WHomework set 6_solns_2011

# WHomework set 6_solns_2011 - Homework set # 6 Problems:...

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Homework set # 6 Problems: 17.29, 34; 18.2, 4-6, 9, 12 Solutions: 17.29 The maximum power that can be dissipated in the circuit is      3 max max 120 V 15 A 1.8 10 W VI    P Thus, one can operate at most 18 bulbs rated at 100 W per bulb. 17.34 (a) At the operating temperature,      120 V 1.53 A 184 W P (b) From   00 1 R RT T    , the temperature T is given by 0 0 0 R R TT R  . The resistances are given by Ohm’s law as   120 V 1.53 A V R I  , and   0 0 0 120 V 1.80 A V R I Therefore, the operating temperature is        1 3 120 1.53 120 1.80 20.0 C 461 C 0.400 10 C 120 1.80 T   18.2 (a) 123 4.0 8.0 12 24 eq RR R R       (b) The same current exists in all resistors in a series combination. 24 V 1.0 A 24 eq V I R

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(c) If the three resistors were connected in parallel, 1 1 123 111 1 1 1 2.18 4.0 8.0 12 eq R RRR     
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## This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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WHomework set 6_solns_2011 - Homework set # 6 Problems:...

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