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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 . ( 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 = 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 . 5. The impedance is Z = X L = L = 2p fL .- + ` - + ` Chapter 31 page 2 6. We find the impedance from Z = X L = L = 2p fL = 2p(33.3 kHz)(36.0 10 3 H) = 7.53 k . For the rms current we have I rms = V rms / X L = (750 V)/(7.53 k ) = 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 . We find the inductance from X L = 2p fL ; 50 = 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 = 7.4 k . ( 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 ) = 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 t , the charge on the capacitor is Q = CV = CV sin t . Thus the current is I = d Q /d t = CV cos t = CV sin ( t + 90). ( b ) If the voltage is V = V sin t , for the circuit we have V = V sin t = L d I /d t , or d I /d t = ( V / L ) sin t . When we integrate this we get I = ( V / L ) cos t = ( V / L ) sin ( t 90). 10. The average power dissipation is P = I rms 2 R = ! I peak 2 R = ! (1.80 A) 2 (260 ) = 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 = 7.58 k ....
View Full Document

Page1 / 16

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online