# Small drop in voltage change in power?

1. Sep 11, 2006

### boderam

In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?

2. Sep 11, 2006

### Andrew Mason

Your idea is right. Use $P_2 = V_2^2/R$ where $P_2 = 5P_1/6 =\frac{5}{6}V_1^2/R$ and find the ratio of V2 to V1.

AM