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Oxana [17]
3 years ago
9

Suppose a large power plant generates electricity at 12.0 kV. Its old transformer once converted this voltage to 315 kV. The sec

ondary coil of this transformer is being replaced so that its output can be 730 kV for more efficient cross-country transmission on upgraded transmission lines.
Randomized Variables

Vi = 315 kV

V2 = 730 kV


(a) What is the ratio of turns in the new secondary to the number of turns in the old secondary?

(b) What is the ratio of new current output to the old current output for the same power input to the transformer?
Physics
1 answer:
SpyIntel [72]3 years ago
7 0

Answer:

  • 2.32
  • 0.43

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

\frac{Vn}{Vo} = \frac{Nn}{No} = \frac{730}{315}  therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }     equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

\frac{In}{Io}  = \frac{Vo}{Vn} =  315 / 730 = 0.43

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Answer:

D) This is the correct answer

Explanation:

In this exercise the two ball loads are suspended by a thread.

To answer this exercise, let us remember that charges of the same sign repel and charges of a different sign attract.

Therefore, for the system to maintain equilibrium, the two charges must be of the same sign.

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A) in this case, as a sphere has no charge, there is no electric force and the induced charge is of the opposite sign, so the spheres attract each other

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D) This is the correct answer, since the charges have the same magnitude and are of the same sign, so the force is repulsive and is counteracted by the weight component

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A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro
Amanda [17]

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

\phi_{1}=BA\cos\theta

Put the value into the formula

\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0

\phi_{1}=0.01021\ Wb

We need to calculate the magnetic flux

Using formula of flux

\phi_{2}=BA\cos\theta

Put the value into the formula

\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0

\phi_{2}=0.00127\ Wb

We need to calculate the magnitude of the average induced emf

Using formula of emf

\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})

Put the value into t5he formula

\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})

\epsilon=0.016867\ V

\epsilon=16.87\ mV

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

I=\dfrac{\epsilon}{R}

Put the value into the formula

I=\dfrac{0.016867}{2.5}

I=0.00675\ A

I=6.75\ mA

Hence, (A). The direction of the induced current will be clockwise.

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3 years ago
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Answer:

F=1.13\,*\,10^{15} (answer a)

Explanation:

Recall the formula for the Coulomb force:

F=k\frac{q_1*q_2}{d^2}

which in our case gives:

F=k\frac{q_1*q_2}{d^2} \\F=9*10^9\,\frac{63*45}{0.15^2} \\F\approx 1134000\,*\,10^9\\F\approx 1.13\,*\,10^{15}

which agrees with answer a)

6 0
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