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Power transmission

  1. Feb 1, 2007 #1
    As a general rule, if a company was to transmit a certain amount of electrical power than transmit at the highest voltage hence lowest current with P=IV. This is done in order to reduce waste power given by PowerWaste=I^2R. However voltage and current are proportional and related by V=IR. Therefore at a certain power transmission, if one was to maximise V and minimise I than the transmission line must have a very high resistance (although given from the PowerWaste equation high I is much worse than high R). Is my conclusion correct?
  2. jcsd
  3. Feb 1, 2007 #2


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    Transformers are used to maximize V (IV is constant). The point of this process is that energy loss is proportional to I2R. Therefore to minimize loss, keep I and R as low as possible.
  4. Feb 1, 2007 #3
    Last edited by a moderator: Apr 22, 2017
  5. Feb 1, 2007 #4


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    No. V=IR doesn't really apply here because the R isn't the real total resistance of the circuit. 90+% of the resistance is in the usage point in your house.
  6. Feb 1, 2007 #5


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    Careful here - the voltage drop from one end of the line to the other is related to IR, which ideally is very low. This is not the line voltage.
  7. Feb 1, 2007 #6
    Thanks for reminding me. I forgot what the thread was called so started a new one. It seems I haven't learnt my lesson from that thread.

    So think of the transformer not as something that compartmentalises the connection but a continuation of the connection which ultimately finishes in the household and starts at the electrical company. High V and low I is allowed because most of the resistance is in the household, on the other side of the transformer. Without the resistance in the household, however, high V and low I would be hard to achieve? Or would it be the case that the transmission lines would first blow up due to so much power in such low reistance transmission lines. This offcourse is a hypothetical situtation because the company would design the system by first calculating the amount of load in the households and then by transmitting the recquired power. But I would like to know what would happen anyway to test my knowledge. Is what I suggested correct?
    Last edited by a moderator: Apr 22, 2017
  8. Feb 1, 2007 #7


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    If you short-circuit the transmission lines, yeah, you'll probably blow something up. If there is simply no load, however, there is no draw and the amperage is low. An unloaded transformer does not act like a short.
    Since the load varies, obviously the power company can't simply choose the right generator and forget about it. A generator supplies the power required or fails trying. This manifests as a voltage drop and/or phase shift (not entirely clear on how that works). The power company compensates by increasing the mechanical power to spin the generators.
  9. Feb 2, 2007 #8
    Halliday, Resnik showed

    The generator’s equation is Vp=IpRp is calculated with Rp=(Np/Ns)^2Rs where Np is the number of coils on the generator’s side or primary side, Ns is the number of coils on the secondary side. Rs is the resistance on the secondary side. So most of the load does indeed come from the secondary side. The resistance in the transmission line is small and the smaller the better to reduce wastage.

    Consider a simple series circuit, if the load is removed than R is significantly reduced and I must increase with V=IR. This fundalmental principle should also apply to this more complex system we are discussing? More information down below.

    What does
    mean? The company can detect if load is low so when it detects that, it can automatically program its machine to reduce current production. Although for large companies, the usage of power is pretty predictable during what time period and dosen't vary much from year to year in general i.e. considering a large population.

    Here my take on removing the load.
    With the hypothetical experiment, If we reduce the resistance on the secondary side then Rp is reduced a lot. Which would mean Ip increases a lot with Vp constant (assuming the production of current doesn’t change). Since the transformer and coils don’t change, Vs will not change.

    By conservation of energy

    IpVp = IsVs

    If Ip increase. Vp and Vs stay put than Is also increase. So the result is more current in the primary and secondary circuit than before. Too much current may lead to damage to the transmission lines or they may blow up? Certainly more power is dissipated as waste power, P=I^2R. since I is increased and R reduced in both compartments. The squared term still implies more power is dissipated now. Another thing is that electrical power has increased in both compartments with increases in Ip and Vp. Did the extra power in the system come as a result of removing the load? In other words, when the load Rs was present, the power was shared between the load and the transmission line, now with the load removed most of the power is in the transmission line (a mini load can still be assumed to exist due to intrinsic resistance of the wire). Although a lot of the power in the line may be going to the production of heat as shown. Is this all correct?
    Last edited: Feb 2, 2007
  10. Feb 2, 2007 #9


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    Power companies don't rely on a single generator. Nor do they rely only on their own generators. In the USA and Canada at least, the various companies' distribution systems are linked together so that power produced by one company can be consumed by another company's customers.
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