22_InstSolManual_PDF_Part10

22_InstSolManual_PDF_Part10 - 5 106 V 200 V 5 0.530 A. I V...

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22.26. Set Up: At resonance, Solve: (a) (b) so (c) 22.27. Set Up: From Problem 22.14, At resonance, At resonance, Solve: (a) (b) The phasor diagram is sketched in Figure 22.27. Figure 22.27 (c) reads reads reads reads zero because and these voltages are out of phase. reads 106 V, the source rms voltage and also the rms voltage across the resistor. (d) Changing R doesn’t affect and (e) Reflect: The potential difference between points c and b is zero at every value of t. The voltages across L and C always add to give zero. I rms 5 V rms Z 5 V rms R 5 106 V 100 V 5 1.06 A. v 0 5 745 rad / s. v 0 , v 0 5 1 " LC . V 5 180° V L 5 V C V 4 V C ,rms 5 I rms X C 5 I rms v C 5 0.530 A 1 745 rad / s 21 2.00 3 10 2 6 F 2 5 356 V. V 3 V L ,rms 5 I rms X L 5 I rms v L 5 1 0.530A 21 745 rad / s 21 0.900 H 2 5 356 V. V 2 V R ,rms 5 I rms R 5 1 0.530 A 21 200 V 2 5 106 V. V 1 I rms 5 V rms Z 5 V rms R
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Unformatted text preview: 5 106 V 200 V 5 0.530 A. I V L V R V V C v 5 1 " LC 5 1 "1 0.900 H 21 2.00 3 10 2 6 F 2 5 745 rad / s V rms 5 V " 2 5 150 V " 2 5 106 V. f 5 0. X L 5 v L . X C 5 1 v C . Z 5 R . I rms 5 V rms Z . v 5 1 " LC . C 5 2.00 m F. L 5 0.900 H, R 5 200 V , V R 5 IR 5 1 0.600 A 21 400 V 2 5 240 V. V C 5 V L 5 30.0 V. V L 5 IX L 5 1 0.600 A 21 50.0 V 2 5 30.0 V. X C 5 X L 5 50.0 V . X L 5 v L 5 1 250 rad / s 21 0.200 H 2 5 50.0 V . R 5 V I 5 240 V 0.600 A 5 400 V . V 5 IZ 5 IR v 5 1 " LC 5 1 "1 0.200 H 21 80.0 3 10 2 6 F 2 5 250 rad / s Z 5 R . V 5 IZ . v 5 1 " LC . 22-10 Chapter 22...
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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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