PHYS
Chapter 32

# 53 solve the cost of energy of a 60 w incandescent

• Notes
• 83

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32.53. Solve: The cost of energy of a 60 W incandescent bulb over its lifetime is 1 kW \$0.10 60 W 1000 h \$6.00 1000 W 1 kWh × × × = This means the life-cycle cost of the incandescent bulb is \$6.50. The cost of energy of a 15 W compact fluorescent bulb over its lifetime is 1 kW \$0.10 15 W 10,000 h \$15.00 1000 W 1 kWh × × × = With the bulb’s cost of \$15, the life-cycle cost is \$30.00. To make a comparison of the cost effectiveness of the two bulbs, we note that you need ten incandescent bulbs to last as long as one fluorescent bulb. Thus, it will cost a consumer \$65.00 to use incandescent bulbs for the same time span as a single fluorescent bulb that will cost only \$30.00. The fluorescent bulb is cheaper.

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32.54. Solve: (a) The cost per month of the 1000 W refrigerator is 1 kW \$0.10 1000 W 30 24 h 0.20 \$14.40 1000 W 1 kWh × × × × × = (b) The cost per month of a refrigerator with a 800 W compressor is \$11.52. The difference in the running cost of the two refrigerators is \$2.88 per month. So, the number of months before you recover the additional cost of \$100 (of the energy efficient refrigerator) is \$100/\$2.88 = 34.7 months.
32.55. Visualize: Please refer to Figure P32.55. Solve: (a) Only bulb A is in the circuit when the switch is open. The bulb’s resistance R is in series with the internal resistance r , giving a total resistance R eq = R + r . The current is bat 1.50 V 0.231 A 6.50 I R r = = = + Ω E This is the current leaving the battery. But all of this current flows through bulb A, so I A = I bat = 0.231 A. (b) With the switch closed, bulbs A and B are in parallel with an equivalent resistance R eq = 1 2 R = 3.00 Ω . Their equivalent resistance is in series with the battery’s internal resistance, so the current flowing from the battery is bat eq 1.50 V 0.428 A 3.50 I R r = = = + Ω E But only half this current goes through bulb A, with the other half through bulb B, so 1 A bat 2 0.214 A I I = = . (c) The change in I A when the switch is closed is 0.017 A. This is a decrease of 7.4%. (d) If r = 0 Ω , the current when the switch is open would be I A = I bat = 0.250 A. With the switch closed, the current would be I bat = 0.500 A and the current through bulb A would be 1 A bat 2 0.250 A I I = = . The current through A would not change when the switch is closed.

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32.56. Model: The battery and the connecting wires are ideal. Visualize: The figure shows the two circuits formed from the circuit in Figure P32.56 when the switch is open and when the switch is closed. Solve: (a) Using the rules of series and parallel resistors, we have simplified the circuit in two steps as shown in figure (a). A battery with emf E = 24 V is connected to an equivalent resistor of 3 Ω . The current in this circuit is 24 V 3 8 A Ω = . Thus, the current that flows through the battery is I bat = 8 A. To determine the potential difference Δ V ab , we will find the potentials at point a and point b and then take their difference. To do this, we need the currents I a and I b . We note that the potential difference across the 3 Ω -3 Ω branch is the same as the potential difference across the 5 Ω -1 Ω branch. So, E = 24 V = I a (3 Ω + 3 Ω ) I a = 4 A = I b Now, V c I a (3 Ω ) = V a , and V c I b (5 Ω ) = V b . Subtracting these two equations give us Δ V ab : V a V b = I b (5 Ω ) – I a (3 Ω ) =
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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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