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UK diy (uk.d-i-y) For the discussion of all topics related to diy (do-it-yourself) in the UK. All levels of experience and proficency are welcome to join in to ask questions or offer solutions. |
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Johnny B Good Wrote in message:
On Thu, 26 Apr 2018 23:58:17 +0100, Dave Plowman (News) wrote: In article , Johnny B Good wrote: The point is when calculating near anything car electrics wise is you have to make allowances for the voltage when running being rather higher than the nominal 12v. As very little is only ever used with the engine stopped. And when starting the engine, you also need to allow for the voltage dropping considerably due to the starter load. This where the positive coefficient of the 2.2W ignition warning lamp's filament resistance might prove useful[1] since even a good condition 12v car battery's terminal voltage will fall to around the 8 to 10 volt mark from the starter cranking load (once the motor has accelerated to cranking speed - the voltage can easily dip to less than 6 volts on the initial starting surge of 300 odd amps or more before it settles down to the 100 or so amps once up to cranking speed). Doesn't really matter, since it would be silly for the alternator to be attempting to charge the battery while the starter is operating. Hence note[1]. :-) Appendix shurely? -- Jim K |
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