# Volt drop in a cable

Joined May 19, 2013
214
Consider a constant supply voltage VS and a consumer with a power rating of P watts.
Then they calculate the voltage drop on the cable connecting the supply with the consumer using Ohm's law like this:

1. Calculate current drawn by consumer: I = P/VS
2. Calculate voltage drop on cable: VD = R*I, where R = resistivity*length/area

So they assume that the consumer will draw that current. But the current through the circuit is I = VS/(RC+R) and not I=VS/RC = P/VS, RC being the resistance of the consumer.

Why they first assume that the current will go through the consumer like all the supply voltage will fall on it? The voltage that the consumer will see is less than the supply voltage. Thus the current drawn is smaller too.

#### Papabravo

Joined Feb 24, 2006
21,308
The resistance of the cable in both directions is added to the consumers load and the current draw is limited by the magnitude of three resistances in series. As in all series circuits, the voltage drop across the load will be less than, sometime much less than, the VS of the constant voltage supply.

#### wayneh

Joined Sep 9, 2010
17,504
Why they first assume that the current will go through the consumer like all the supply voltage will fall on it? The voltage that the consumer will see is less than the supply voltage. Thus the current drawn is smaller too.
That is a simplifying assumption, that the drop across the supply wiring will be very low compared to the drop at the load. Like all simplifying assumptions, they are useful but not always true.

Joined May 19, 2013
214
The resistance of the cable in both directions is added to the consumers load and the current draw is limited by the magnitude of three resistances in series. As in all series circuits, the voltage drop across the load will be less than, sometime much less than, the VS of the constant voltage supply.
I understand that.

I will just go with the simplifying assumption.

#### MrAl

Joined Jun 17, 2014
11,715
Hi,

The power company also loses money when the line voltage drops.

Joined May 19, 2013
214
In the book it also says:

"The choice of cable size depends on the current drawn by the consumer. The larger the cable used then the smaller the volt drop in the circuit, but the cable will be heavier. This means a trade-off must be sought between the allowable volt drop and maximum cable size. "

The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?

Joined May 19, 2013
214
I think it refers to the cross-sectional area, not its length. The choice of words is very poor.

#### MrAl

Joined Jun 17, 2014
11,715
In the book it also says:

"The choice of cable size depends on the current drawn by the consumer. The larger the cable used then the smaller the volt drop in the circuit, but the cable will be heavier. This means a trade-off must be sought between the allowable volt drop and maximum cable size. "

The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?
Hi,

The key is in the first sentence. You dont adjust the *length* of the cable based on the current drawn, you adjust the diameter. Therefore that idea carries over to the next sentence where 'larger' would mean a larger diameter. They could have made it more clear though for someone who doesnt already know how this works.

#### profbuxton

Joined Feb 21, 2014
421
"The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?"

WRONG: Larger cable is LESS resistance=Lower volt drop. Look up cable supplier tables of resistance per metre.
Usually maximum allowable volt drop is 10%.(in OZ).

Joined May 19, 2013
214
Hi,

The key is in the first sentence. You dont adjust the *length* of the cable based on the current drawn, you adjust the diameter. Therefore that idea carries over to the next sentence where 'larger' would mean a larger diameter. They could have made it more clear though for someone who doesnt already know how this works.
I know but english is not my first language. "Larger" in my language means both "longer" and "larger in diameter", but mainly "longer". Hence my confusion.

Joined May 19, 2013
214
"The larger the cable, the bigger its resistance is. And thus the higher the volt drop will be. Why the volt drop gets smaller?"

WRONG: Larger cable is LESS resistance=Lower volt drop. Look up cable supplier tables of resistance per metre.
Usually maximum allowable volt drop is 10%.(in OZ).
In that post, for me "larger" meant "longer". The bigger the length, the bigger the resistance.

#### wayneh

Joined Sep 9, 2010
17,504
It gets worse. A "larger" (greater diameter) cable will have a smaller gauge number. A 10 gauge wire is much fatter than a 20 gauge wire.

#### profbuxton

Joined Feb 21, 2014
421
Understood. Here we mean bigger cross-sectional area when we say "larger". When we say "longer" is means just that., longer length.

#### MrAl

Joined Jun 17, 2014
11,715
I know but english is not my first language. "Larger" in my language means both "longer" and "larger in diameter", but mainly "longer". Hence my confusion.
Well see if you trusted what you knew already you would know they were not talking about the length. You already knew that an increase in length would cause a higher resistance, thus they could only be talking about the diameter.
But for someone who did not already know that they could be confused, so the writer could have done better.

I see this kind of thing quite often on the web. The writers are not very succinct.
The best way to write a technical piece is to try to think of all the ways it might be interpreted and then make it clear enough so that it can only be taken one way. This sometimes requires redundancy which means more typed characters for the authors which they try to avoid most likely. What i like to do is start with a little thing like, "In other words..." and then restate the information in a different way. This helps to avoid a misinterpretation.
For example, "In other words, when the diameter increases the resistance goes down.".

...going back to sleep now

Last edited:

Joined May 19, 2013
214
Well see if you trusted what you knew already you would know they were not talking about the length. You already knew that an increase in length would cause a higher resistance, thus they could only be talking about the diameter.
But for someone who did not already know that they could be confused, so the writer could have done better.

I see this kind of thing quite often on the web. The writers are not very succinct.
The best way to write a technical piece is to try to think of all the ways it might be interpreted and then make it clear enough so that it can only be taken one way. This sometimes requires redundancy which means more typed characters for the authors which they try to avoid most likely. What i like to do is start with a little thing like, "In other words..." and then restate the information in a different way. This helps to avoid a misinterpretation.
For example, "In other words, when the diameter increases the resistance goes down.".

...going back to sleep now
The best way for me is saying those things in terms of formulas. And after the formulas are stated, words can be used to describe those formulas.
Using only words can be annoying, especially if I do not know the formulas before. Especially if I do not know the whole subject before. You try to understand those words and what the author wants to say and somewhere else you see someone explaining the same thing in terms of formulas and everything clicks. And then you reread the words that you could not understand initially and say: 'yeah, it is easy".

#### MrAl

Joined Jun 17, 2014
11,715
Hi,

Yes, formulas are always nice, and ones that show relative relationships are nice too.
For example, the resistance goes down as the square of the diameter but goes up in direct proportion to the length:
R=K1*(d1/d2)^2
R=K2*Length