tag:blogger.com,1999:blog-9193284457263034796.post8029034158228809089..comments2017-04-09T11:48:35.864+01:00Comments on Waterlily: Waterlily is for sale Nev Wellshttp://www.blogger.com/profile/07047359519459723079noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-9193284457263034796.post-6715355772766577052013-01-28T17:27:04.293+00:002013-01-28T17:27:04.293+00:00Thanks Halfie, I have posted it up with an explana...Thanks Halfie, I have posted it up with an explanation. Always learning ...thanks for taking the time to explain,<br /><br />Nev<br /><br />Nev Wellshttps://www.blogger.com/profile/07047359519459723079noreply@blogger.comtag:blogger.com,1999:blog-9193284457263034796.post-72930227203879829772013-01-28T14:12:27.567+00:002013-01-28T14:12:27.567+00:00Nev - I tried to leave this comment on your latest...Nev - I tried to leave this comment on your latest post on Percy, but it doesn't seem to have worked:<br /><br /><br />Nev, there are only two circumstances where the voltage at the batteries would be exactly the same as the voltage at the alternator. One: no current is flowing; two: there is zero resistance between the batteries and the alternator.<br /><br />Case one could theoretically arise if the batteries just happened to be at the same voltage as the alternator - unlikely if you are measuring 14.5V.<br /><br />Case two would imply that the connections and the cables connecting the batteries and alternator together have zero resistance, which is impossible at normal temperatures.<br /><br />Ohm's Law is essential: V = IR (voltage equals current times resistance). The voltage here is that across the resistance in question, i.e. the cables/connectors. If we assume a reasonable alternator output of 20A, then a resistance as low as 0.01 ohms would give a voltage drop of 20 x 0.01 = 0.2V. A measurable voltage drop. If the resistance of the cables/connections were to rise to 0.05 ohms then you'd get a significant drop of one volt. <br /><br />As the batteries charge up, the current flowing from the alternator falls, so that the effect of the resistance diminishes. In the example above with resistance of 0.01 ohms and a reduced current of 2A, the voltage drop is now only 2 x 0.01 = 0.02V, barely noticeable.<br /><br />Introducing a forward-biassed diode, as in a split-diode-charging system, into the circuit immediately gives you an extra voltage drop of 0.5V - 0.7V, reasonably irrespective of the amount of current flowing.<br /><br />This gives an indication of the importance of good clean connections, and as thick and short a cable as practicable (the thicker the cable the less resistance; the shorter the cable the less resistance too).<br /><br />I hope this is helpful, if not to you, Nev, then to anyone else who might chance upon this!Halfiehttps://www.blogger.com/profile/00167481543065324357noreply@blogger.com