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Exercise 25.13 - Enhanced - with Solution < & A 1.50 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 17.0V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0C) the ammeter reads 18.4 A, while at 92.0" treads 172 A. You can ignore any thermal expansion of the rod. Review IC IDENTIFY: First use Ohm's law to find the resistance at 20.0°C; then calculate the resistivity from the Finally use the dependence of resistance on temperature to calculate the temperature coefficient of resit SET UP: Ohm's law is R-VIR L /A, and the radius is one-half the diameter. EXECUTE: At 20.0°C, R=V/I= (17.0 V)/(18.4 A) = 0.924 . Using R pL/A and solving for p gives p=RA/L = R (D/2)°/L (0.924 ) ((0.450 x 10 m)/2]°/(1.50m) - 9.80 x 10-60 For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Temperature dependence of resistance. Part B Find the temperature coefficient of resistivity at 20.0°C for the material of the rod. Express your answer in reciprocal degrees Celsius to two significant figures. 10 AERO? Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining