930_Physics ProblemsTechnical Physics

# 930_Physics - Chapter 33 e je j X L = L = 2 50.0 s 1 250 10 3 H = 78.5 XC =(c Z = R 2 XL XC(d I max =(e P33.22(a(b P33.21 271 = tan 1(a Z = R 2 X L

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Chapter 33 271 P33.21 (a) XL L == × = −− ωπ 25 0 0 2 5 01 0 7 8 5 13 .. s H ej e j (b) X C C × = 1 0 0 2 0 0 1 5 9 16 1 ω π . F k e j (c) ZR XX LC =+ = 2 2 152 bg . k (d) I V Z max max × = 210 138 V 1.52 10 mA 3 (e) φ = L N M O Q P =− = ° tan tan . . 11 10 1 84 3 R af P33.22 (a) ZR XX = +− = 2 2 2 2 68 0 16 0 101 109 X C L C = × = 100 0 160 16 0 100 99 0 10 101 6 a fa f a f . (b) I V Z max max . . === 40 0 0367 V 109 A (c) tan . . . = = R 16 0 101 68 0 125 : ° 0896 513 rad I max . = 0367 A = 100 rad s ° rad P33.23 Xf L L = 2 2 60 0 0 460 173 ππ a f X fC C × = 1 2 1 26 0 02 1 0 126 6 a f (a) tan . = = = R 173 126 0314 150 ΩΩ ° 0304 174 rad (b) Since > , is positive; so voltage leads the current . *P33.24 For the source-capacitor circuit, the rms source voltage is VX sC = 25 1 . mA . For the circuit with resistor, VR X X C = 15 7 25 1 22 mA mA . This gives
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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