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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
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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Example 27.7 Power in an Electric 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. Solution Because AV = IR, we have the following. AV I 120 V R 9.00 N X A Your 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 W Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. What If? What if the heater were accidentally connected to a 240 V supply? (This is difficult to do because the shape and orientation of the metal contacts in 240 V plugs are different from those in 120 V plugs.) How would this 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.
Transcribed Image Text:Example 27.7 Power in an Electric 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. Solution Because AV = IR, we have the following. AV I 120 V R 9.00 N X A Your 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 W Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. What If? What if the heater were accidentally connected to a 240 V supply? (This is difficult to do because the shape and orientation of the metal contacts in 240 V plugs are different from those in 120 V plugs.) How would this 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.
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