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Determine the power used by a light bulb with a potential difference of 120 V and a resistance of 5 Ω.
in Grade 12 Physical Sciences by Diamond (50,514 points) | 43 views

4 Answers

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Best answer
P=V^2/R

=120^2/5

=2880 W

= 2.8 KW
by Wooden (1,121 points)
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P= V^2 / R

= (120)^2 / 5

=14400 / 5

P= 2880 watts
by Wooden (1,088 points)
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P=V^2/R

P= 120^2/5

P=2880watts
by Wooden (474 points)
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P = ?                    P = V² / R

V = 120V                = (120)² / 5

R = 5 Ω                   = 2880 W
by Wooden (786 points)

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