22_InstSolManual_PDF_Part7 - V 2 318 V 2 2 5 369 V X C 5 1...

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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 f 5 1 57.2°. tan f 5 X L 2 X C R 5 628 V 2 318 V 200 V 5 1 1.55 Z 5 " 1 200 V 2 2 1 1 628 V 2
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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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