P27_062.PDF

62. (a) We use P = V 2 /R

∝

V 2 ,whichgives∆ P

∝

∆ V 2

≈

2 V ∆ V . The percentage change is roughly

8 . 6% .

(b) A drop in V causes a drop in P , which in turn lowers the temperature of the resistor in the coil. At

a lower temperature R is also decreased. Since P

−

115) / 115=

−

R − 1 a decrease in R will result in an increase

in P , which partially oﬀsets the decrease in P due to the drop in V .Thus,theactualdropin P

will be smaller when the temperature dependency of the resistance is taken into consideration.

∝

∆ P/P =2∆ V/V =2(110