0
\$\begingroup\$

Given all other factors are same, with only the wire length increased (for example, 0.5meter increase to 50 meters,) will the current consumption/usage increase as well? Can anyone explain?

\$\endgroup\$
7
  • 1
    \$\begingroup\$ You need to show a circuit and point out what the wire is that you are talking about. \$\endgroup\$
    – Andy aka
    Nov 5 '19 at 12:53
  • 1
    \$\begingroup\$ Do you know Ohm's Law? \$\endgroup\$
    – HandyHowie
    Nov 5 '19 at 12:54
  • \$\begingroup\$ the wire size is exactly the same, just the length is different, as indicated, from 0.5meter to 50meters. \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:01
  • \$\begingroup\$ ohm's law, do you mean v=ir ? \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:01
  • 2
    \$\begingroup\$ Voltage drop, the voltage will be lower at the end of the cable because of resistance. \$\endgroup\$
    – Codebeat
    Nov 5 '19 at 13:12
6
\$\begingroup\$

Given all other factors are same, with only the wire length increased (for example, 0.5 meter increase to 50 meters,) will the current consumption/usage increase as well?

Generally there are two types of load to consider.

Dumb loads

This includes things like lightbulbs, heaters and motors whose current draw varies with voltage. As the voltage drops the current will fall too. Adding extra cable will cause a voltage drop at the load so the current drawn will fall.

Smart loads

This includes things like computer / TV power supplies, motor speed controllers, LED lighting PSUs, etc. These differ in that they regulate the output they provide and when the voltage decreases they increase the current draw from the mains to provide the required output power. Adding extra cable will cause a voltage drop at the load so the current drawn will increase.

\$\endgroup\$
3
  • \$\begingroup\$ just three light bulbs currently as loads, with distance of 50meters, the 12v battery voltage drop faster in comparison to short distance, 0.5meter. But this is just visual reading on daily basis though. comment and/or thoughts? \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:20
  • \$\begingroup\$ Get all your numbers: (1) Work out the current from the bulb wattage (P = VI). (2) Work out the cable resistance from the wire gauge and length. Don't forget to include the return leg. There are many online calculators for metric and AWG. (3) Calculate the voltage drop on the cable from Ohm's Law (V = IR). I have no idea what "this is just visual reading on daily basis" means. \$\endgroup\$
    – Transistor
    Nov 5 '19 at 13:25
  • \$\begingroup\$ visual reading as in read the voltage drop periodically by measuring the 12v battery using multimeter \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:27
2
\$\begingroup\$

Anything is possible, depending on the type of load: current may decrease, increase or stay the same.
Let \$R_{line}\$ be the resistance of the line which increases with length; here are examples for each case:

  • If the load is a constant resistance load (e.g. a good resistor) or a constant voltage load (e.g. Z-diode in reverse direction or a LED in forward direction (both idealized)) current will decrease if resistance in line increases: \$I=\frac{V}{R_{load}+R_{line}}\$ or \$I=\frac{V-V_{diode}}{R_{line}}\$

  • If the load is a constant power load, e.g. a DC/DC converter with constant load or a SMPS for a laptop computer that provides constant power over a voltage range (assuming effciency will stay about the same), current will increase if resistance in line increases (of course only to a certain extend until minimum operating voltage at load is reached):
    \$I=\frac{P}{V-V_{line\_drop}}=\frac{P}{V-IR_{line}}\$

  • If the load is a constant current supply, e.g. for a LED array, of course current will stay the same if the line resistance increases (of course only to a certain extend until minimum operating voltage at load is reached):
    \$I=I_{const}\$

\$\endgroup\$
1
  • \$\begingroup\$ thank you for valuable inputs. \$\endgroup\$
    – Jasonw
    Nov 6 '19 at 12:30
1
\$\begingroup\$

The resistance of any conductor is given by the formula: R=rho*(L/A).

Rho is the property of the material. L is the length of the conductor and A is the cross sectional area of the conductor. If everything is kept constant here, increasing the length would increase the resistance of the wire.

Increased resistance means decrease in current flow. Ohm's Law!!

\$\endgroup\$
2
  • \$\begingroup\$ Please let me know what's wrong with this answer. \$\endgroup\$
    – G-aura-V
    Nov 5 '19 at 13:10
  • \$\begingroup\$ downvote was not from me. \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:11
1
\$\begingroup\$

Current flow will decrease. Wire has resistance. The longer the wire, the higher the resistance, the less current will flow.


If you connect a load right at the battery and measure the voltage at the load then the voltage will drop very little.

If you connect the load to the battery with very long wires and measure the voltage at the load, then you will measure a lower voltage.

If you simultaneously measure the voltage at the load and at the battery, then you will find the voltage at the battery to be higher than at the load. This is due to the resistance of the wires.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ What if the wire is part of a 4-20mA control loop? What if it is connected to a constant current source for LEDs? What if it carries a 100MHz signal? \$\endgroup\$ Nov 5 '19 at 13:10
  • \$\begingroup\$ @JRE , thanks for your concise response. With the load further away (in this example 50meters length of wire), it seem the 12v battery voltage drop faster in comparison with short distance wire, 0.5meter. But this is just a feeling and reading on daily basis. \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:17
  • 3
    \$\begingroup\$ @ElliotAlderson: Given the basic level of the question, I didn't think it appropriate to muddy the waters with concepts the OP probably hasn't encountered yet. If you feel that those cases need to be addressed, then you are welcome to post your own answer and expand on the concepts you've mentioned. \$\endgroup\$
    – JRE
    Nov 5 '19 at 13:22
  • \$\begingroup\$ Given the basic level of the question, I think it's vital that you make it clear what assumptions you have made. \$\endgroup\$ Nov 5 '19 at 13:24
  • 1
    \$\begingroup\$ The person asking the question isn't in a position to understand those assumptions. \$\endgroup\$
    – JRE
    Nov 5 '19 at 13:27
1
\$\begingroup\$

$$ R_F = \frac {\rho \ell} {A}$$

Resistance of feeder will increase as length increases. \$R_F\ \alpha\ \ell\$

Regardless of the circuit (AC/DC, 3-phase, single-phase), material and wire size, voltage at the load will be less than the supply voltage (Basic KVL where load and feeder form a series circuit).

If a constant power load is connected to your increasing length load, resistance/impedance increases, current increases because voltage drop to feeder increases.

Impact depends on relationship between feeder resistance and load. In a lab, a short feeder is irrelevant when powering a 1kΩ. But if the feeder is not sized correctly (cross-sectional area), say supplying a 10hp motor at 250ft, the load may be affected to an extent that the load does not operate correctly. Too much voltage is lost to the feeder. Motor may start at no-load, but fail under load.

\$\endgroup\$
1
  • \$\begingroup\$ thank you for valuable inputs. \$\endgroup\$
    – Jasonw
    Nov 6 '19 at 12:31
0
\$\begingroup\$

A longer wire has a proportionally larger resistance, so the current usage will not increase. Rather would it decrease as the total load increases, but for most circuits the resistance of the connected wire is negligible compared to the "load" itself so that you won't see any change in power consumption or drawn current.

\$\endgroup\$
5
  • \$\begingroup\$ "as the total load increases" - why would the load increase? \$\endgroup\$
    – HandyHowie
    Nov 5 '19 at 13:08
  • \$\begingroup\$ If you consider the wire as part of the supplies load \$\endgroup\$
    – po.pe
    Nov 5 '19 at 13:09
  • \$\begingroup\$ What if the wire is part of a 4-20mA control loop? What if it is connected to a constant current source for LEDs? What if it carries a 100MHz signal? \$\endgroup\$ Nov 5 '19 at 13:10
  • 2
    \$\begingroup\$ I see you're going for some edge cases here. I was assuming by "given all factors are the same" the author was talking about a normal voltage supplied circuit and not a 4-20mA control loop or a constant current source where the name already implies that the current would be constant. I think this anwer is appropirate enough for this question but you're free to give your own. \$\endgroup\$
    – po.pe
    Nov 5 '19 at 13:14
  • \$\begingroup\$ @ElliotAlderson nothing fancy, just three lightbulbs as the loads and am new to electrical, would be glad to read response and learn knowledge from various people. \$\endgroup\$
    – Jasonw
    Nov 5 '19 at 13:22

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