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
changes.

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!

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

I think the delta in front of voltage is just there because measured voltage is the potential difference, ie. delta

You're on the right track with your calculations for part (a). To calculate the power consumed by a "100 W" lightbulb when the voltage drops to 108 V, you correctly used the formula P = IV. Given that P = 100 W and V = 108 V, you found I = P/V = 100/108 = 0.93 A. And then you multiplied I = 0.93 A by V = 108 V to get P = IV = 0.93 A * 108 V = 100.44 W. Therefore, the power consumed by the lightbulb during the brownout is approximately 100.44 W.

Now, let's move on to part (b) of the question. You are correct that when a filament is not as hot, the light emitted by the lightbulb will not be as bright, indicating a decrease in power. To understand how the power drop changes, we need to consider the relationship between power, voltage, and resistance. In the case of a lightbulb filament, its resistance increases as it gets hotter.

If we assume that the resistance of the lightbulb filament remains constant during the brownout (which is an approximation in this scenario), then using the formula P = IV, where I is the current and V is the voltage, we can rearrange the formula as I = P/V.

Based on our previous calculations, when the voltage drops to 108 V, I = 0.93 A. Now, if the resistance of the lightbulb filament decreases due to it not being as hot as usual during the brownout, then the current will increase for the same voltage. Ultimately, a higher current would lead to a higher power consumption.

So, in summary, if the filament is not as hot during the brownout, the resistance of the filament decreases, causing an increase in current. As a result, the power drop would be less than what you calculated in part (a) because the increase in current would compensate for the lower voltage.

To validate this conclusion, you can perform a comparison between the power calculated in part (a) (100.44 W) and the power obtained by using the increased current due to decreased resistance.