ch31-p059

# ch31-p059 - 59 We shall use 2 mR Pavg = 2Z 2 b where Z = R...

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() 2 avg 30.0V 90.0W. 2 5.00 P == (d) At maximum power, the reactances are equal: X L = X C . The phase angle φ in this case may be found from tan , = = XX R LC 0 which implies = 0 ° . (e) At maximum power, the power factor is cos = cos 0° = 1. (f) The minimum average power is P avg = 0 (as it would be for an open switch). (g) On the other hand, at minimum power X C 1/ C is infinite, which leads us to set tan =−∞ . In this case, we conclude that = –90°. (h) At minimum power, the power factor is cos = cos(–90°) = 0. 59. We shall use 22 avg 2 2 2 . 2 21 / mm dd RR P Z RLC εε ωω +− where ZR L C =+ 2 2 1 / b g is the impedance. (a) Considered as a function of C, P avg has its largest value when the factor 2 2 1/ R has the smallest possible value. This occurs for
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