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Unformatted text preview: 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 Ω . ( 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 Ω . 2. We find the frequency from X L = ω L = 2p fL ; 660 Ω = 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 Ω = 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 1000 f (Hz) X C (k Ω ) 1.5 500 5. The impedance is Z = X L = ω L = 2p fL . f (Hz) 10,000 5000 150 300 X L ( Ω ) Chapter 31 page 2 6. We find the impedance from Z = X L = ω L = 2p fL = 2p(33.3 kHz)(36.0 × 10 –3 H) = 7.53 k Ω . For the rms current we have I rms = V rms / X L = (750 V)/(7.53 k Ω ) = 99.6 mA . 7. If there is no current in the secondary, there will be no induced emf from the mutual inductance. We find the impedance from Z = X L = V rms / I rms = (110 V)/(2.2 A) = 50 Ω . We find the inductance from X L = 2p fL ; 50 Ω = 2p(60 Hz) L , which gives L = 0.13 H . 8. ( a ) We find the impedance from Z = X C = 1/2p fC = 1/2p(600 Hz)(0.036 × 10 –6 F) = 7.37 × 10 3 Ω = 7.4 k Ω . ( b ) We find the peak value of the current from I peak = v2 I rms = v2( V rms / Z ) = v2(22 kV)/(7.37 k Ω ) = 4.2 A . The frequency of the current will be the frequency of the line: 600 Hz . 9. ( a ) If the voltage is V = V sin ω t , the charge on the capacitor is Q = CV = CV sin ω t . Thus the current is I = d Q /d t = ω CV cos ω t = ω CV sin ( ω t + 90°). ( b ) If the voltage is V = V sin ω t , for the circuit we have V = V sin ω t = L d I /d t , or d I /d t = ( V / L ) sin ω t . When we integrate this we get I = – ( V / ω L ) cos ω t = ( V / ω L ) sin ( ω t – 90°). 10. The average power dissipation is P = I rms 2 R = ! I peak 2 R = ! (1.80 A) 2 (260 Ω ) = 421 W . 11. Because the capacitor and resistor are in parallel, their currents are I C = V / X C , and I R = V / R . The total current is I = I C + I R . ( a ) The reactance of the capacitor is X C1 = 1/2p f 1 C = 1/2p(60 Hz)(0.35 × 10 –6 F) = 7.58 × 10 3 Ω = 7.58 k Ω . For the fraction of current that passes through C , we have fraction1 = I C1 /( I C1 + I R ) = (1/ X C1 )/[(1/ X C1 ) + (1/ R )] = R /( R + X C1 ) = (0.400 k Ω )/(0.400 k Ω + 7.58 k Ω ) = 0.050 = 5.0%. ( b ) The reactance of the capacitor is X C2 = 1/2p f 2 C = 1/2p(60,000 Hz)(0.35 × 10 –6 F) = 7.58 Ω . For the fraction of current that passes through C , we have fraction2 = I C2 /( I C2 + I R ) = (1/ X C2 )/[(1/ X C2 ) + (1/ R )] = R /( R + X C2 ) = (400 Ω )/(400 Ω + 7.58 Ω ) = 0.98 = 98%....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics

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