Date:         Sun, 5 Apr 2009 19:12:28 -0300
Reply-To:     David Beierl <dbeierl@ATTGLOBAL.NET>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         David Beierl <dbeierl@ATTGLOBAL.NET>
Subject:      Learning Electricity Part Four-a
Content-Type: text/plain; charset="us-ascii"; format=flowed

So y'all don't get bored, here's the beginning of the next
bit...maybe more tonight, maybe not.

Now what can we do with all that?

Here's a typical circuit, we'll say it's for a headlight.  The small
letters represent test points.  The battery itself is behind the page
where you can't see it, just the terminals poking through at the edges.

+BAT--a---------b--fuse--c---------d--switch--e----------f--bulb--g--------h--ground--i--BAT-

We've got a monster bulb in there, 100 watts @ 13.5v, and we measure
13.5 volts from a to f.  What can we find out?

First, how many amps (ideally) will flow?  Well, we know E and P, and
we want I.  From the formulas, I=P/E = 100/13.5 = 7.4
amps.  Interesting; if we run four such bulbs that's about thirty
amps.  And a bit more than half a horsepower.

What voltage will we find (ideally) between points on the circuit?
a-f =  0 volts.
f-g = 13.5 volts.
g-i = 0 volts.

What's the working resistance of the bulb?  R=E/I = 13.5/7.4 = 1.82 ohms.

We were ignoring everything but the bulb.  What happens if there's
half an ohm of resistance in the switch?
Rtotal = 1.82+0.5 = 2.32 ohms.
Current is I=E/R = 13.5/2.32 = 5.82 amps.
Total power P=EI = 13.5 x 5.82 = 78.56 watts.
Power developed by the bulb? P=I^2R = 5.82^2x1.82 = 61.65 watts.
By the switch? P=I^2R = 5.82^2x0.5 = 16.94 watts.
Will the switch melt?  You bet.
Before it melts, what voltages will we see in the circuit?
E=IR so a-d = 5.82x0 = 0.  d-e = 5.82x0.5 = 2.91 volts.  f-g =
5.82*1.82 = 10.59 volts.

+BAT--a---------b--fuse--c---------d--switch--e----------f--bulb--g--------h--ground--i--BAT-



  --
David Beierl - Providence RI USA -- http://pws.prserv.net/synergy/Vanagon/
'89 Po' White Star "Scamp"