ch10_solutions

ch10_solutions - CHAPTER 10 10.1 A/C temperature Automobile...

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10-316 CHAPTER 10 10.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage amplitude voltage phase current amplitude current phase power power factor spectrum Fan speed Humidity Lawn mower speed Light intensity Oven temperature Refrigerator temperature Sewing machine speed Stereo volume Stove temperature Time TV picture brightness TV sound level Wind velocity 10.2 ( a ) 20 log (125) = 41.9 dB | 20 log (50) = 34.0 dB | 20 log (50000) = 94.0 dB 20 log(100000) =100 dB | 20 log(0.85) = - 1.41 dB ( b ) 20 log (600) = 55.6 dB | 20 log (3000) = 69.5 dB | 20 log (10 6 ) = 120 dB 20 log(200000) =106 dB | 20 log(0.95) = - 0.446 dB ( c ) 10 log (2x10 9 ) = 93.0 dB | 10 log (4x10 5 ) = 56.0 dB 10 log (6x10 8 ) = 87.8 dB | 10 log(10 10 ) =100 dB 10.3 -4 -2 0 2 4 0 0.5 1 1.5 2 2.5 3 3.5 4 x10 -3 v S v O
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10-317 b ( 29 500 Hz : 1 0 o | 1500 Hz : 0.333 0 o | 2500 Hz : 0.200 0 o c ( 29 500 Hz : 2 30 o | 1500 Hz : 1 30 o | 2500 Hz : 1 30 o d ( 29 500 Hz : 2 30 o | 1500 Hz : 3 30 o | 2500 Hz : 5 30 o e ( 29 Yes 10.4 V s = 0.005 V | P O = 40 W | V o = 2 P O R L = 2 40 ( 29 8 ( 29 = 25.3 V A v = 25.3 .005 = 5060 | 20 log 5060 ( 29 = 74.1 dB I s = 0.005 V 5 k Ω+ 50 k = 90.91 nA | I o = V o 8 = 25.3 V 8 = 3.162 A A i = 3.162 A 90.91 nA = 3.48 x 10 7 | 20 log 3.48 x 10 7 ( 29 = 151 dB A p = 40 W .005 V 90.91 nA ( 29 2 = 1.76 x 10 11 | 10 log 1.76 x 10 11 ( 29 = 112 dB 10.5 V s = 0.01 V | P O = 20 mW | V o = 2 P O R L = 2 .02 ( 29 8 ( 29 = 0.566 V A v = 0.566 .01 = 56.6 | 20 log 56.6 ( 29 = 35.0 dB I s = 0.01 V 2 k Ω+ 50 k = 192 nA | I o = V o 8 = 0.566 V 8 = 70.8 mA A i = 70.8 mA 192 nA = 3.68 x 10 5 | 20 log 3.68 x 10 5 ( 29 = 111 dB A p = 0.02 W .01 V 192 nA ( 29 2 = 2.08 x 10 7 | 10 log 2.08 x 10 7 ( 29 = 73.2 dB 10.6 + - R th v th a ( 29 v th = v oc = 0.768 2 = 1.09 V v o = R L R th + R L v th R th = R L v th - v o v o = 430 0.768 - 0.721 0.721 = 28.0 b ( 29 v th = v oc = 0.760 2 = 1.08 V v o = R L R th + R L v th R th = R L v th - v o v o = 1040 0.760 - 0.740 0.740 = 28.1 c ( 29 1.09 V and 1.08 V 9% error and 8% error 10.7 G4 laptop – 1 V, 28 .
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10-318 10.8 a ( 29 v i = 1 V 20000 20000 Ω+ 33 = 0.998 V | v o = 2 R L P O = 2 16 ( 29 5 ( 29 = 12.7 V P i = v i 2 2 R i = 1 2 40066 = 25.0 m W | i i = 1 V 20000 Ω+ 33 = 49.9 m A | i o = 12.7 V 16 = 0.794 A A v = 12.7 0.998 = 12.7 | A P = 5 W 25 mW = 2.00 x 10 5 | A i = 0.794 A 49.9 mA = 1.59 x 10 4 b ( 29 v o = 12.7 V ; recommend ± 15 - V supplies 10.9 v = 2 PR | i = 2 P R -- The 24- case represents a good trade off between voltage and current. R(
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ch10_solutions - CHAPTER 10 10.1 A/C temperature Automobile...

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