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High voltage input question?

If you double voltage, what happens to power?

  • The full question: Power is being transmitted through normal transmission lines from a generator to a substation. If the input voltage is doubled and the current halved, how will the power input and the useful power output change? a) Both the power input and the useful power output will decrease. b) Both the power input and the useful power output will increase. c) The power input will not change but the useful power output will decrease. d) The power input will not change but the useful power output will increase.

  • Answer:

    P supply= IV input P waste= V^2/R If you double voltage, your power increases by a power of 4. Plug in 240V and 120V as your values and you'll see how it turns out with 4 :)

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P supply= IV input P waste= V^2/R If you double voltage, your power increases by a power of 4. Plug in 240V and 120V as your values and you'll see how it turns out with 4 :)

Alex

d) The power input will not change but the useful power output will increase. P = V^2/R P = I^2 /R Increasing voltage squares the value which gives more useful power output Decreasing current by half also cuts power by a 1/ square of the current And this is generally what happens with real substations. Voltage is sent at high as possible ranges and amperage is sent as low as possible so that a smaller wire can deliver the equivalent amount of power. It is a lot more expensive to send, transmit and deliver amps

biire2u

For the same power input, the power output will increase, as the losses along the resistance of the line are lessened by halving the current (doubling the voltage halves the current). Power loss = I^2R. Note the SQUARING of the current, so halving it means 1/4 of the losses. The R doesn't change. e.g. Current = 20A., line resistance 5 ohms. 20^2 x 5 ohms = 2,000W. Now, halve the current, = 10A. 10^2 x 5 ohms = 500W. That's 1/4 the loss, OK? Ans = D.

Technobuff

P =V* I You figure it out.

Sexy Homer

d) The power input will not change but the useful power output will increase. P = V^2/R P = I^2 /R Increasing voltage squares the value which gives more useful power output Decreasing current by half also cuts power by a 1/ square of the current And this is generally what happens with real substations. Voltage is sent at high as possible ranges and amperage is sent as low as possible so that a smaller wire can deliver the equivalent amount of power. It is a lot more expensive to send, transmit and deliver amps

biire2u

For the same power input, the power output will increase, as the losses along the resistance of the line are lessened by halving the current (doubling the voltage halves the current). Power loss = I^2R. Note the SQUARING of the current, so halving it means 1/4 of the losses. The R doesn't change. e.g. Current = 20A., line resistance 5 ohms. 20^2 x 5 ohms = 2,000W. Now, halve the current, = 10A. 10^2 x 5 ohms = 500W. That's 1/4 the loss, OK? Ans = D.

Technobuff

P =V* I You figure it out.

Sexy Homer

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