Basic electrics, 1965 B40F

[Phil C]

Okay, thanks AWJDT, that makes sense, even to me. So the wire NU can have current flowing both ways in it at the same time (the two-way street Bess described.)  And the current for the lights etc always flows from the battery, never straight from the alternator/rectifier. Please tell me that's right!

Also Julian I get the bit about the alternator output control via the switches (kind of!) - thanks for that.

By the way, just before this I thought I had posted a message asking three questions, but I don't see it posted - I hope it didn't come across wrong in any way and so was removed  - can't think how it could have, and certainly didn't mean it to.

Thanks everyone for your efforts and patience.

Phil (novice)

[Bess]

Hi Phil,
          Hope all works out for you, it's great to see another person keen to keep a BSA on the road.

ps: please remove the novice tag, you're in the fold


Best wishes..........

[AWJDThumper]

Okay, thanks AWJDT, that makes sense, even to me. So the wire NU can have current flowing both ways in it at the same time (the two-way street Bess described.)  And the current for the lights etc always flows from the battery, never straight from the alternator/rectifier. Please tell me that's right!

Unfortunately, the answer to both questions is NO!

Please see diagram which explains things graphically. From this you should be able to see that the current in wire NU can flow in either direction but in only one direction at a given instance in time.  You should also be able to see that the current to the lights is either straight from the alternator/rectifier, or from a combination of the current from the alternator/rectifier and the battery, or from just the battery if the engine isn't running.

[Phil C]

Thanks very much indeed, AWJD Thumper - I get it! Thanks very much - that's made my weekend! 

Phil.

PS Bess suggested I drop the (novice) bit after my name. I thought about changing it to Phil (clueless), but thought I would try to build my confidence by leaving it out.  Thanks again!

[Phil C]

Good morning everyone,

When I looked at AWJDThumper's diagram (over a year ago now) it made perfect sense, and I printed it off for future reference if needed.  I just happened to come across it again this morning, and it is now not quite so clear to me. 

As I see it, AWJDT's equation Ic = Ir - Ii - Il  (sorry, lower case rather than smaller capitals, as I don't know how to do the smaller capitals)means that the alternator/rectifier send a certain number of amps to the lights etc, and any not needed for that go to charge the battery. Correct?

AWJDT's final equation, It = Is - Ic, made sense to me a year ago. Why not now, I wonder. I thought the amps returning to the battery should equal the amps leaving it. In which case It would = Is + Ic?  Just out of interest, could someone explain please?

Phil.

[Phil C]

As AWJDThumper is now around again, I thought I would test people's patience and reintroduce this. Any explanation, please?    Phil

[AWJDThumper]

Blimey - did I draw that diagram a year ago??

I think the answer to the last question needs to take into account the fact that Ic can flow in two possible directions: when the alternator is not supplying enough current (eg headlight on and bike ticking over), Ic will flow out of the battery; however, when the alternator is providing sufficient current to supply all the bike's electrics (eg headlight off and bike revving at 5,000 rpm), Ic will flow the opposite way into the battery. The diagram shows arrows for Ic in both directions.

Therefore, strictly, we have It = Is + Ic for the first scenario but It = Is - Ic for the second which is normally the way the system should be operating.

[Phil C]

Okay, yes, thanks AWJDT. I guess the equation It = Is + Ic holds true either way, because if c is negative then in the equation  the plus will effectively change to a minus. Thanks again for be all your help with this.  Phil