h U kW h W 4 34 W h 0003 kW h Cost per kW h 2663 Recharging cost 0027 cents 59

H u kw h w 4 34 w h 0003 kw h cost per kw h 2663

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h) × U (kW . h) W = 4 × 3/4 W . h = 0.003 kW . h Cost per kW . h = $2663 Recharging cost = 0.027 cents 59 ∙∙ A 12-V automobile battery with negligible internal resistance can deliver a total charge of 160 A h. ( a ) What
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Chapter 26 Electric Current and Direct-Current Circuits is the total stored energy in the battery? ( b ) How long could this battery provide 150 W to a pair of headlights? ( a ) U = E It ( b ) t = U / P U = 12 × 160 W . h = 1.92 kW . h = 6.91 MJ t = 1920/150 = 12.8 h 60 ∙∙ A space heater in an old home draws a 12.5-A current. A pair of 12-gauge copper wires carries the current from the fuse box to the wall outlet, a distance of 30 m. The voltage at the fuse box is exactly 120 V. ( a ) What is the voltage delivered to the space heater? ( b ) If the fuse will blow at a current of 20 A, how many 60-W bulbs can be supplied by this line when the space heater is on? (Assume that the wires from the wall to the space heater and to the light fixtures have negligible resistance.) ( a ) Find R Cu = ρ L / A , L = 60 m; V = E IR Cu ( b ) Find R bulb = E 2 / P = 240 ; I bulb = V / R bulb N = ( I max – 12.5)/ I bulb R Cu = 0.308 ; V = (120 – 0.308 × 12.5) V = 116 V I bulb = 116/240 A = 0.483 A N = 15 bulbs 61* ∙∙ A lightweight electric car is powered by ten 12-V batteries. At a speed of 80 km/h, the average frictional force is 1200 N. ( a ) What must be the power of the electric motor if the car is to travel at a speed of 80 km/h? ( b ) If each battery can deliver a total charge of 160 A h before recharging, what is the total charge in coulombs that can be delivered by the 10 batteries before charging? ( c ) What is the total electrical energy delivered by the 10 batteries before recharging? ( d ) How far can the car travel at 80 km/h before the batteries must be recharged? ( e ) What is the cost per kilometer if the cost of recharging the batteries is 9 cents per kilowatt-hour? ( a ) P = Fv ( b ) Q = It ( c ) W = Q E ( d ) W = Fd ( e ) Cost = $0.09( E It) /1000 P = (1200 × 22.2) W = 26.7 kW Q = (160 × 10 × 3600) C = 5.76 MC W = 69.1 MJ d = 57.6 km Cost/km = (0.09 × 120 × 160/10 3 )/57.6 = $0.03/km 62 ∙∙∙ A 100-W heater is designed to operate with an emf of 120 V. ( a ) What is its resistance, and what current does it draw? ( b ) Show that if the potential difference across the heater changes by a small amount V , the power changes by a small amount P , where P / P 2 V / V . ( Hint: Approximate the changes with differentials.) ( c ) Find the approximate power dissipated in the heater if the potential difference is decreased to 115 V. ( a ) I = P / V ; R = V / I ( b ) P = V 2 / R ; P = ( dP / dV ) V ( c ) Find P ; P = P 0 + P I = 100/120 A = 0.833 A; R = 144 P = (2 V / R ) V = 2 P ( V / V ); P / P = 2 V / V P –8.33 W; P 91.7 W 63 Two resistors are connected in parallel across a potential difference. The resistance of resistor A is twice that of resistor B. If the current carried by resistor A is I , then what is the current carried by B? ( a ) I ( b ) 2 I ( c ) I /2 ( d ) 4 I ( e ) I /4 ( b ) 64 Two resistors are connected in series across a potential difference. Resistor A has twice the resistance of resistor B. If the current carried by resistor A is I , then what is the current carried by B? ( a ) I ( b ) 2 I ( c ) I /2 ( d ) 4 I ( e ) I /4 ( a )
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Chapter 26 Electric Current and Direct-Current Circuits 65* ∙∙
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