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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