28_Chapter 21 HomeworkCH21 Alternating Current Circuits and Electromagnetic Waves

28_Chapter 21 HomeworkCH21 Alternating Current Circuits and Electromagnetic Waves

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online