A) If the internal resistance of the batteries is negligible,what power is deliv

Question

A) If the internal resistance of the batteries is negligible,what power is delivered to the bulb? I thought the answer was 2V2/R, but that was wrong.I also tried 2V2 x R and that doesn't work either. I can't figure this one out without part A. C) The resistance of real batteries increases as they rundown. If the initial internal resistance is negligible, what is thecombined internal resistance of both batteries when the power tothe bulb has decreased to half its initial value? (Assume that theresistance of the bulb is constant. Actually, it will changesomewhat when the current through the filament changes, becausethis changes the temperature of the filament and hence theresistivity of the filament wire.)

Explanation / Answer

A) P= V2/R B) E= P t C) Since P' = 0.5P we have assuming the voltage or emf is the same V2/(R +Rb) =0.5V2/R and 2R= (R + Rb) and finally Rb = R (you are correct the resitane of the filament would getlower but we ignore it. Oterwice we would have to consider BBradiation and all that) Have fun (you are correct the resitane of the filament would getlower but we ignore it. Oterwice we would have to consider BBradiation and all that) Have fun

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