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# Relationship between power loss in wires and voltage watch

1. Hi,

I'm wondering is there an equation to link voltage generated at for example power stations and the power loss in cables.

I understand that:
Take P as power, V as voltage and I as current.

P(generation)= V(gen) x I(gen)

P(cable loss) = V(cable) x I(gen)

P(cable) = I(gen)^2 x R(cable)

P(cable) = V(cable)^2 / R(cable)

But what is the relationship between P(cable loss) and V(at generation)???

Surely if the answer isn't at a university level will come in handy when explaining why dropping the current will reduces power loss. At the moment I can only explain it using P(cable) = I(gen)^2 x R(cable), but no way of explaining the relationship of P(cable loss) and V(at generation).
2. (Original post by WannabeScientist)
Hi,

I'm wondering is there an equation to link voltage generated at for example power stations and the power loss in cables.

I understand that:
Take P as power, V as voltage and I as current.

P(generation)= V(gen) x I(gen)

P(cable loss) = V(cable) x I(gen)

P(cable) = I(gen)^2 x R(cable)

P(cable) = V(cable)^2 / R(cable)

But what is the relationship between P(cable loss) and V(at generation)???

Surely if the answer isn't at a university level will come in handy when explaining why dropping the current will reduces power loss. At the moment I can only explain it using P(cable) = I(gen)^2 x R(cable), but no way of explaining the relationship of P(cable loss) and V(at generation).
Hello and welcome to the TSR physics forum.

The cable has resistance and any current flowing in that cable will cause work to be done which is lost as heat.

The heating effect is a function of current (P= V x I), and since it's the voltage pressure at the generator which provides the force to drive current around the circuit we have:

P = V x I

I = V/R and substituting back into the first eq

P = V x (V/R) = V2/R

also

V = I x R and again substituting back into the first eq

P = (I x R) x I = I2R

Since R is the cable resistance, the power lost in the cable is by definition a function of the voltage dropped across the cable. i.e. generator source voltage - p.d. at load together with that cable resistance.

Pcable = (Vgen - Vload)2 / Rcable

The current in the cable is a function of the total load presented to the generator. i.e. the sum of the cable and load resistances or impedances.

And since

Igen = Icable = Vgen / (Rcable + Rload)

then increasing the load resistance or reducing the cable resistance will result in a lower voltage dropped across the cable and hence reduce the power loss in the cable.
3. (Original post by WannabeScientist)

x.
Are you OK with this now?
4. Thank you for the reply,

If P(cable) = V(voltage drop in cable) x I(gen). If I(gen) is increased, will V(cable) not decrease proportionally?
If P(cable) is proportional to V(cable)^2. Would that not cancel the effect of P(cable) = I(gen)^2/R ?

Not sure if I'm missing something here, with P(cable) = I(gen) x V(cable), if you increase the I(gen) then voltage drop in cable must go up accordingly? As the heating effect(power loss) is a function of current though the cable.

What about the power generated, P(gen) = I(gen) x V(gen), If I(gen) is increased then V(gen) and P(gen) must go up?

Thank you
5. (Original post by WannabeScientist)

If P(cable) = V(voltage drop in cable) x I(gen). If I(gen) is increased, will V(cable) not decrease proportionally?
No. Vcable will still increase because it's still ohms law V = I x R

P = V2 / R

The current supplied by the generator will only increase if a) the generator voltage is increased or b) the load presented to the generator is increased. i.e. total load resistance is decreased.

But because the resistance of the cable has not changed, then an increase in current flowing through the cable will under all circumstances, produce a larger voltage drop across the cable and hence the power loss will still increase exponentially because of that V2 term.

The cable together with the load resistances, forms a potential divider.

(Original post by WannabeScientist)
If P(cable) is proportional to V(cable)^2. Would that not cancel the effect of P(cable) = I(gen)^2/R ?
No. Think about what causes the current to increase.

Note that Rcable is present in both the numerator and divisor of that potential divider equation above. As the load resistance falls, the current supplied by the generator will increase but more voltage is dropped across the cable resistance. When the load resistance is zero, all of the generator voltage is lost across the cable resistance. i.e. the cable forms the total load presented to the generator.

Hence more power is lost in the cable.

Hence as the load resistance falls, because of the static cable resistance, the p.d. developed across the load resistance must decrease while the p.d. across the cable increases.

(Original post by WannabeScientist)

Not sure if I'm missing something here, with P(cable) = I(gen) x V(cable), if you increase the I(gen) then voltage drop in cable must go up accordingly? As the heating effect(power loss) is a function of current though the cable.
Yes. Exactly.

(Original post by WannabeScientist)
What about the power generated, P(gen) = I(gen) x V(gen), If I(gen) is increased then V(gen) and P(gen) must go up?

Thank you
The generator is the source of energy.

If Igen has increased, that can only be because Vgen is increased (motor speed turned up) or the load resistance has decreased.

The power supplied by the generator is dependent on both of these variables, it not something that is independent of them.

Do not confuse the power capability of the generator (i.e. the power the generator is able to supply before running out of puff) with the power demanded by the load.
6. Sorry I forgot to specify my intended question is from a transformer topic,

From you explanation, I now understand, in exam format that:

Step up transformer helps to reduce power loss in cables by increasing the voltage, P=I x V, so current is decreased. P = I^2 x R therefore power loss in cable is decreased.

or

To explain it in terms of voltage drop: V=I x R, current is decreased, voltage drop is also decreased, P = V^2 / R, therefore power loss in cable is decreased.

7. (Original post by WannabeScientist)
Sorry I forgot to specify my intended question is from a transformer topic,

From you explanation, I now understand, in exam format that:

Step up transformer helps to reduce power loss in cables by increasing the voltage, P=I x V, so current is decreased. P = I^2 x R therefore power loss in cable is decreased.

or

To explain it in terms of voltage drop: V=I x R, current is decreased, voltage drop is also decreased, P = V^2 / R, therefore power loss in cable is decreased.

You are very welcome.

I had an inkling that is where you were trying to get to.

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