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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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Please help me with this please <br><br> I’ll mark you Brainly
ivolga24 [154]
I think it is option (C).

If the answer is helpful then mark me as brainly.
4 0
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FREE BRAINIEST IF YOU ANSWER THIS
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The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

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<u>Input our values:</u>

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<u>Divide both sides by 400 to isolate t:</u>

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<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

<u></u>

If you have any questions on how I got to the answer, just ask!

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