A 95-W lightbulb connected to a 120-V source experiences a voltage surge that produces 131 V for a moment. By what percentage does its power output increase? Assume its resistance does not change.

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Best answer

The formula for power can be written in terms of voltage and resistance:

P = v^2 / R (where P=power, v=voltage, R=resistance)

So by establishing the ratio the original power, to the new power, we can find the percentage change as follows.

P1 = v1^2 / R

P2 = v2^2 / R

P1 / P2 = (v1^2 / R)(v2^2 / R) and so the R's will cancel, and we're left with

P1 / P2 = (v1^2 )(v2^2)

now plug in values for v1 and v2:

v1 = 120 V

v2 = 131V

so:

P1 / P2 = (120 V^2 )(131V^2) = 1.192,

which is equivalent to and increase of 19.2%

P = v^2 / R (where P=power, v=voltage, R=resistance)

So by establishing the ratio the original power, to the new power, we can find the percentage change as follows.

P1 = v1^2 / R

P2 = v2^2 / R

P1 / P2 = (v1^2 / R)(v2^2 / R) and so the R's will cancel, and we're left with

P1 / P2 = (v1^2 )(v2^2)

now plug in values for v1 and v2:

v1 = 120 V

v2 = 131V

so:

P1 / P2 = (120 V^2 )(131V^2) = 1.192,

which is equivalent to and increase of 19.2%

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