bkkkpewsey
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Post by bkkkpewsey on Mar 3, 2014 12:57:00 GMT
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Deleted
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Post by Deleted on Mar 3, 2014 14:00:48 GMT
It will tell you the voltage drop within the mod (contacts/tube/switch) when you subtract the reading from that across the battery itself and while an in-line voltmeter would be ideal, not everyone has one which is why I suggested a multimeter.
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erik514
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Post by erik514 on Mar 3, 2014 14:09:56 GMT
fred, yes, vamo in RMS... but using the same tanks etc the mech doesn't come close hence my concerns...... @yinyang, protank @ 2.1ohms, only use 3.4-ish volts on the vamo hence expecting the mech to vape similar, but it just doesn't If you lower your resistance to between 1 and 1.5 ohms your mech will perform much better. My experience of mechs is that anything above around 1.6 is better in a vv/vw mod
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bkkkpewsey
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Post by bkkkpewsey on Mar 3, 2014 14:48:05 GMT
It will tell you the voltage drop within the mod (contacts/tube/switch) when you subtract the reading from that across the battery itself and while an in-line voltmeter would be ideal, not everyone has one which is why I suggested a multimeter. I am afraid you signature is correct as in this case you are wrong. A bit of understanding of Ohm's Law wouldn't go amiss e.g. V(drop)= IxR As a multimeter draws virtually no current (I) - then there will be virtually no volt drop. As I said before to measure voltage drop in any mech mod, it must be under load.
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geordie_vaper
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Post by geordie_vaper on Mar 3, 2014 15:06:12 GMT
no need to be like that people are trying to help somone no need for the attitude at all
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Post by Deleted on Mar 3, 2014 16:14:53 GMT
Thank you for that explanation bkkkpewsey - I am always ready to be educated.
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Post by Deleted on Mar 3, 2014 17:32:58 GMT
I wouldn't use a UDT-L mod... Breaks too easily
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spikus
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Post by spikus on Mar 3, 2014 21:38:11 GMT
Could I as a daft question ?.... With my old A level physics being just a fading memory of my ancient past I was trying to make sense of the above convo .... Now surely the difference between the voltage of the battery and the voltage at the 510! Connector is the voltage drop... I don't understand why putting the battery under load will measure anything other than the difference between the battery and the load ... What we're trying to ascertain is how much our super shiny battery holders lose in transmitting power to our coils ... Isn't it ?..... Or am I totally wrong and need to go back to school .. ?
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