30_Chapter 18 HomeworkCH18 Direct-Current Circuits

30_Chapter 18 HomeworkCH18 Direct-Current Circuits - 128...

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128 CHAPTER 18 (d) The power dissipated in each resistor is found from () 2 VR =∆ P with the following results: ( ) 2 2 6.0 V 6.0 W 6.0 ac ac ac V R == = P ( ) 2 2 1.2 V 0.60 W 2.4 ce ce ce V R == = P ( ) 2 2 1.8 V 0.54 W 6.0 ed ed ed V R P 2 2 1.8 V 0.36 W 9.0 fd fd fd V R P ( ) 2 2 3.0 V 1.5 W 6.0 cd cd cd V R P ( ) 2 2 6.0 V 6.0 W 6.0 db db db V R P 18.48 (a) From 2 P , the resistance of each of the three bulbs is given by 22 120 V 240 60.0 W V R = P 23 and RR 1 R As connected, the parallel combination of is in series with . Thus, the equivalent resistance of the circuit is 1 1 11 1 1 0 36 240 240  =+ + + + = Ω  ΩΩ  1 eq 24 = The total power delivered to the circuit is
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