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    Please help me with this GCSE question. Many thanks

    According to V = IR, increasing the voltage increases the current (provided resistance is constant.

    However, in the national grid, it is said and also true that high voltage results in low current resulting in less heat lost. For this reason, voltage is stepped up in national grid to reduce heat lost.

    Why doesn't the current increase when voltage increases in the national grid? Because according to V=IR, current must increase when voltage increases.

    Appreciate your answers
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    (Original post by Usernametaken)
    Please help me with this GCSE question. Many thanks

    According to V = IR, increasing the voltage increases the current (provided resistance is constant.

    However, in the national grid, it is said and also true that high voltage results in low current resulting in less heat lost. For this reason, voltage is stepped up in national grid to reduce heat lost.

    Why doesn't the current increase when voltage increases in the national grid? Because according to V=IR, current must increase when voltage increases.

    Appreciate your answers
    The idea that increasing voltage decreases current comes from the requirement to deliver the same power.
    Power = VI
    so you can deliver, say, 1,000,000 watt either by having a voltage of 100,000V and a current of 10A or a current of 100,000A and a voltage of 10V

    It's not a case of "increasing voltage and decreasing current as a result", it's about making a decision about how you set up the circuit/system to achieve this power transfer.
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    (Original post by Stonebridge)
    The idea that increasing voltage decreases current comes from the requirement to deliver the same power.
    Power = VI
    so you can deliver, say, 1,000,000 watt either by having a voltage of 100,000V and a current of 10A or a current of 100,000A and a voltage of 10V

    It's not a case of "increasing voltage and decreasing current as a result", it's about making a decision about how you set up the circuit/system to achieve this power transfer.
    Thanks for the answer.
    I have understood that.
    But why doesn't increasing V result in increasing I: V=IR
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    (Original post by Usernametaken)
    Thanks for the answer.
    I have understood that.
    But why doesn't increasing V result in increasing I: V=IR
    you appear not to have understood.

    The distribution cables are not the load.

    If you could get everything electrical in the city switched off at the same time, the long distance cables leading to the city would still be at a high voltage, they wouldn't have any current in them at all though.
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    In addition to what Joindeup has said, you may find this diagram useful.

    On the left is a power source which you want to deliver 1000W to the load on the right.
    The resistance of the cable is fixed at 10 Ohm
    In the 1st case you arrange the circuit so that the current is 1A
    If you do the calculations you can see that the power lost in the cables will be I2R and equals 10W, meaning 990W is delivered to the load.
    In the second case you arrange it so that the current is 2A
    If you do the calculations again you see that you now only deliver 960W to the load.
    Clearly, in order to do this, the resistance of the load + cable has changed, but the resistance of the cable is constant.
    Increasing V allows you to decrease I
    Decreasing I means you lose less power in the cables and deliver more to the load.
    The load resistance is not the same in the two cases so when you apply V=IR you have to think exactly what you mean by R in this formula, and realise that it isn't the same in the two cases. So there is no contradiction in the statement that increasing V means you can decrease I.

 
 
 
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