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RoseWind [281]
3 years ago
6

5. An electrical power plant generates electricity with a current of 50 A and a potential difference of 20 000 V. In order to mi

nimize the power losses due to the high current heating up the line, a transformer steps up the potential difference to 500 000 V before it is transmitted. What is the current in the transmission lines?
Physics
1 answer:
lakkis [162]3 years ago
3 0

Answer: Current = 2 A

Explanation:

Given that an electrical power plant generates electricity with a

current I = 50 A

Potential difference V = 20 000 V

The resistance R will be achieved by Ohms law formula which state that

V = IR

But the power generated will be the product of potential difference and the current

Power P = IV

P = 50 × 20000

P = 1, 000000 W

When the transformer steps up the potential difference to 500 000 V before it is transmitted

Power is always constant.

Using the formula for power again with

V = 500000

1000000 = 500000× I

Make I the subject of formula

Current I = 1000000/500000

Current I = 2 A

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Answer:

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Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

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M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

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The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

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A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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