19_InstSolManual_PDF_Part6

19_InstSolManual_PDF_Part6 - / h 2 5 6320 J P 5 1 9.0 V 21...

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19.31. Set Up: When current passes through a battery in the direction from the terminal toward the terminal, the terminal voltage of the battery is Solve: (a) gives (b) so Reflect: The total resistance in the circuit is which agrees with the value specified in the problem. 19.32. Set Up: The graph shows when Solve: (a) When (b) When As A qualitative sketch of versus R is sketched in Figure 19.32a. Figure 19.32 (c) If then The graph of I versus R is sketched in Figure 19.32b. when and when is constant, independent of R. The graph of versus R is sketched in Figure 19.32c. 19.33. Set Up: For a resistor, and Solve: (a) (b) 19.34. Set Up: For a resistor, and Solve: (a) (b) 19.35. Set Up: Solve: (a) (b) 19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is Solve: (a) (b) Energy 5 Pt 5 1 300 W 21 3.0 3 10 2 3 s 2 5 0.90 J P 5 1 25 V 21 12 A 2 5 300 W. P 5 V ab I . Energy 5 1 1.17 W 21 1.5 h 21 3600 s
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Unformatted text preview: / h 2 5 6320 J P 5 1 9.0 V 21 0.13 A 2 5 1.17 W Energy 5 Pt . P 5 VI . P 5 V 2 R 5 1 120 V 2 2 9.0 3 10 3 V 5 1.6 W V 5 IR 5 1 0.0183 A 21 15 3 10 3 V 2 5 275 V. I 5 P R 5 5.0 W 15 3 10 3 V 5 0.0183 A. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI R 5 V I 5 15.0 V 21.8 A 5 0.688 V I 5 P V 5 327 W 15.0 V 5 21.8 A V 5 IR . P 5 VI V ab V ab V ab 5 E ; R S ` . I S R S I S ` I 5 E R . r 5 V ab R E (a) R I (b) V ab R E (c) V ab V ab S E . R S ` , V ab 5 0. R 5 0, V ab 5 IR 5 E R R 1 r 5 E 1 1 r / R . r 5 E I 5 5.00 V 2.5 A 5 2.00 V R 5 0, R 5 0. I 5 2.5 A V ab 5 E 2 Ir . I 5 E R 1 r . I 5 24.0 V 6.00 V 5 4.00 A, R 1 r 5 6.00 V . R 5 V ab I 5 21.2 V 4.00 A 5 5.30 V . V ab 2 IR 5 r 5 E 2 V ab I 5 24.0 V 2 21.2 V 4.00 A 5 0.700 V . V ab 5 E 2 Ir V ab 5 E 2 Ir . V ab 1 2 19-6 Chapter 19...
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