Chap21Sol

Chap21Sol - SOLUTIONS TO PROBLEMS Easy P21.1 I=!Q!t N=!Q =...

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SOLUTIONS TO PROBLEMS Easy P21.1 I = ! Q ! t ! Q = I ! t = 30.0 " 10 # 6 A ( ) 40.0 s ( ) = 1.20 " 10 # 3 C N = Q e = 1.20 ! 10 " 3 C 1.60 ! 10 " 19 C electron = 7.50 ! 10 15 electrons P21.6 I = ! V R = 120 V 240 " = 0.500 A = 500 mA P21.7 ! V = IR and R = ! l A : A = 0.600 mm ( ) 2 1.00 m 1 000 mm ! " # $ % & 2 = 6.00 ' 10 ( 7 m 2 ! V = I " l A : I = ! VA l = 0.900 V ( ) 6.00 # 10 $ 7 m 2 ( ) 5.60 # 10 $ 8 %& m ( ) 1.50 m ( ) I = 6.43 A P21.14 I = P ! V = 600 W 120 V = 5.00 A and R = ! V I = 120 V 5.00 A = 24.0 " . *P21.18 The energy taken in by electric transmission for the fluorescent lamp is P ! t = 100 h ( ) 3 600 s 1 h " # $ % & ' = 3.96 ( 10 6 J cost = 3.96 ( 10 6 J $0.08 kWh " # $ % & ' k 1 000 " # $ % & ' W ) s J " # $ % & ' h 3 600 s " # $ % & ' = $0.088 For the incandescent bulb, P ! t = 40 W 100 h ( ) 3 600 s 1 h " # $ % & ' = 1.44 ( 10 7 J cost = 1.44 ( 10 7 J $0.08 3.6 ( 10 6 J " # $ % & ' = $0.32 saving = $0.32 ) $0.088 = $0.232
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P21.21 You pay the electric company for energy transferred in the amount E = P ! t (a) P ! t = 40 W 2 weeks ( ) 7 d 1 week " # $ % & ' 86 400 s 1 d " # $ % & ' 1 J 1 W ( s " # $ % & ' = 48.4 MJ P ! t = 40 W 2 weeks ( ) 7 d 1 week " # $ % & ' 24 h 1 d " # $ % & ' k 1 000 " # $ % & ' = 13.4 kWh P ! t = 40 W 2 weeks ( ) 7 d 1 week " # $ % & ' 24 h 1 d " # $ % & ' k 1 000 " # $ % & ' 0.12 $ kWh " # $ % & ' = $1.61 (b) P ! t = 970 W 3 min ( ) 1 h 60 min " # $ % & ' k 1 000 " # $ % & ' 0.12 $ kWh " # $ % & ' = $0.005 82 = 0.582¢ (c) P ! t = 5 200 W 40 min ( ) 1 h 60 min " # $ % & ' k 1 000 " # $ % & ' 0.12 $ kWh " # $ % & ' = $0.416 P21.25 (a) P = ! V ( ) 2 R becomes 20.0 W = 11.6 V ( ) 2 R so R = 6.73 ! . (b) ! V = IR so 11.6 V = I 6.73 ! ( ) and I = 1.72 A e = IR + Ir so 15.0 V = 11.6 V + 1.72 A ( ) r r = 1.97 ! . FIG. P21.25
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P21.27 (a) R p = 1 1 7.00 ! ( ) + 1 10.0 ! ( ) = 4.12 ! R s = R 1 + R 2 + R 3 = 4.00 + 4.12 + 9.00 = 17.1 ! (b) ! V = IR 34.0 V = I 17.1 ! ( ) I = 1.99 A for 4.00 ! , 9.00 ! resistors. Applying ! V = IR , 1.99 A ( ) 4.12 ! ( ) = 8.18 V 8.18 V = I 7.00 ! ( ) so I = 1.17 A for 7.00 ! resistor 8.18 V = I 10.0 ! ( ) so I = 0.818 A for 10.0 ! resistor. FIG. P21.27 P21.30 (a) Since all the current in the circuit must pass through the series 100 ! resistor, P = I 2 R P