Lecture11_Kim

# Lecture11_Kim - Lecture 11-1 PHYS241 Question 1 September...

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Lecture 11-1 PHYS241 – Question 1 – September 27, 2011 To you, the Exam I was a. Much more difficult than expected b. More difficult than expected c. About as expected d. Easier than expected e. Much easier than expected

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Lecture 11-2 PHYS241 – Question 2 – September 27, 2011 On the Exam I, you think you did a. Very well b. Well c. Average d. Poorly e. Very poorly
Lecture 11-3 Energy in Electric Circuits (Joule’s Heating) (review) dU P IV dt   So, Power dissipation = rate of decrease of U = Steady current means a constant amount of charge Q flows past any given cross section during time t , where I= Q / t . Energy lost by Q is () ab U Q V V I t V        22 / I R V R  => Heat (Joule’s Heating) V

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Lecture 11-4 emf – Electromotive Force • An emf device is a charge pump that can maintain a potential difference across two terminals by doing work on the charges when necessary. Examples: battery, fuel cell, electric generator, solar cell, thermopile, … Converts energy (chemical, mechanical, solar, thermal, …) into electrical energy . Within the emf device, positive charges are lifted from lower to higher potential. If work dW is required to lift charge dq ,   dW Volt dq dq dW   emf
Lecture 11-5 Circuit with an emf and resistors 12 0 iR i R  R 1 R 2 iR 2

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Lecture 11-6 Energy Conservation Ideal battery (B) with emf , resistor R , and wires of negligible resistance. Energy conservation Work done by battery is equal to energy dissipated in resistor 2 () , B R B Q P P P i t i i R or iR  
Lecture 11-7

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## This note was uploaded on 12/07/2011 for the course PHYS 241 taught by Professor Wei during the Fall '08 term at Purdue University-West Lafayette.

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Lecture11_Kim - Lecture 11-1 PHYS241 Question 1 September...

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