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.
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my2cts
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Hufaiza Hufaiza
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4Does this answer your question? High voltage power lines - clarification of energy loss – Farcher Feb 23 '24 at 22:01
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@Farcher No my question is different. It isn't about the amount of powerloss – Hufaiza Hufaiza Feb 23 '24 at 23:25
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Power loss, energy loss all quite the same. – Albertus Magnus Feb 24 '24 at 00:12
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@AlbertusMagnus I know, but that isn't my question – Hufaiza Hufaiza Feb 24 '24 at 00:16
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1Are 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. – The Photon Feb 24 '24 at 03:46
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In your first equation $V$ is the voltage drop across the cables and in your second equation $I^2R$ is the power (rate of energy loss) in the cables. As the current drawn from the power station increases due to an increased demand at the consumer end the energy loss increases. If the transmission cables ate longer then so is the resistance of the cables which means that there is a greater loss of electrical energy. – Farcher Feb 24 '24 at 09:36
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@Farcher Yes! got it! Thank you – Hufaiza Hufaiza Feb 24 '24 at 10:33
2 Answers
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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.
Albertus Magnus
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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? – Puk Feb 24 '24 at 01:00
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1@Puk Not really. I see what you are saying; I will edit that part out, thanks. – Albertus Magnus Feb 24 '24 at 01:33
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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.
mond
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