Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power Dissipation along power lines and Resistance value

  1. Nov 23, 2014 #1
    Hello Forum,

    It is well known that the electrical power dissipated (in the form of heat) by a resistor of resistance R connected to a battery of voltage V is:

    P_diss = (V^2)/ R or equivalently P= (I^2)*(R)

    Clearly the smaller the resistance R, the more the dissipated power because R is at denominator in the first equation and in the second equation a small R causes a large current I which dominates being squared.
    That said, the resistance of wires in power lines is sought to be as small as possible "to reduce losses". But that seems to be against what those two equations above would predict....
    Does anyone have any clarification on why the small resistance in power line wires is desirable and can cause the least dissipation?
    I know that power lines carry AC power (P=I*V) at a high voltage and lower current. Maybe the fact that the current is low, together with the small R, help produce small losses.

    However, I have heard that there is also DC power line transmission which seems to be very efficient.....
  2. jcsd
  3. Nov 23, 2014 #2


    User Avatar
    Science Advisor

    The power used in an electrical network depends on the resistance of all the loads on it (lights, TVs cookers etc) and also the resistance of the wires in series with the loads.

    If the wire resistance is low ( ie thicker wires) then there will be less voltage dropped across them and less power lost as heat.

    So, if power is I*I * R and I is approximately constant, then the power loss in the wires will be lower if R is smaller.
  4. Nov 23, 2014 #3


    User Avatar
    Science Advisor
    Gold Member

    This resistive loss is generally much lower than reactive power loss.

  5. Nov 23, 2014 #4
    A very basic model of a transmission line is like that of a resistive divider. If we look at two examples I think it will be clear.


    First Example:

    V1 = 120 Volts.
    R_Line = 0.01 Ohms.
    R_Load = 2.00 Ohms.

    There's only one current to calculate since everything is in series:

    I = V1/(R_Line+R_Load) = 120/(0.01+2.00) = 59.70 Amps.

    We can calculate the power loss in the transmission line using either P = (V^2)/ R or P = (I^2)*(R). Let's do both. The important thing to note when using (V^2)/ R is that "V" is the voltage across the resistor, not V1.

    Using (V^2)/R, we first need to find out V:

    V = V1 - V_Load. (V_Load is between the two yellow dots).

    We know V = I*R, and we know R_Load and I (current). So V_Load = 59.70*2.00 = 119.4 V.

    Then V = V1 - V_Load = 120 - 119.4 = 0.6 V

    So P_Line = (0.6^2)/0.01 = 36 W.

    Now using (I^2)*R to find power:

    (I^2)*R_Line = (59.70^2)*0.01 = 35.65 W

    A slight difference due to rounding but practically the same. A power loss of ~ 36 W due to the transmission line resistance.

    If we repeat this in another example with a higher R_Line then you will see that there is in fact higher power losses. This is because R_Load is the main determinant of current, and R_Line is more or less an obstacle. In addition, we'd like V_Load to be the same as V1, 120 Volts. That could only happen if R_Line was zero.

    Second Example:

    V1 = 120 Volts.
    R_Line = 0.1 Ohms.
    R_Load = 2.0 Ohms.

    There's only one current to calculate since everything is in series:

    I = V1/(R_Line+R_Load) = 120/(0.1+2.0) = 57.14 Amps.

    We can calculate the power loss in the transmission line using either P = (V^2)/ R or P = (I^2)*(R). Let's do both. The important thing to note when using (V^2)/ R is that "V" is the voltage across the resistor, not V1.

    Using (V^2)/R, we first need to find out V:

    V = V1 - V_Load. (V_Load is between the two yellow dots).

    We know V = I*R, and we know R_Load and I (current). So V_Load = 57.14*2.00 = 114.29 V.

    Then V = V1 - V_Load = 120 - 114.29 = 5.71 V

    So P_Line = (5.71^2)/0.1 = 326 W.

    Now using (I^2)*R to find power:

    (I^2)*R_Line = (57.14^2)*0.1 = 326.50 W

    So with a transmission line with ten times the resistance, the losses are about ten times as much.

    A few important notes:
    In reality, transformers are used so that the V1 voltage is very high (tens of thousands Volts or higher), and then bucked back down to a more usable voltage for home (~120 V here in the US) & industry (usually ~480 V three phase in US). The higher voltage helps reduce the size of transmission lines since they will not have to carry as much current. Capacitance and inductance are also important factors to deal with (and ignored here to make it more simple). I imagine the answer to your last sentence about DC lines is that power factor (due to capacitance and inductance in lines and loads) is not an issue for power losses. Someone else could probably elaborate more on that, though.
  6. Nov 23, 2014 #5


    User Avatar
    Science Advisor

    HVDC lines are very efficient at long distances as the lines don't suffer from skin effect as the cable size is increased (the local intertie from Oregon to California runs at 1000 kV pole-to-pole) plus they have the advantage of not needing rotational synchronization of distant generators to control power flow and frequency. (Like the interconnection of different frequency 50/60hz grids)
  7. Nov 24, 2014 #6


    Staff: Mentor

    It is true, HVDC has all the advantages thst nsaspook said. In addition there is the advantage of avoiding excess reactive power generated by the capacitance between the line and ground. If you double the voltage, and hold current (and thus losses) roughly constant, the you double the power capacity. But you also create the problem of more excess reactive power. HVDC avoids that.

    But the costs of HVDC are high because the power must be converted from AC-DC at one end of the line and DC-AC at the other end. If you want to tap the line in the middle to serve local loads, another AC-DC converter is needed. Costs of these conversions can run as high as one billion USD. That is far too much to be motivated by losses alone.

    HVDC is used where the other advantages outweigh the cost.
  8. Nov 24, 2014 #7
    Cost benefit of HVDC over HVAC is for transmission distance > 600-700 km
  9. Nov 24, 2014 #8
    Thanks everyone.
    Great answers. Here some comments:

    1) Reactive (inductive or capacitative) losses are a type of loss which is very different from resistive loss. In reactive losses energy does not get dissipated, it only gets undelivered to the load and "trapped", stored undesirably along the line, correct?

    2) A transformer converts power at voltage V1 and current I1 into power at voltage V2 and current I2. An ideal transformer has zero power loss so the input power is equal to the output power. I have read about the maximum power transfer theorem


    which appears to state that in the best case only 50% of the input power from a source can be delivered to the load. If we look at the transformer as a load, how does its behavior with the max power transfer theorem?

    I think the confusion is about power efficiency and max power transfer, two different concepts. The power actually delivered to the load/ possible max power deliverable to the load is achieved when the load and source are impedance matched (conjugate). Transformers deal with power efficiency. Power efficiency is only 50% when max power transfer condition is achieved. As power efficiency approaches 100%, the power transfer approaches 0.
    Can we not look at a voltage transformer as a load which receives power from a source and then transfers that power to another load.

    3) HVDC lines don't suffer from skin effect. DC current travels and distributes itself uniformly inside the conducting wire while AC tends to concentrate only on the surface layer of the wire. Wires are covered by plastic insulation. Do the E field and magnetic field penetrate across that insulation or do they get attenuated? I think they get attenuated only a little bit. What is the purpose of that insulation? Mechanical protection only? I have read in several authoritative books that current travels inside the wires but energy travels OUTSIDE the wires from the source to the load in the form of electromagnetic fields. The E and B fields composing the EM field are outside the wires. I don't think the wire insulation affects them.

    Thanks for your help and patience,
  10. Nov 24, 2014 #9


    Staff: Mentor

    Don't feel bad. Before Thomas Edison, almost all power engineers were deceived by the maximum power transfer relationship. Edison realized that the optimum is to have zero internal resistance and to raise the voltage to whatever level you need. Maximum power transfer for a battery/load just does not apply for generators/lines/loads.

    Designers of bulk power transmission grids (think big power and long distances) have a long list of things to worry about. Losses are so small that they are way down their priority list. Losses are more significant for low voltage distribution systems (I.e. neighborhood scale)
  11. Nov 24, 2014 #10
    For EHV and UHV transmission, reduction of corona loses is of more importance than conduction loses.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook