voltage drop

[Under_The_Surface]

hey all
Ive pretty much got the design for my rov figured out
but... I was wondering if someone could tell me how many volts wold be left at the load (Thrusters)
if I was using three 12v bilge pumps drawing 5 amps each over a 50 or 70 metre tether using one common 16 gauge phone wire for power?
Any reply is appreciated

[rossrov]

Hi Todd. I can post a voltage drop with calculations, but your cable description is confusing. 16 gauge phone cable? Maybe describe in a bit more detail what you are planning re the controls too.

Ross

[Under_The_Surface]

not sure what u mean but id be running 50 or 70 metres (depending on the voltage drop) of 16 gauge (AWG) wire as the power source for everything on the rov and then using a separate cat 6e wire to control relays to control the thrusters. using a 12v battery
I just want to know how much voltage I have at the end of the circuit if:
50 or 70 metres tether
16 gauge (AWG) wire for power
12volt battery
thrusters draw 3amps

One more question, If I have three thrusters drawing 3 amps each, does that mean its drawing 9 amps or 3 amps altogether?
Thanks again

[fluxno]

3amps
50meter = 8volts on end of cable
70meter = 6,3volts

3 thusters active = 9 amps
50meter = nothing left..

With 2 thusters (6 amps)
50meter = 4volts
70meter = 0,6volts

Calculated with electrodroid app for Android devices.

[rossrov]

16 AWG has a resistance of approximately 0.013 ohms per meter (from a table off the internet)
50 metres of two-conductor cable has a total circuit length of 100 metres
Total resistance = 0.013 ohms X 100 =1.3 ohms
Voltage drop = current X resistance, or V=IR
=3 amps X 1.3 ohms
=3.9 volts
So if running off 12 volt battery on the boat, the motor will get 8.1 volts. Same as fluxno's Android app, but explains the theory.

The short answer to the second question is 9 amps, because the motors would be connected in parallel. In practice things get more complicated because as the voltage on the motor terminals gets lower so does the current draw.

Even with thicker cable you still have the problem of volts dropping by double for two motors running or triple for all 3, assuming no voltage regulator at the motor end of the cable. Just not practical. The simplest way to do it would be a separate two-conductor cable for each motor, then the motors will not affect each other. Then you could do away with the relays by connecting the tether to the motors direct and switch with centre-off DPDT toggle switches. If you plan on ROVing in waters with little or no current, then the bulky tether might be OK. Cable is not normally cheap though.

Depending on the camera cable you are using, having those long motor wires running alongside your camera cable could possibly lead to interference on the camera side of things, but you'd really need to try it to find out.

[joey]

16 AWG has a resistance of approximately 0.013 ohms per meter (from a table off the internet)
50 metres of two-conductor cable has a total circuit length of 100 metres
Total resistance = 0.013 ohms X 100 =1.3 ohms
Voltage drop = current X resistance, or V=IR
=3 amps X 1.3 ohms
=3.9 volts
So if running off 12 volt battery on the boat, the motor will get 8.1 volts. Same as fluxno's Android app, but explains the theory.

Don'f forget there is more to this formula. As voltage drops, current increases, it is ohms law.

So if you start with 12vDC@3A on the wire, and add the resistance of the cable to come to the conclusion that the voltage drop will result in 8.1vDC at the end...

8.1vDC / 1.3 ohms
= 6.2A

So if the motor is not rated for 6.2 amps you will burn it up.

[rossrov]

Joey, the original poster here has asked a common "beginners" question, which I (and fluxno) have answered with technical accuracy, and effort given to make that answer as clear as possible, so he can move on through some of the other topics and learn more about basic electronics and amateur ROV building. Then you comment with something which is not only totally inaccurate but that will also totally confuse anyone actually trying to learn.

16 AWG has a resistance of approximately 0.013 ohms per meter (from a table off the internet)
50 metres of two-conductor cable has a total circuit length of 100 metres
Total resistance = 0.013 ohms X 100 =1.3 ohms
Voltage drop = current X resistance, or V=IR
=3 amps X 1.3 ohms
=3.9 volts
So if running off 12 volt battery on the boat, the motor will get 8.1 volts. Same as fluxno's Android app, but explains the theory.

Don'f forget there is more to this formula. As voltage drops, current increases, it is ohms law.

So if you start with 12vDC@3A on the wire, and add the resistance of the cable to come to the conclusion that the voltage drop will result in 8.1vDC at the end...

8.1vDC / 1.3 ohms
= 6.2A

So if the motor is not rated for 6.2 amps you will burn it up.

[joey]

Joey, the original poster here has asked a common "beginners" question, which I (and fluxno) have answered with technical accuracy, and effort given to make that answer as clear as possible, so he can move on through some of the other topics and learn more about basic electronics and amateur ROV building. Then you comment with something which is not only totally inaccurate but that will also totally confuse anyone actually trying to learn.

Explain to me how Ohm's Law is completely inaccurate?

What you are telling is that is as V=IxR, therefore I=V/R (ohms law)

If any of those number are altered, the values all change... is that not true? Since resistance increased, altering voltage (ohms law), why would amperage not be altered as well?

Simple question, explain.

[rossrov]

Never said Ohm's Law was inaccurate. You've just plugged the wrong numbers in. Draw yourself a diagram of the circuit and you will see. You've divided the voltage across the motor by the resistance of the cable....