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Post by Clinton Cool on Dec 8, 2019 18:12:26 GMT
A question about volt drop. If a power supply (in this case a wind turbine) was delivering 10 amps of power but an overly thin wire resulted in a 20% drop in volts: would the current going into the batteries be 8 amps, or a different figure?
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Post by phil70 on Dec 8, 2019 19:24:30 GMT
A different figure Phil
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Post by Deleted on Dec 8, 2019 19:28:28 GMT
It depends where the 10 amps was measured.
It's an interesting question. If you had a 10amp charge (measured at the charger output wires) going into a battery via a long thin extension wire and you had a DC clamp meter for testing would the amp ie current readings vary along the wire. I guess they would.
Maybe it's best to think in watts rather than amps as that takes account of the voltage drop with the W=VA formula triangle.
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Post by faffer on Dec 8, 2019 19:29:20 GMT
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Post by Mr Stabby on Dec 8, 2019 19:31:10 GMT
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Post by Telemachus on Dec 8, 2019 20:29:51 GMT
A question about volt drop. If a power supply (in this case a wind turbine) was delivering 10 amps of power but an overly thin wire resulted in a 20% drop in volts: would the current going into the batteries be 8 amps, or a different figure? There are some problems with your question - “10 amps of power” doesn’t make any sense because amps isn’t a unit of power. Anyway, current cannot be “dropped” in a circuit so if there are 10A at one end of a wire, there must be 10A at the other end. Electrons don’t leak out and fall into the canal! However power is the product of volts times amps and whilst the amps remains constant through your thin wire, the volts drop. So in terms of power, you lose power in the thin wire. But you don’t lose current, only voltage.
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Post by bodger on Dec 8, 2019 20:39:06 GMT
A question about volt drop. If a power supply (in this case a wind turbine) was delivering 10 amps of power but an overly thin wire resulted in a 20% drop in volts: would the current going into the batteries be 8 amps, or a different figure? There are some problems with your question - “10 amps of power” doesn’t make any sense because amps isn’t a unit of power. Anyway, current cannot be “dropped” in a circuit so if there are 10A at one end of a wire, there must be 10A at the other end. Electrons don’t leak out and fall into the canal! However power is the product of volts times amps and whilst the amps remains constant through your thin wire, the volts drop. So in terms of power, you lose power in the thin wire. But you don’t lose current, only voltage. wot 'e said. ............... but the power available for charging also depends on the voltage generated by the wind turbine. whatever voltage is available at the end of the line may be insufficient to charge the battery. if the turbine generates 15v and you lose 20% you'll get nowt into a 12v battery, and no current will flow. .... that poses the question - what happens to all those excitable electrons flying around in the generator but that have nowhere to go.
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Post by Deleted on Dec 8, 2019 20:42:29 GMT
They get blown away.
Oh dear I want to listen to Neil Young now !
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Post by Gone on Dec 8, 2019 21:37:06 GMT
There are some problems with your question - “10 amps of power” doesn’t make any sense because amps isn’t a unit of power. Anyway, current cannot be “dropped” in a circuit so if there are 10A at one end of a wire, there must be 10A at the other end. Electrons don’t leak out and fall into the canal! However power is the product of volts times amps and whilst the amps remains constant through your thin wire, the volts drop. So in terms of power, you lose power in the thin wire. But you don’t lose current, only voltage. wot 'e said. ............... but the power available for charging also depends on the voltage generated by the wind turbine. whatever voltage is available at the end of the line may be insufficient to charge the battery. if the turbine generates 15v and you lose 20% you'll get nowt into a 12v battery, and no current will flow. .... that poses the question - what happens to all those excitable electrons flying around in the generator but that have nowhere to go. Not quite so simple. The volt drop in the cable will depend on the current flow in the wire, so the low voltage at the battery will reduce the current flow, which will reduce the loss in the cable increasing the battery voltage. So it will fully charge the battery BUT it will take a lot longer. Of course if you have some load on the battery eg an inverter in sleep mode or the occasional running of the bilge pump, then the power going in might be less than power coming out, it which case the battery will eventually go flat.
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Post by phil70 on Dec 8, 2019 22:59:16 GMT
I don't understand the science but when faced with potential voltage drop I've always gone BIG and have ever had any problems Phil
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Post by samsam on Dec 9, 2019 2:26:41 GMT
It depends where the 10 amps was measured. It's an interesting question. If you had a 10amp charge (measured at the charger output wires) going into a battery via a long thin extension wire and you had a DC clamp meter for testing would the amp ie current readings vary along the wire. I guess they would. Maybe it's best to think in watts rather than amps as that takes account of the voltage drop with the W=VA formula triangle. Sorry, but complete rubbish. The current is the same anywhere on the wire if there is a circuit.
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Post by Deleted on Dec 9, 2019 6:49:49 GMT
Yes that's been explained further up. Thank you.
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Post by Clinton Cool on Dec 9, 2019 7:10:43 GMT
Wow I'm even more confused than I was earlier!
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Post by Jim on Dec 9, 2019 7:14:02 GMT
Yes that's been explained further up. Thank you. What? Wasn't it about voles eating currents, unhampered, the fatter the vole the more currents.
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Post by Deleted on Dec 9, 2019 7:21:31 GMT
Surely you meant unhamstered.
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