Transmission line loss

Question

Let us suppose that a transmission line is carrying power to a resistor $R$ The resistor $R$ requires a power of $P=IV$.

Applying the most simple case where a transmission line is supplying a current $I$. The Ohmic loss is $P_c=I^2R_c$ for the cable.

Putting $I=P/V$

$P_c=(P^2/V^2)R_c$ So to decrease Ohmic loss we need to increase $V$ but this is the potential difference across the resistor, right? I have read other questions here, but it still doesn't make sense. Please help.

What my textbook says

Consider a resistor R to which a power P is to be delivered via a transmission cable having res. $R_c$ to be dissipated finally. If $V$ is voltage across $R$ and $I$ the current then

$P=VI$

The connecting wires from the power station to the resistor has res. $R_c$. The Ohmic loss is $P_c=I^2R_c$

$P_c=(P^2/V^2)R_c$

Thus to drive a device with $P$ the power wasted in the connecting wires is inversely proportional to $V^2$.

A few paragraphs later...

The transmission cables from power stations are hundreds of miles long and their resistance $R_c$ is considerable. To reduce $P_c$, these wires carry current at enormous values of $V$

This is a duplicate in that the voltage $V$ is ill defined and it is this lack of clarity which causes the ambiguity. Have a read of https://physics.stackexchange.com/a/248233/104696 — Farcher, Dec 27 '17 at 18:55

score 0 · Accepted Answer · answered Dec 27 '17 at 17:26

0

to decrease Ohmic loss we need to increase V but this is the potential difference across the resistor, right?

One option is to transmit the power over long distances at high voltage, then use a transformer near the load to reduce the voltage to whatever is required by the load.

For example, power may be transmitted across continents at 100's of kV, then reduced by a sequence of transformers to 110 or 220 V to power your home.

If your power source is DC rather than AC, then you could use a switching converter to do something similar. You'd first want to evaluate whether it's cheaper to include that switching converter in your system, or just use a larger diameter wire to reduce the $R_c$ term.

answered Dec 27 '17 at 17:26

The Photon

27,393

I agree, but my textbook mentions that since we increase $V$. The Ohmic loss is reduced in the cable, which I don't think is true as $V$ is meant for the resistor – Daivya Jadeja Dec 27 '17 at 17:29
Without seeing exactly what your textbook says, I can only guess what's meant. Maybe they meant increase V and (load) R both so that the same power is delivered? – The Photon Dec 27 '17 at 17:30
They made $R$ constant so yeah – Daivya Jadeja Dec 27 '17 at 17:31
Maybe they just meant increase V by $I R_c$ so that the load gets the voltage that was originally intended? Including a complete quote (with context) from your book into your question might make things more clear. – The Photon Dec 27 '17 at 17:32
I think it would be an error? – Daivya Jadeja Dec 27 '17 at 17:37
@DaivyaJadeja, I'm not going to say it's an error if you don't share the exact quote with enough context to make clear what they meant. – The Photon Dec 27 '17 at 17:37
"Consider a device R to which a power P is delivered via transmission cables having res. $R_c$ to be dissipated finally . – Daivya Jadeja Dec 27 '17 at 17:39
@DaivyaJadeja, please edit your question to include all the information needed to answer it. – The Photon Dec 27 '17 at 17:39
Just a minute.. – Daivya Jadeja Dec 27 '17 at 17:40
The text you quoted doesn't say anything about increasing V to reduce loss. It just says power loss is proportional to $1/V^2$. And the sentence that says it says P is fixed, not R. I don't see any error there. – The Photon Dec 27 '17 at 17:50
But then increasing $V$ shouldn' affect the power loss in the cables as they are independent? – Daivya Jadeja Dec 27 '17 at 17:52
For a fixed resistive load, it will increase the power lost in the line, because the line loss is $I^2 R_c$ and I will be increased by increasing V – The Photon Dec 27 '17 at 17:54
Unless, like I said in my answer, you can increase the transmission voltage without increasing the voltage at the load. – The Photon Dec 27 '17 at 17:55
Edited once more – Daivya Jadeja Dec 27 '17 at 18:09
My answer already addresses that. They use transformers near the load to reduce the voltage before delivering power to the load. – The Photon Dec 27 '17 at 18:22

score -1 · Answer 2 · answered Dec 27 '17 at 17:45

-1

The V in Ohm's law is the voltage drop across the load.

You are trying to reduce the voltage drop and so the power lost along the line.

answered Dec 27 '17 at 17:45

Martin Beckett

30,766

the load includes the wires and the resistor? – Daivya Jadeja Dec 27 '17 at 17:50
This is a very common error, but I don't see where OP made it. They carefully distinguished $R_c$ (line resistance) from $R$ (load resistance). – The Photon Dec 27 '17 at 17:51
@ThePhoton, all 'V' are really dV, there are two dV's the one along the transmission lone and the one across the load. Or if you prefer there is a single I and two R's and P = I^2 R – Martin Beckett Dec 27 '17 at 18:40
@MartinBeckett, I still don't see where OP made any mistake. They consistently used $V$ to be the voltage across the load and never used it where they should have used the voltage across the line. – The Photon Dec 28 '17 at 00:59
@ThePhoton, the only 'losses' are across the line if we assume the load at the end is all useful work. If you ignore voltage drop along the line then there is no benefit in a high voltage, power is just =IV. If you want to increase efficency then the higher voltage allows a lower current in the line (where the P=IV is wasted) for the same P=IV at the load, that is why losses are reduced by V^2 – Martin Beckett Dec 28 '17 at 18:40
@MartinBeckett, the losses are due to the resistance of the line. But OP has written a correct expression for those losses without using a term for the voltage across the line (hint: they wrote the current through the line in terms of the power and voltage delivered to the load). So your answer does not address what they were asking about. – The Photon Dec 28 '17 at 21:10

Transmission line loss

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