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Question

We normally give "brightest" of a house-hold light bulb by giving the amount of power that light bulb uses. A house-hold bulb that discipates 100W of electrical power puts out more light that one that discipates 60 Watts of electrical power. A standard light bulb operates on a potential drop of around 95 V.

1) For a 75 watt light bulb, what is the current through the bulb?

i =

2) What is the resistance of this bulb?

R =

3) Now assume you have a light bulb that has half the resistance of the one in part (2) and the same potential drop. What is the is the power rating of this bulb?

P=

Explanation / Answer

power = P = V*I

1)

75 = 95*I

I = 0.79 A

2)

from ohms law

R = V/I = 95/0.79 = 120.25 ohms

3))

R' = R/2 = 60.125

P' = v^2/R'

P' = 95^2/60.125 = 150.1 W

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