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The electric power P is dissipated by a light bulb of resistance R is P=v^2/R, where V represents the line voltage. During a brownout, the
The electric power P is dissipated by a light bulb of resistance R is P=v^2/R, where V represents the line voltage. During a brownout, the line voltage is 10.0% less than its normal value. How much power is drawn by a light bulb during the brownout if it normally draws 100.0 W (watts)? Assume that the resistance does not change. The answer is 81.0 W but I don't know how to get the
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College Physics A Strategic Approach Volume 1
Authors: Randall D. Knight, Brian Jones, Stuart Field
2nd Edition
0321602285, 978-0321602282
