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wolverine [178]
3 years ago
12

Help ASAP

Physics
1 answer:
SIZIF [17.4K]3 years ago
4 0
I think all of those are examples
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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer
SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The  Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75MVA

b

The efficiency is 99.4%

Explanation:

From the question  we are given are given that

 The transformer has Mega Volt Amp rating of 25MVA

                          The frequency is 60-Hz

                           Voltage rating 8.0kV : 78kV

   The short circuit test gives : 453kV,321A,77.5kW

   The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the  Low Voltage Rating for Auto - Transformer is 78kV

 Now to obtain the current flowing through the 8kV  coil in the Auto-transformer we have

             \frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}

The volt will cancel each other

             \frac{25*10^6}{8*10^3} =  3125\ A

 Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

MVA \ rating = (86*10^3)(3125) =268.75

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil I_2 as shown in the circuit diagram can be obtained as

       I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321

Now the efficiency can be obtained as thus

           \frac{(operational \ MVA )*(Power factor \pf))}{(operational\  MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}

             =99.941%

8 0
3 years ago
How much work is done (in Joules) by a weightlifter in raising a 60-kg barbell from the floor to a height of 2m? Work done =
Vika [28.1K]

Answer:

1176 Nm or J

Explanation:

W = F*d

F = 60kg * 9.8 kgm/s^2 = 588 N

W = 588 N * 2m = 1176 N*m

7 0
3 years ago
Read 2 more answers
A positive charge is fixed at point (−4,−3) and a negative charge − is fixed at point (−4,0). Determine the net electric force ⃗
Masteriza [31]

The net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

<h3>Net electric force on the charges</h3>

The net electric force on the charges is calculated as follows;

F = kq₁q₂/r²

where;

  • k is coulomb's constant
  • q₁ and q₂ are the charges
  • r is the distance between the charges
<h3>Distance between the charges</h3>

|r| = \sqrt{(x_2-x_1)^2+ (y_2-y_1)^2} \\\\|r| = \sqrt{(-4--4)^2+ (0--3)^2} \\\\|r| = \sqrt{(0)^2 + (3)^2} \\\\|r| = 3 \ units

F_{net} = \frac{kq_1q_2}{3^2} \\\\F_{net} = \frac{1}{9} (kq_1q_2) , \  N

Thus, the net electric force acting on a positive test charge at the origin is determined as ¹/₉(kq₁q₂).

Learn more about electric force here: brainly.com/question/17692887

#SPJ1

3 0
2 years ago
What best describes the speed of light waves in solids, liquids, and gases?
hammer [34]

Answer:

C

Explanation:

Generally, the speed of light slows down when passing through a medium that is not a vacuum. This is not always the case, but I will be ignoring the rare/exotic exceptions. Light has a harder time traveling through solids and liquids than it does with gases.

3 0
2 years ago
What are the units for gravitational field strength ?
Jobisdone [24]

Answer:

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

7 0
3 years ago
Read 2 more answers
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