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Unit 13 Solutions_Page_12

# Unit 13 Solutions_Page_12 - (C 2010 McGraw—Hill The peak...

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Unformatted text preview: (C) 2010 McGraw—Hill The peak voltages are given in the table below. 0.10 0.83 0.50 4.2 (b) Compute the peak current. 1:1: V ##- X (111, 2:1f(L1+L2)291(126H2)(0.10H+0.50H) L eq 30. Strategy Since the capacitor and inductor are in series, the same current flows through both. Solution (a) V0 (t) lags 1'01) by 90° and V]. (t) leads 1(1) by 90°, so the phase difference is M =¢L —¢c 2900—(—900) = -. (b) VL (I) leads VC (I) by 180°, which means that they are opposite in sign. The ac voltmeter reads the difference, -- Problems: 21.36, 21.41, 21.53, [21.54], 21.56, [21.57] 36. Strategy Use 8 : 12 where Z :‘IRZ +Xg. Solution Find R. R 0.80 11F z—%—,/R2+:§ I I 2 8] l 3 [71: 2R2 +X§= w2C2 110 v 1 ‘l2 4:: f —C2:2 {0.— 0284 A 432(600 Hz)2(0.80><10’ﬁ 1:)2 41. Strategy The average power is given by PaY — I 8 00595 where Inns :gms/Z and COS g5 : R/Z. rms rms Solution Find the average power dissipated. 2 2 2209 B... =Im8m cosng—m8m32{g—m] RZM (g Z Z Z R2+(a}L—ﬁ) 0.1511111 (12 W2(220 Q) 80 as 2—: 0.33W 2 22092 2 2500Hz 015 10-3 —1 2 - ( ) +[ j“ X ' X mikﬁﬂSﬂDl-{zXSDXlD—G a] 782 ...
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