22_InstSolManual_PDF_Part7

22_InstSolManual_PDF - V 2 318 V 2 2 5 369 V X C 5 1 2 p fC 5 1 1 2 p 21 1000 Hz 21 0.500 3 10 2 6 F 2 5 318 V X L 5 2 p fL 5 1 2 p 21 1000 Hz 21

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22.15. Set Up: Solve: (a) and The source voltage lags the current. The phasor diagram is sketched in Figure 22.15a. Figure 22.15 (b) and The source voltage leads the current. The phasor diagram is sketched in Figure 22.15b. Reflect: When f increases, increases and decreases. When the phase angle is negative and when the phase angle is positive. 22.16. Set Up: 20.0 W is the average power P. For a pure resistance, Solve: (a) The average power is one-half the maximum power, so the maximum instantaneous power is 40.0 W. (b) (c) R 5 P I rms 2 5 20.0 W 1 0.167 A 2 2 5 720 V I rms 5 P V rms 5 20.0 W 120 V 5 0.167 A P 5 V rms I rms 5 I rms 2 R . X L . X C X C . X L X C X L f51 57.2°. tan f5 X L 2 X C R 5 628 V2 318 V 200 V 51 1.55 Z 5 "1 200 V 2 2 1 1 628
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Unformatted text preview: V 2 318 V 2 2 5 369 V . X C 5 1 2 p fC 5 1 1 2 p 21 1000 Hz 21 0.500 3 10 2 6 F 2 5 318 V . X L 5 2 p fL 5 1 2 p 21 1000 Hz 21 0.100 H 2 5 628 V . I V L V R V V L 2 V C V C f ( b ) I V L V R V V C 2 V L V C f ( a ) f 5 2 58.3°. tan f 5 X L 2 X C R 5 314 V 2 637 V 200 V 5 2 1.62 Z 5 "1 200 V 2 2 1 1 314 V 2 637 V 2 2 5 380 V . X C 5 1 2 p fC 5 1 1 2 p 21 500 Hz 21 0.500 3 10 2 6 F 2 5 637 V . X L 5 2 p fL 5 1 2 p 21 500 Hz 21 0.100 H 2 5 314 V . tan f 5 X L 2 X C R . X C 5 1 v C 5 1 2 p fC . X L 5 v L 5 2 p fL . Z 5 " R 2 1 1 X L 2 X C 2 2 . Alternating Current 22-7...
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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.

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