/sci/ - Science & Math

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Anonymous Tue Aug 10 07:12:12 2010 No.1577105 [Reply] [Original]

gais any advice on this problem?

--A 6 V
ashlight battery will deliver 5 mA for 40 hours of usage. During that time the voltage will drop from 6 V to 5 V . The
drop in voltage is linear with time. How much energy is delivered by the battery in this
40 hour interval?

In return, you become better than everone else automatically.

>>	Anonymous Tue Aug 10 07:19:12 2010 No.1577110 5.5V0.005A5hours*(3600 seconds/hour) =

>>	Anonymous Tue Aug 10 07:20:12 2010 No.1577112 >>1577110 oops, that should be 40 hours.

>>	Anonymous Tue Aug 10 07:28:12 2010 No.1577119 5.5v average 5.5v x 5mA = 27.5mW times that by number seconds to give total energy!

>>	David !4changtcqk Tue Aug 10 07:29:12 2010 No.1577122 3960 J

>>	Anonymous Tue Aug 10 07:32:12 2010 No.1577128 >>1577110 Winner! Can I ask what equation is behind the energy consumption? I assume it's average voltage * ampage* time(seconds), does this sound right?

>>	Anonymous Tue Aug 10 07:33:12 2010 No.1577129 >>1577119 oh wait, never mind, cheers guys!

>>	David !4changtcqk Tue Aug 10 07:34:12 2010 No.1577130 >>1577128 P=VI Pt=E VIt=E

>>	Anonymous Tue Aug 10 07:38:12 2010 No.1577135 >>1577128 maybe if you bothered to learn the simple relationships between amps,voltage,resistance and time and what they each one is actually describing then you could formulate your own!

>>	Anonymous Tue Aug 10 07:44:12 2010 No.1577144 5mA40h=0.2Ah (6V+5V)/2=5.5V 5.5V0.2Ah=1.1Wh

>>	Anonymous Tue Aug 10 07:46:12 2010 No.1577145 >>1577144 haha lets see how many diferent ways it can be done and confuse the hell out of him

>>	Anonymous Tue Aug 10 07:52:12 2010 No.1577153 >>1577145 Well... integral from 0 to 40 of function ((-1/40t+6)0.005)

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