19_InstSolManual_PDF_Part7

19_InstSolManual_PDF_Part7 - 2 2 1 1.0 V 2 5 4.0 W P E 5 VI...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
19.37. Set Up: Solve: (a) (b) P increases when R decreases, so the 100 W bulb has less resistance. Reflect: The bulb with smaller R draws more current and this more than compensates for smaller R in 19.38. Set Up: and Solve: (a) Yes, this is very dangerous. (b) We can also calculate P as or VI. These expressions also give 14.4 W. 19.39. Set Up: and energy is the product of power and time. Solve: Reflect: The energy delivered depends not only on the voltage and current but also on the length of the pulse. 19.40. Set Up: The heater consumes 540 W when Solve: (a) so (b) so (c) The cost is (d) Assuming that R remains P is smaller by a factor of 19.41. Set Up: Solve: (a) The energy supplied is (b) Reflect: The battery stores a large amount of energy and it takes time to return this energy to the battery in order to recharge it. 19.42. Set Up: For an emf, For a resistor, Solve: (a) and The rate at which chemical energy is being converted to electrical energy within the battery is (b) (c) Note that P E 5 P r 1 P R . P R 5 I 2 R 5 1 2.0 A 2 2 1 5.0 V 2 5 20.0 W P r 5 I 2 r 5 1 2.0 A
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 2 1 1.0 V 2 5 4.0 W P E 5 VI 5 1 12.0 V 21 2.0 A 2 5 24.0 W. I 5 2.0 A. 12.0 V 2 I 1 1.0 V 1 5.0 V 2 5 P 5 I 2 R . P 5 VI . t 5 energy P 5 2.7 3 10 6 J 0.45 3 10 3 J / s 5 6.0 3 10 3 s 5 100 minutes VIt 5 1 12.6 V 21 60 A 21 3600 s 2 5 2.7 3 10 6 J. P 5 VI . Energy 5 Pt . 1 110 / 120 2 2 . P 5 V 2 R 5 1 110 V 2 2 26.7 V 5 453 W. 26.7 V , 1 0.540 kWh 21 7.2 cents / kWh 2 5 3.9 cents. Energy 5 1 0.540 kW 21 1 h 2 5 0.540 kWh. I 5 P V 5 540 W 120 V 5 4.50 A P 5 VI R 5 V 2 P 5 1 120 V 2 2 540 W 5 26.7 V P 5 V 2 R Energy 5 Pt . V 5 120 V. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI . Energy 5 Pt 5 1 40 W 21 10 3 10 2 3 s 2 5 0.40 J. P 5 1 500 V 21 80 3 10 2 3 A 2 5 40 W. P 5 VI I 2 R P 5 V 2 R 5 1 120 V 2 2 1.0 3 10 3 V 5 14.4 W. I 5 V R 5 120 V 1.0 3 10 3 V 5 0.12 A. P 5 V 2 R . V 5 IR I 2 R . P 5 I 2 R . R 5 V 2 P 5 1 120 V 2 2 100 W 5 144 V R 5 V 2 P 5 1 120 V 2 2 60 W 5 240 V P 5 V 2 R Current, Resistance, and Direct-Current Circuits 19-7...
View Full Document

This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

Ask a homework question - tutors are online