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Ch31Word - Chapter 31 page 1 CHAPTER 31 AC Circuits 1(a The...

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Chapter 31 page 1 CHAPTER 31 – AC Circuits 1. ( a ) The reactance of the capacitor is X C1 = 1/2p f 1 C = 1/2p(60 Hz)(7.2 × 10 –6 F) = 3.7 × 10 2 . ( b ) For the new frequency we have X C2 = 1/2p f 2 C = 1/2p(1.0 × 10 6 Hz)(7.2 × 10 –6 F) = 2.2 × 10 –2 . 2. We find the frequency from X L = ω L = 2p fL ; 660 = 2p f (22.0 × 10 –3 H), which gives f = 4.77 × 10 3 Hz = 4.77 kHz . 3. We find the frequency from X C = 1/2p fC ; 6.70 × 10 3 = 1/2p f (2.40 × 10 –6 F), which gives f = 9.90 Hz . 4. The impedance is Z = X C = 1/2p fC . 3.0 0 0 1000 f (Hz) X C (k ) 1.5 500 5. The impedance is Z = X L = ω L = 2p fL . f (Hz) 10,000 5000 0 0 150 300 X L ( )
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