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Deffense [45]
3 years ago
10

An empty capacitor is connected to a 12.0 V battery and charged up. The capacitor is then disconnected from the battery, and a s

lab of dielectric material (dielectric constant "k" = 2.8) is inserted between the plates. Find the amount by which the potential difference accross the plates changes. Specify whether the change is an increace or a decreace.
Physics
1 answer:
Temka [501]3 years ago
3 0

Answer:

7.71 V

Explanation:

When the charge remains constant, the potential difference decrease by the factor of the dielectric constant K

V new = V initial / K = 12 V / 2.8 = 4.29 V

ΔV ( change in potential) = V initial - V new ( after dielectric has been added) = 12 V - 4.29 V =      7.71 V

The potential difference decreases by 7.71 V

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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
Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons
gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

8 0
3 years ago
What is the connection between electrons and protons
TEA [102]

Answer:

Electrons and protons are connected with one another in term of charge and size.

Explanation:

  • The size of electron and proton is always same in any atom but they possess opposite charge.
  • Electron in any atom carries negative charge where as proton carries the positive charge.
  • In any neutral atom the charge between electron and proton is balanced along with the size.
  • The nucleus of any atom bounds only proton and neutron but the electron is present revolving around the nucleus.
7 0
3 years ago
HELP ASAP <br> Describe when contact metamorphism occurs?
diamong [38]

Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole

5 0
3 years ago
Read 2 more answers
two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma
nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force =  3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees  

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos\theta

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

 35 =√ 9x² + 25x² + 15x²    (cos 60° = 0.5)

35 = √49 x²

 x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0
3 years ago
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