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snow_tiger [21]
2 years ago
9

Might a synchronous motor equally well be called a synchronous inductor

Engineering
1 answer:
sergiy2304 [10]2 years ago
3 0

In Engineering, a synchronous motor is sometimes referred to as synchronous capacitor but not a synchronous inductor.

<h3>What is a synchronous motor?</h3>

A synchronous motor can be defined as an alternating current (AC) electric motor in which the rotation speed of the shaft is directly proportional to the frequency of the supply current, especially at steady state.

Basically, a synchronous motor is sometimes referred to as a synchronous capacitor because it can be used to correct the power factor (PF) of a lagging load such as an induction motor.

However, a synchronous motor cannot be called a synchronous inductor because the rotation speed of the shaft is not proportional to the synchronous speed of the motor. Thus, they are called an synchronous motor.

Read more on synchronous motor here: brainly.com/question/12975042

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A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu
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Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

emf= N\frac{d \phi}{dt}

where;

ΔФ is change in magnetic flux

ΔФ  = BA = B x πr²

ΔФ  = 0.573 x π(0.0395)² = 0.002809 T.m²

emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0
3 years ago
The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m
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Answer:

The circulation around the cylinder is 0.163 \frac{m^{2} }{s}

Explanation:

Given :

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Sea level span L = 6 \frac{N}{m}

Lift per unit circulation is given by,

  L = \rho v c

Where c = circulation around cylinder

   c = \frac{L}{\rho v}

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Therefore, the circulation around the cylinder is 0.163 \frac{m^{2} }{s}

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