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" > |
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