You said there wouldn't be any math.
--
Mike "Rocket J Squirrel" Elliott
84 Westfalia: Mellow Yellow ("The Electrical Banana")
74 Utility Trailer. Ladybug Trailer, Inc., San Juan Capistrano
Bend, OR
KG6RCR

On 4/5/2009 3:12 PM David Beierl wrote:

> 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"
>