How to calculate voltage drop across multiple wire gauges?










3












$begingroup$


This seems like a googleable question, but I simply cannot come across anything.



I'm working on a 12v system (aluminum boat), trying to supply power to a motor 12ft away from the circuit breaker. Through online sources, I was able to determine that at the 45amp max, 6awg wire could handle it, with a 5% voltage drop.



The confusion enters when I try to calculate the voltage drop with this 6awg wire, plus the potentially 8awg wire that runs 1.5 feet from the battery to the circuit breaker.



How does one correctly calculate the voltage drop over two different gauged pieces of wire?










share|improve this question









$endgroup$











  • $begingroup$
    It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
    $endgroup$
    – mkeith
    Aug 27 '18 at 3:59






  • 1




    $begingroup$
    Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
    $endgroup$
    – mkeith
    Aug 27 '18 at 4:00










  • $begingroup$
    @mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
    $endgroup$
    – xtraorange
    Aug 27 '18 at 19:25










  • $begingroup$
    If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
    $endgroup$
    – mkeith
    Aug 28 '18 at 0:27










  • $begingroup$
    @mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
    $endgroup$
    – xtraorange
    Aug 28 '18 at 1:45















3












$begingroup$


This seems like a googleable question, but I simply cannot come across anything.



I'm working on a 12v system (aluminum boat), trying to supply power to a motor 12ft away from the circuit breaker. Through online sources, I was able to determine that at the 45amp max, 6awg wire could handle it, with a 5% voltage drop.



The confusion enters when I try to calculate the voltage drop with this 6awg wire, plus the potentially 8awg wire that runs 1.5 feet from the battery to the circuit breaker.



How does one correctly calculate the voltage drop over two different gauged pieces of wire?










share|improve this question









$endgroup$











  • $begingroup$
    It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
    $endgroup$
    – mkeith
    Aug 27 '18 at 3:59






  • 1




    $begingroup$
    Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
    $endgroup$
    – mkeith
    Aug 27 '18 at 4:00










  • $begingroup$
    @mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
    $endgroup$
    – xtraorange
    Aug 27 '18 at 19:25










  • $begingroup$
    If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
    $endgroup$
    – mkeith
    Aug 28 '18 at 0:27










  • $begingroup$
    @mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
    $endgroup$
    – xtraorange
    Aug 28 '18 at 1:45













3












3








3





$begingroup$


This seems like a googleable question, but I simply cannot come across anything.



I'm working on a 12v system (aluminum boat), trying to supply power to a motor 12ft away from the circuit breaker. Through online sources, I was able to determine that at the 45amp max, 6awg wire could handle it, with a 5% voltage drop.



The confusion enters when I try to calculate the voltage drop with this 6awg wire, plus the potentially 8awg wire that runs 1.5 feet from the battery to the circuit breaker.



How does one correctly calculate the voltage drop over two different gauged pieces of wire?










share|improve this question









$endgroup$




This seems like a googleable question, but I simply cannot come across anything.



I'm working on a 12v system (aluminum boat), trying to supply power to a motor 12ft away from the circuit breaker. Through online sources, I was able to determine that at the 45amp max, 6awg wire could handle it, with a 5% voltage drop.



The confusion enters when I try to calculate the voltage drop with this 6awg wire, plus the potentially 8awg wire that runs 1.5 feet from the battery to the circuit breaker.



How does one correctly calculate the voltage drop over two different gauged pieces of wire?







wire






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Aug 27 '18 at 2:49









xtraorangextraorange

1183




1183











  • $begingroup$
    It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
    $endgroup$
    – mkeith
    Aug 27 '18 at 3:59






  • 1




    $begingroup$
    Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
    $endgroup$
    – mkeith
    Aug 27 '18 at 4:00










  • $begingroup$
    @mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
    $endgroup$
    – xtraorange
    Aug 27 '18 at 19:25










  • $begingroup$
    If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
    $endgroup$
    – mkeith
    Aug 28 '18 at 0:27










  • $begingroup$
    @mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
    $endgroup$
    – xtraorange
    Aug 28 '18 at 1:45
















  • $begingroup$
    It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
    $endgroup$
    – mkeith
    Aug 27 '18 at 3:59






  • 1




    $begingroup$
    Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
    $endgroup$
    – mkeith
    Aug 27 '18 at 4:00










  • $begingroup$
    @mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
    $endgroup$
    – xtraorange
    Aug 27 '18 at 19:25










  • $begingroup$
    If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
    $endgroup$
    – mkeith
    Aug 28 '18 at 0:27










  • $begingroup$
    @mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
    $endgroup$
    – xtraorange
    Aug 28 '18 at 1:45















$begingroup$
It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
$endgroup$
– mkeith
Aug 27 '18 at 3:59




$begingroup$
It might be a good idea to replace the 1.5 feet of 8AWG wtih 6AWG while you are at it. Then you have a continuous 6AWG path from battery to motor.
$endgroup$
– mkeith
Aug 27 '18 at 3:59




1




1




$begingroup$
Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
$endgroup$
– mkeith
Aug 27 '18 at 4:00




$begingroup$
Copper wire, plus batteries, plus an aluminum hull in water can lead to some pretty interesting failures. Be careful!
$endgroup$
– mkeith
Aug 27 '18 at 4:00












$begingroup$
@mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
$endgroup$
– xtraorange
Aug 27 '18 at 19:25




$begingroup$
@mkeith - Is there a significant advantage there? Without getting into detail, the 8awg wire saves me quite a bit of money.
$endgroup$
– xtraorange
Aug 27 '18 at 19:25












$begingroup$
If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
$endgroup$
– mkeith
Aug 28 '18 at 0:27




$begingroup$
If the motor current flows through the 8AWG wire (which is what I think based on your question) then it will produce a voltage drop between battery and breaker. I am envisioning that the breaker is in a small panel with other stuff. Everything connected to that panel will experience that voltage drop. If the only consumer of power is the motor, then it probably doesn't matter as long as the 8AWG can handle 45 Amps without overheating.
$endgroup$
– mkeith
Aug 28 '18 at 0:27












$begingroup$
@mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
$endgroup$
– xtraorange
Aug 28 '18 at 1:45




$begingroup$
@mkeith - that circuit breaker is separate from my main fuse panel, which has its own connection to the battery. However, they do both share a common ground... would that be an issue at all? If not, it sounds like I can use the 8awg for the short run, which would be great new for me, haha.
$endgroup$
– xtraorange
Aug 28 '18 at 1:45










2 Answers
2






active

oldest

votes


















6












$begingroup$

Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.



You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.



Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.






share|improve this answer









$endgroup$








  • 1




    $begingroup$
    How did this get downvoted? This is correct.
    $endgroup$
    – K H
    Aug 27 '18 at 3:33










  • $begingroup$
    If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
    $endgroup$
    – K H
    Aug 27 '18 at 3:40










  • $begingroup$
    @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
    $endgroup$
    – DoxyLover
    Aug 27 '18 at 6:21










  • $begingroup$
    Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
    $endgroup$
    – K H
    Aug 27 '18 at 6:32


















1












$begingroup$

From google:
https://www.calculator.net/voltage-drop-calculator.html



You can just first calculate the drop over the 6AWG wire:
45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.



Then for the extra:
45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.



If your load is not DC you can recalculate.






share|improve this answer









$endgroup$












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.



    You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.



    Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      How did this get downvoted? This is correct.
      $endgroup$
      – K H
      Aug 27 '18 at 3:33










    • $begingroup$
      If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
      $endgroup$
      – K H
      Aug 27 '18 at 3:40










    • $begingroup$
      @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
      $endgroup$
      – DoxyLover
      Aug 27 '18 at 6:21










    • $begingroup$
      Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
      $endgroup$
      – K H
      Aug 27 '18 at 6:32















    6












    $begingroup$

    Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.



    You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.



    Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      How did this get downvoted? This is correct.
      $endgroup$
      – K H
      Aug 27 '18 at 3:33










    • $begingroup$
      If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
      $endgroup$
      – K H
      Aug 27 '18 at 3:40










    • $begingroup$
      @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
      $endgroup$
      – DoxyLover
      Aug 27 '18 at 6:21










    • $begingroup$
      Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
      $endgroup$
      – K H
      Aug 27 '18 at 6:32













    6












    6








    6





    $begingroup$

    Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.



    You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.



    Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.






    share|improve this answer









    $endgroup$



    Each piece of wire is a resistor. Add up the resistance of all the wire that the current to the motor runs thru. Remember to consider both directions.



    You find the resistance of each piece of wire by looking in a wire chart to get the resistance per unit length. Multiply that by the actual length of the wire segment to get the resistance of that segment.



    Add up all the individual resistances of the individual wire segments. That's the total wire resistance in your installation. Multiply that by the current, and you have the voltage drop caused by the wire.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Aug 27 '18 at 3:10









    Olin LathropOlin Lathrop

    283k30340794




    283k30340794







    • 1




      $begingroup$
      How did this get downvoted? This is correct.
      $endgroup$
      – K H
      Aug 27 '18 at 3:33










    • $begingroup$
      If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
      $endgroup$
      – K H
      Aug 27 '18 at 3:40










    • $begingroup$
      @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
      $endgroup$
      – DoxyLover
      Aug 27 '18 at 6:21










    • $begingroup$
      Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
      $endgroup$
      – K H
      Aug 27 '18 at 6:32












    • 1




      $begingroup$
      How did this get downvoted? This is correct.
      $endgroup$
      – K H
      Aug 27 '18 at 3:33










    • $begingroup$
      If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
      $endgroup$
      – K H
      Aug 27 '18 at 3:40










    • $begingroup$
      @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
      $endgroup$
      – DoxyLover
      Aug 27 '18 at 6:21










    • $begingroup$
      Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
      $endgroup$
      – K H
      Aug 27 '18 at 6:32







    1




    1




    $begingroup$
    How did this get downvoted? This is correct.
    $endgroup$
    – K H
    Aug 27 '18 at 3:33




    $begingroup$
    How did this get downvoted? This is correct.
    $endgroup$
    – K H
    Aug 27 '18 at 3:33












    $begingroup$
    If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
    $endgroup$
    – K H
    Aug 27 '18 at 3:40




    $begingroup$
    If the motor is not a balanced multiphase load, make sure you take into account the resistance both on the way to and back from the motor.
    $endgroup$
    – K H
    Aug 27 '18 at 3:40












    $begingroup$
    @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
    $endgroup$
    – DoxyLover
    Aug 27 '18 at 6:21




    $begingroup$
    @KH The OP is talking about a 12 volt boat. There’s little doubt it’s a 2 wire DC motor.
    $endgroup$
    – DoxyLover
    Aug 27 '18 at 6:21












    $begingroup$
    Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
    $endgroup$
    – K H
    Aug 27 '18 at 6:32




    $begingroup$
    Figured it was possible anyway tiny hobby motors go to 3 phase and boats sometimes vary on the side of fancy.
    $endgroup$
    – K H
    Aug 27 '18 at 6:32













    1












    $begingroup$

    From google:
    https://www.calculator.net/voltage-drop-calculator.html



    You can just first calculate the drop over the 6AWG wire:
    45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.



    Then for the extra:
    45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.



    If your load is not DC you can recalculate.






    share|improve this answer









    $endgroup$

















      1












      $begingroup$

      From google:
      https://www.calculator.net/voltage-drop-calculator.html



      You can just first calculate the drop over the 6AWG wire:
      45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.



      Then for the extra:
      45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.



      If your load is not DC you can recalculate.






      share|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        From google:
        https://www.calculator.net/voltage-drop-calculator.html



        You can just first calculate the drop over the 6AWG wire:
        45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.



        Then for the extra:
        45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.



        If your load is not DC you can recalculate.






        share|improve this answer









        $endgroup$



        From google:
        https://www.calculator.net/voltage-drop-calculator.html



        You can just first calculate the drop over the 6AWG wire:
        45A, (assume DC) 12V, 6AWG, 12ft = 11.57 V.



        Then for the extra:
        45A, (assume DC) 11.57V, 8 AWG, 1.5ft = 11.517 V.



        If your load is not DC you can recalculate.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Aug 27 '18 at 6:59









        jerlichjerlich

        111




        111



























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