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Hiii, do you think you could help me on this problem? :P A 20-ohm resistor is connected to four 1.6v batteries. What is the joule heat loss per minute in the resistor if the batteries are connected (a) in series and (b) in parallel?
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@wolfe8
Well, do you know how to add voltages in parallel and in series?
Power in Watts = Joules/s so J=Ws
\[\large P={V^2\over R}\]
now just add your batteries and substitute
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Did: (1.6x4)^2/20 = 2.048J/min... doesn't seem right? @roadjester
just a minor correction; the math is right but Watts are J/s; next you need to multiply by 60
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