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Gold Prospecting
Alluvial Gold Prospecting
DC voltage drop over long wires to bilge pump
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<blockquote data-quote="AngerManagement" data-source="post: 296641" data-attributes="member: 3760"><p>Yes but only a very light load so much so that it is not an actual load of any significance...</p><p></p><p>I have plenty of stuffed batteries that will measure close to 12V but even running a low voltage lamp will drop due to their high internal resistance. </p><p></p><p>In reality the only way other than to do the maths and assume min losses in the power supply; is to measure at the motor when it is running. And then is the pump Current Sensitive or Voltage Dependant (both when under load <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /> ) </p><p></p><p>Thus you have a Supply V then a drop across one supply line (R1) the Load (R2) and the return line (R3) is the total Resistance of the cct. </p><p></p><p>V = I x (R1 + R2 + R3) Or V = (I x R1) + (I x R2) + (I x R3 ) </p><p></p><p>And unless you have a Quality Meter your not going to be able to accurately measure the resistance. </p><p></p><p>But excluding the DunningKruger mob even a small resistance in the leads will have an impact on Voltage available to the Pump. Then add in Pump and Supply efficiency OR lack of and you have a pigs breakfast. </p><p></p><p>For DC and any load over any distance - Go a big as you can and I mean big <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /></p></blockquote><p></p>
[QUOTE="AngerManagement, post: 296641, member: 3760"] Yes but only a very light load so much so that it is not an actual load of any significance... I have plenty of stuffed batteries that will measure close to 12V but even running a low voltage lamp will drop due to their high internal resistance. In reality the only way other than to do the maths and assume min losses in the power supply; is to measure at the motor when it is running. And then is the pump Current Sensitive or Voltage Dependant (both when under load :-) ) Thus you have a Supply V then a drop across one supply line (R1) the Load (R2) and the return line (R3) is the total Resistance of the cct. V = I x (R1 + R2 + R3) Or V = (I x R1) + (I x R2) + (I x R3 ) And unless you have a Quality Meter your not going to be able to accurately measure the resistance. But excluding the DunningKruger mob even a small resistance in the leads will have an impact on Voltage available to the Pump. Then add in Pump and Supply efficiency OR lack of and you have a pigs breakfast. For DC and any load over any distance - Go a big as you can and I mean big :-) [/QUOTE]
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Gold Prospecting
Alluvial Gold Prospecting
DC voltage drop over long wires to bilge pump
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