28_Chapter 21 HomeworkCH21 Alternating Current Circuits and Electromagnetic Waves

# 28_Chapter 21 HomeworkCH21 Alternating Current Circuits and Electromagnetic Waves

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228 CHAPTER 21 21.60 L XL ω = , so L = Then, () 11 C L X CCXL == which gives ( ) 8.0 C Ω Ω   12 LC LXXC =⋅ = or ( ) 2 96 L =Ω C (1) From 00 1 2 f LC ωπ , we obtain 2 0 1 2 LC f π = Substituting from Equation (1), this becomes 22 96 C 2 0 1 2 f Ω= or 5 2.6 10 F 26 F 6 2 2 0 0 0 H z 9 6 0 29 C f µ ππ = × = ΩΩ Then, from Equation (1), ( )( ) 25 3 10 F 2.5 10 H 2.5 mH −− × = 96 2.6 L 21.61 (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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