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AC vs DC power loss across a resistor

  1. Nov 20, 2012 #1
    I was recently informed that AC is the preferred method of transmitting power because it suffers significantly less power loss going from the generating station to where ever it needs to go when compared to DC losses.

    In my introductory electricity class we learned that power through a given circuit follows P=IV where P is power, I is current, and R is the resistance.

    So why would the power loss be different for an AC vs DC transmission?

    For AC I know there would be a low amperage and a high voltage so that you could later step the current up through a transformer so that means a low current and high voltage.

    DC would just be a high voltage and low current so either way we're stuck with the same power delivery.

    Even going through the substitution of P=IV=I2R the last term across a load is still P=V2/R so I just don't see how there is any difference mathematically between AC and DC power losss across a resistor.

    However, conceptually it does make sense. With more current, I conceptually see why more "collisions" would occur in the wire and cause more power loss but I can't wrap my head around it mathematically.

    Any help is appreciated!
  2. jcsd
  3. Nov 20, 2012 #2
    It is easy to convert AC to high voltage for transmission and back to lower voltages for local distribution, doing the same thing with DC is much more complicated and expensive.
    Transmission losses are calculated as I*I*R if you halve the current you get one quarter of the transmission losses. as to V*V/R or I*V remember you dont drop your total voltage over the transmission lines.
  4. Nov 20, 2012 #3


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    AC is better over short distances (<1000km), where the better transformators matter. If you want to transmit power over large distances, DC is better, and the losses in the transformation (they do not depend on the distance) are acceptable.

    In both cases, high voltage (between the cables, not IN the cables!) and low current are important, as Jobrag showed.
  5. Nov 20, 2012 #4
    The logistics part I completely get but could you elaborate on

    because in my mind I still see I2* R = V2/R.

    What do you mean by you don't drop the total voltage of the transmission lines.
  6. Nov 20, 2012 #5

    Resistance is the same whether AC or DC. AC might be preferred for other reasons like ease of changing voltages with a transformer, but not for resistance. Resistance is determined by the physical shape and materials of the conductors, not by what voltage or current are sent through the wires.

    Those formulas are correct for both AC and DC, provided that the current and voltage are in phase in AC.

    Ask the person who told you that.

    That sentence does not make any sense to me.

    For long distance transmission, high voltage and low current is preferred for both AC and DC to minimize IR losses.

    More current causes higher IR losses in both AC and DC, so that does not make any difference.

    DC power had one advantage and one disadvantage. The DC current is steady, while the AC goes to zero and reverses. That means that AC current has an RMS value that is 30% less than DC for the same wire diameter. So DC transmission can get by with smaller diameter wire, which saves a lot of money. The disadvantage is that inverters to change the voltage of DC are more expensive.

  7. Nov 20, 2012 #6
    I have no issues with the logistics of AC and DC but would you please elaborate on your last sentence? I can't see the mathematical difference in P=I*V vs P=I2*R vs P= V2/R which is my big issue.
  8. Nov 20, 2012 #7

    OK, according to P=I*V, if you double the voltage, the transmission line needs to carry half the current to deliver the same power to a load somewhere. The transmission line resistance is constant at R, so if the current is halved, then according to (I^2)*R, the power loss in the transmission line is one quarter of what it was before.

  9. Nov 20, 2012 #8


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    You have to consider two different voltages:

    V between the cables, and V' as voltage drop in the cables itself.
    P=V*I is the total transmitted power, and P'=V'*I is the power lost in the cables.
    V'=I*R, so P'=I^2*R
    The fraction which is lost in the cable is then given by (I^2*R)/(VI) = IR/V = PR/V^2 = I^2R/P

    As you can see, a higher voltage between the cables and a lower current in the cables is better.
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