935_Physics ProblemsTechnical Physics

935_Physics ProblemsTechnical Physics - 276 P33.40...

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276 Alternating Current Circuits P33.40 L = 20 0 . mH , C 100 10 7 . , R = 20 0 . , V max = 100 V (a) The resonant frequency for a series –RLC circuit is f LC == 1 2 1 356 π k H z . (b) At resonance, I V R max max . 500±A . (c) From Equation 33.38, Q L R ω 0 22 4 .. (d) VX I L I LL ,m a x m a x . max kV = 0 224 P33.41 The resonance frequency is 0 1 = LC . Thus, if ωω = 2 0 , XL LC L L C L F H G I K J = 2 2 and X C LC C L C C = 1 2 1 2 ZR XX R L C LC =+ F H G I K J 2 2 2 225 bg .s o I V Z V RL C rms rms rms + ∆∆ 2 . and the energy delivered in one period is Et = P : E VR C C RC L LC C L C L = + F H G I K J = + = + rms rms rms ej 2 2 2 2 2 2 2 4 49 0 0 . . With the values specified for this circuit, this gives: E = ×× ×+ × = −− 4 50 0 10 0 100 10 10 0 10 4 10 0 100 10 9 00 10 0 10 242 2 6 32 3 12 2 63 . . V F H F H mJ af e j e j . P33.42 The resonance frequency is 0 1 = LC . Thus, if = 2 0 , LC L L C L F H G I K J = 2 2 and X C LC
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