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In DC, for example at 12V, will a longer power wire (lets say 10 meters vs 10 centimeters) cause a high current consumption?

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  • \$\begingroup\$ You should probably specify the nature of the load (the thing at the other end). Wires that are not connected on both ends are typically insignificant (at DC). \$\endgroup\$
    – uint128_t
    Jun 26 '16 at 0:57
  • \$\begingroup\$ Longer wires generate higher voltage drops, some loads will try and pull more current to compensate, but lots of things will just run slower, dimmer or quieter etc. Only stuff with switching or constant power supplies will try and pull more current. But unless you're pulling a lot of current down long, small leads, it's probably not going to be a noticable problem. \$\endgroup\$
    – Sam
    Jun 26 '16 at 1:07
  • \$\begingroup\$ You're asking two different questions and should probably read up on the relationship between power and current. Do they consume more power? Yes. Do they cause a higher current to run? No, in fact the reverse is true. At 12 V DC, though, you will not notice either effect unless we are talking a very large current. \$\endgroup\$
    – Jacob
    Jun 26 '16 at 15:58
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Power can be expressed as \$P = IV\$. We also know that Ohms law says \$V = IR\$. Putting these two together, we get the expression:

$$P = \frac{V^2}{R}$$

In your case you have a fixed voltage source - so \$V=12\mathrm{V}\$. Put that in to the equation we get:

$$P = \frac{12^2}{R} = \frac{144}{R} = \frac{144}{R_{wire}+R_{load}}$$

As your wire gets longer, the resistance of it increases. As a result, the total power dissipated in the entire circuit is going to decrease. The power dissipated in the wire itself may go up, or it may go down.


How the current is affected will depend on what the circuit is. For example if you have a simple resistive load, then from ohms law we can find out how the current will be affected:

$$I = \frac{V}{R} = \frac{12}{R_{wire} + R_{load}}$$

If you increase the wire resistance, the total current will decrease. Depending on the ratio between wire resistance and load resistance, the power dissipated in the wire itself may go up, or it may go down.


However, if your load was, say, a constant current load, then things get a little more tricky. So long as there is enough voltage across the load to maintain the constant current, then we can say for sure that the power dissipated in the wire will increase. Going back to the first two equations, we can also derive that power is:

$$P = I^2R$$

So if the current stays the same, but the resistance of the wire is increased, then the power dissipated in the wire is going to be larger.

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  • \$\begingroup\$ To make it a bit more intuitive, would perhaps V^2/(Rload+Rwire) be a better demonstration assuming the wires aren't the final load? \$\endgroup\$
    – Sam
    Jun 26 '16 at 1:09
  • \$\begingroup\$ @Tom See my edit, took a little while to write :) \$\endgroup\$ Jun 26 '16 at 1:14
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A real example of where a longer wire will cause greater current consumption is a switching DC to DC converter - it may work across a range of input voltages such as 5V to 20V to produce a constant (say) 10V on the output.

When it is fed from 20V and the final load is (say) 10 ohms, the power demand is 10 watts (100 % efficiency) and this will therefore take 0.5 amps from the 20V supply. However if the converter feed cable resistance became significant, the voltage at the converter's input terminals would reduce and, to maintain 10 watts into the load, a current greater than 0.5 amps is required to flow down the cable. As the cable got longer more current would be required and taking this to it's natural conclusion, eventually the cable resistance would not be able to support 5V at the input to the converter and the converter output would begin to fall-off.

So yes, there are circumstances when a longer cable causes more current flow.

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