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Ipatiy [6.2K]
3 years ago
8

Technician A says that current is created in the stator windings of an alternator. Technician B says that the voltage regulator

controls the current flow through the rotor. Who is right
Physics
1 answer:
drek231 [11]3 years ago
4 0

Answer:

Technician A and Technician B both are right.

Explanation:

In an AC alternator,  there are two windings

1. Stator winding (stationary)

2. Rotor winding (rotating)

The current is induced in the stationary coils due to the magnetic field produced by the rotor. The DC suppy is provided to the rotor winding via slip rings and brushes and a voltage regulator precisely controls this supply to control the current flow through the rotor.

Therefore, both technicians are right.

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What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?
IRISSAK [1]

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, \vec{F_{up}} = 3.50\times 10^{- 5} N

Charge, Q = - 1.55\micro C = - 1.55\times 10^{- 6} C

Now, we know by Coulomb's law,

F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}

Also,

Electric field, E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

\vec{E} = \frac{\vec F_{up}}{Q}

\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}

\vec{E} = - 22.58 N/C

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

|\vec{E}| = 22.58\ N/C

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