In transmission cables, why does power loss increase when length of conductor is increased? According to the formulas V=IR and P=I²R, When we increase the length, the resistance increases, while the potential difference is same, the power loss decreases as current decreases. So why does the increase in length of wire, hence the increase in resistance, cause the transmission line loss? The power should decrease as per my understanding.

  • 4
    $\begingroup$ Does this answer your question? High voltage power lines - clarification of energy loss $\endgroup$
    – Farcher
    Feb 23 at 22:01
  • $\begingroup$ @Farcher No my question is different. It isn't about the amount of powerloss $\endgroup$ Feb 23 at 23:25
  • $\begingroup$ Power loss, energy loss all quite the same. $\endgroup$ Feb 24 at 0:12
  • $\begingroup$ @AlbertusMagnus I know, but that isn't my question $\endgroup$ Feb 24 at 0:16
  • 1
    $\begingroup$ Are those the actual words used in your textbook, or is this a translation from another language (mabye a machine translation)? Because if your book was written in English and it said that, you should throw it out. $\endgroup$
    – The Photon
    Feb 24 at 3:46

2 Answers 2


Since you have that $$P=I^2R,$$ it is reasonable to think that because an increase in resistance leads to a drop in current that it might be possible this could happen in a way that doesn't always lead to power loss, the change could be such that there is break even, or even less power loss in higher resistance lines, however, the reality of this is not so in practical applications. In real life practical applications, the current on the transmission lines is largely determined by the consumer power demand and not that of the line resistance, so you get in general, an increases in ohmic losses for an increase in transmission line length. A fact that power plant engineers mitigate for by using AC power transmitting at high voltages.

  • $\begingroup$ I agree with everything except I don't understand the argument about power being quadratic in current. Would any of the conclusions change if it were linear in current? Cubic? $\endgroup$
    – Puk
    Feb 24 at 1:00
  • 1
    $\begingroup$ @Puk Not really. I see what you are saying; I will edit that part out, thanks. $\endgroup$ Feb 24 at 1:33

The power you send in on the source side: $P_{in}=U \cdot I $ The current that flows through the line causes a voltage drop $U_{line}=R_{line} \cdot I$ . Thus the power you get out is: $P_{out}=U_{out} \cdot I = (U-U_{line}) \cdot I = U \cdot I - U_{line} \cdot I = P_{in} - I^2 R_{line}$

So your Power Budget:

  • You send in $P_{in} = U \cdot I $
  • Your loss $P_{loss} = I^2 R $
  • You get out: $P_{out}=P_{in}-P_{loss}$

One easy way to make the loss smaller is to use a higher voltage. With a higher voltage you need a lower current for the same transmitted power. An as the loss is poportional to $I^2$ it is reduced. But as the voltage gets higher you need to make sure the wires are spaced far apart or else you get sparks over your line. This is why high power transmission lines look the way that they are.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.