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The question is related to quadcopters and else. So my quadcopter (brushless dc motors and electronic speed controllers powere by 3cell lithium battery 2700mah 25C) has thin wires on the main power port, Im trying to figure out if it could affect current flow, I know that when gauge is too small it can overheat, burn or melt, but can it just restrict current flow without fires?

I having hard time to figure it out because once I tried to jump start a truck with cables that were too thin and the starter motor just weren't crancking the engine, wires weren't getting hot or anything... connecting another pair of wires in parallel resulted in successful start of the engine, and it was matter of seconds, not like the battery had time to recharge.

Can some one please explain this to me.

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  • \$\begingroup\$ Indeed. This wires have resistance, which both reduces the overall amount of power that can be drawn from a source of a given voltage, and means that some of the power drawn goes to heating up the wires rather than doing useful work. Once the resistance of the wire becomes significant compared to the effective resistance of the load, you typically face not only a heat problem, but a loss of effectiveness. Brushless motors, especially for aircraft, have very low effective load resistance, and need similarly low resistance power feeds. \$\endgroup\$ – Chris Stratton Apr 22 '17 at 18:16
  • \$\begingroup\$ If you know the wire diameter, wire length, and maximum (or average) current, you can calculate the maximum (or average) voltage drop in the wire. \$\endgroup\$ – mkeith Apr 22 '17 at 19:04
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The wire has a certain resistance and also a certain capacity for carrying current, proportional to its cross-sectional area. This is similar to a water pipe, which can only fit a certain quantity of water down its cross-sectional area at any moment.

If you have a circuit that relies on more current flowing down the wire than it can carry, the circuit will produce a voltage drop across the cable. This will dissipate power, and therefore heat, in the cable, following the law P = VI. If the power is sufficient to raise the wire's temperature above its melting point, the wire will burn out. At the other end of the scale, a smaller overamperage will cause the wire to restrict the current flow without noticeable effects on the wire.

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  • \$\begingroup\$ "This is similar to a water pipe, which can only fit a certain quantity of water down its cross-sectional area at any moment." This isn't true. It isn't even true for water pipes except extreme circumstances. \$\endgroup\$ – Austin Apr 22 '17 at 19:36
  • \$\begingroup\$ @Austin, ignoring the waterside, there is a maximum number of free electrons in a conductor's cross section and a terminal (unexceedable) velocity for those electrons - it's not ever-increasing. Thanks for the downvote :-( \$\endgroup\$ – TonyM Apr 22 '17 at 22:14
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Wire has resistance. Thin wire has more resistance. That's also why it dissipates more power than thick wire at the same current.

Assuming that your wires aren't getting hot enough to melt or cause other problems due to the heat, the issue is voltage drop. Ohm's law tells us that the voltage across a resistor is the current thru it times its resistance.

The wire has some finite resistance. When there is current thru it, there is therefore some voltage across it. When in series with a battery, this voltage across the wire subtracts from the battery voltage being delivered to the rest of the system.

How much this matters is up to what you consider matters. For a 12 V battery, for example, a additional 100 mV drop is less than 1% and doesn't matter much. In contrast, a 300 mV drop from a 6 V battery is a 5% reduction, which can matter in some cases.

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There are two separate issues. Will the wire get too hot? That is purely a matter of current and the wire diameter. The wires do not heat up instantly. Brief overloads will not cause the wire temperature to rise much, and the wire may cool down pretty quickly after the current stops. This is what happens with your starter motor in your car.

The second issue is, will the wire prevent power being transferred to the load? This depends on current, system voltage, wire diameter, and wire length. A 1V voltage drop in a 120V system is not a big deal. But a 1V Voltage drop in a 3.3V system is probably catastrophic. Note that wire length is equally important to wire diameter when considering power transmission. Short wires will still allow power to get through, even if they are thin.

For all these reasons, low-voltage, high-current applications (such as quadcopters and such) often need to use surprisingly large diameter wires. People also tend to underestimate how large wire should be when connecting a 12V inverter to a 12V battery.

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So, the key thing that wire thickness changes is the resistance. A thin wire has a higher resistance than a thick wire. This can affect two related things, Power and Current.

If the resistance of a wire is higher, all else being the same, the current going through it will decrease. This is what would have prevented your starter motor from turning - the resistance of the thin wire was too high, so not enough current could flow to make the motor turn.

At the same time, the resistance of the wire can affect the power. Power, in this case, is the amount of energy per second that the wire is converting to heat. This energy is being wasted as heat instead of doing whatever you want it to, like turning your quadcopter blades. The equation for this lost power is \$\ P=I^2R\$, where I is the current flowing through the wire, and R is the resistance. If you try to put current through the wire, a high resistance will mean a larger amount of the energy will be converted to heat.

If the power is too high, the wire will heat up so much that it melts.

It's slightly more complicated than that, as the voltage will also affect things, but what I wrote is basically correct.

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