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DennisDoesPhys.

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**In a "brownout" situation, the voltage supplied by the electric company falls. Assuming the percent drop is small, show that the power output of a given appliance falls by approximately twice that percent (assuming resistance does not change).**

How much of a voltage drop does it take for a 60-W lightbulb to begin acting like a 50-W lightbulb?

Relevant equations:

(Power output/Power input)*100%

P=(v^2/R)

The attempt at a solution:

(50/60)*100%=83%

V[1]=sq.root(60R) v[2]=sq.root(50R)

60R * (0.83)^2 = 50R

60 * ((0.83)^2 )/R = 50

((0.83^2)/(60R) = 50/60

0.83^2=50R

R= ((0.83)^2)/50

R= 0.014

(0.91 V - 0.83 V) *100%= (approximately 8%)

Answer key told me it

Relevant equations:

(Power output/Power input)*100%

P=(v^2/R)

The attempt at a solution:

(50/60)*100%=83%

V[1]=sq.root(60R) v[2]=sq.root(50R)

**sq.root(60R) * 0.83 =****sq.root(50R)**60R * (0.83)^2 = 50R

60 * ((0.83)^2 )/R = 50

((0.83^2)/(60R) = 50/60

0.83^2=50R

R= ((0.83)^2)/50

R= 0.014

**Ω**

**V[1]=sq.root(60(**

**0.014**)) v[2]=sq.root(50(0.014))**V[1]= 0.91 V****v[2]= 0.83 V**(0.91 V - 0.83 V) *100%= (approximately 8%)

Answer key told me it

**was =8.4%,****as i was a bit off, was wondering if i had made a mistake or not.**