Asked by Jon
If a wire of resistance R is stretched uniformly to 2.8 times its initial length, by what factor does the power dissipated in the wire change, assuming it remains hooked up to the same voltage source? Assume the wire's volume and density remain constant. (The answer is not 7.84)
Answers
Answered by
scott
"the wire's volume and density remain constant"
... so the cross-sectional area is decreased
... two cylinders with the same volume , but one is 2.8 times as long
the new resistance is related to the lower cross-section
... as well as the longer length
the new resistance should be ... R * 2.8^2 ... or ... R * 7.84
power = (voltage)^2 / resistance
the power changes by a factor of ... 1 / 7.84
(power new) / (power old) = 1 / 7.84
... so the cross-sectional area is decreased
... two cylinders with the same volume , but one is 2.8 times as long
the new resistance is related to the lower cross-section
... as well as the longer length
the new resistance should be ... R * 2.8^2 ... or ... R * 7.84
power = (voltage)^2 / resistance
the power changes by a factor of ... 1 / 7.84
(power new) / (power old) = 1 / 7.84
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