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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)
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.
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)
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.