228CHAPTER 21 21.60LXLω=, so L=Then, ()11CLXCCXL==which gives ( )8.0 CΩ Ω12LCLXXC=⋅=or ( )296 L=ΩC(1) From 0012fLCωπ, we obtain 2012LCfπ=Substituting from Equation (1), this becomes 2296 C2012fΩ=or 52.610F26 F622000Hz96029Cfµππ−=×=ΩΩThen, from Equation (1), ( )( )25310F2.510H2.5 mH−−×=×=96 2.6L21.61(a) The box cannot contain a capacitor since a steady DC current cannot flow in a series circuit containing a capacitor. Since the AC and DC currents are different, even when a 3.0 V potential difference is used in both cases, the box must contain a
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