is directly proportional to the rate of change of magnetic flux:

V = NABω

Where V is the peak output voltage, N is the number of turns, A is the area of the coil, B is the magnetic field strength, and ω is the angular frequency. Rearranging the formula, we can solve for B:

B = V / (NAω)

Substituting the given values, we get:

B = 120 V / (200 turns x 20.0 cm x 20.0 cm x 3600 rpm x 2π / 60)

B ≈ 0.001 T