Example 27.7 Power in an Electric HeaterProblemAn electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has atotal resistance of 9.00 N. Find the current carried by the wire and the power rating of the heater.SolutionBecause AV = IR, we have the following.AVI120 VR9.00 NX AYour response differs from the correct answer by more than 10%. Double check your calculations.We can find the power rating using the expression P = IR.P = 1²R = 1²(9.00 N)X WYour response differs significantly from the correct answer. Rework your solution from the beginning andcheck each step carefully.What If? What if the heater were accidentally connected to a 240 V supply? (This is difficult to do because theshape and orientation of the metal contacts in 240 V plugs are different from those in 120 V plugs.) How wouldthis affect the current carried by the heater and the power rating of the heater?Answer If we doubled the applied potential difference, Equation 27.8 tells us that the current would double.According to Equation 27.23, P = (AV)²/R, the power would be four times larger.

Answered: Image /qna-images/answer/7578ba54-3ad8-40b2-8aa8-05f2a9a513ce.jpg

Problem An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 9.00 N. Find the current carried by the wire and the power rating of the heater.

Problem An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 9.00 N. Find the current carried by the wire and the power rating of the heater.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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