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Ok, so I made an arithmetic error, and the perpetual motion machine won't work. Darn! 0. Watts are watts. AC / DC / burning paper. Doesn't matter where they come from, they're the same amount of power. You can't create new energy with a transformer or converter. In fact, any *real* device will lose energy (and get warm in the process). So if it's putting out 300 watts, it's gotta be taking in MORE THAN 300 watts. 1. Even though your power supply is rated at "5 amps" it's not actually drawing that much. If it did, you'd be wasting about 250 watts as heat in the power supply, which I think you would notice pretty quickly. It's probably drawing somewhere under 400 watts, or less than 3.3 amps on the input. 2. I think the current draw question for an inverter is answered pretty well by the calculator at the URL someone else posted earlier http://www.skingcompany.com/portable_power/calc2.htm). These guys are in the business of selling inverters, and I have to assume they know their stuff. Their calculator claims that an inverter that puts out 300(real) watts will draw 30 amps from the battery. (Just for grins (check my math!) 30 amps x 12 volts = 360 watts (got it right this time! :-) so the inverter will be putting out about 60 watts as heat, too.) 3. (for geeks only) Why does the Volts * Amps = Watts thing work for AC too? Well, it's because the "120" volt AC current is really delivered as a 60Hz sine wave with a peak voltage of about 170 volts. If we pretend for the moment that it's a DC current at 120 V and 2.5A, then it would dissipate 300W in a 48ohm resistor. If, instead, we apply our 170V peak sine wave to the 48 ohm resistor, we get current varying in a sine wave from -3.5A to +3.5A. Multiplying the current by the voltage, we get a power curve in the shape of cos^2 with a peak power of 600W. (Precisely, we get P(t) = 600*cos^2(t*60*2*pi).) Averaging over a full cycle, we get 300W. (The average power over a period DT is equal to integral(t=0,DT)(600W * cos^2(t*60*2pi)dt) / DT = 600W * (DT/2 + 1/(4*60*2pi) * sin(2 * DT * 60 * 2pi) / DT If DT = 1/60 this reduces to 600W * (1/120 + 1/(240*2pi) * sin(2*2pi)) / (1/60), but sin(240*2pi) is zero, so we get 600W * (1/120) / (1/60) = 300W.) =Spencer homeroast mailing listhttp://lists.sweetmarias.com/mailman/listinfo/homeroast |

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From: "Spencer W. Thomas" Subject: +Re: adapters Date: Wed, 28 Mar 2001 17:05:07 -0500 Although we have more important coffee issues than electrical engineering... < |

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Ahh this brings me back to days spent studying power distribution systems during college. (painful memories I might add) The AC discussion below reminds me of the reason power companies force users who burden the network with highly inductive loads to install large capacitors. Now there is a thought! A few farads of capacitance would do the trick. Solve the current spike issues as well. Alex |

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"Transmission Theory" a class designed to make you appreciate all others - but it didn't help with my feelings about Chem 200+ YUK! |

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From: "Alex McGregor" Subject: RE: +Re: adapters Date: Wed, 28 Mar 2001 16:42:17 -0800 < |

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Suzuki-san, You are absolutely correct! My priorities were wrong when I was in college. I now make time for roasting and it has aided my journey. Reminds me of the following quote. " Like the sharp edge of a razor, the sages say, is the path, Narrow it is, and difficult to tread." Alex |