During a “brownout,” which occurs when the power companies cannot
keep up with high demand, the voltage of the household circuits drops below its normal 120 V. (a) If
the voltage drops to 108 V, what would be the power consumed by a “100 W” lightbulb (that is, a
lightbulb that consumes 100.0 W when connected to 120 V)? Ignore (for now) changes in the
resistance of the lightbulb filament. (You can ignore the fact that household voltage is AC and treat
the problem as if it was normally 120 V DC.) (b) More realistically, the lightbulb filament will not be
as hot as usual during the brownout. Does this make the power drop more or less than what you
calculated in part (a)? Explain your answer by doing a calculation or algebra to show which way it
I have calculated my current as P=IV, when P=100W and V=100V, so I=0.833, and from there, I multipled I=0.88 and V=108 to get P=90W.
I know that when a filament is not as hot, the light will not be as bright, so that means less power, right? I'm confused about the last part of the question, what exactly are they asking? They want to know how the power changes, so would I just do P1=V1I1> P2=V2I2? to show that the power is less when the voltage is less?
Thanks for any help. I hope I'm at least on the right track!
physics - Megan, Wednesday, November 24, 2010 at 2:06pm
I think that power is actually equal to the change in voltage multiplied by the current, not just the voltage itself. Or at least that is what my formula sheet says. Good luck! I'm working on the same question
physics - Trace, Wednesday, November 24, 2010 at 7:48pm
I think the delta in front of voltage is just there because measured voltage is the potential difference, ie. delta