31_InstructorSolutions - ALTERNATING CURRENT 31 31.1....

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31-1 A LTERNATING C URRENT 31.1. IDENTIFY: cos iI t ω = and rms /2 . II = SET UP: The specified value is the root-mean-square current; rms 0.34 A. I = EXECUTE: (a) rms 0.34 A I = (b) rms 2 2(0.34 A) 0.48 A. == = (c) Since the current is positive half of the time and negative half of the time, its average value is zero. (d) Since rms I is the square root of the average of 2 , i the average square of the current is 22 2 rms (0.34 A) 0.12 A . I EVALUATE: The current amplitude is larger than its rms value. 31.2. IDENTIFY and SET UP: Apply Eqs.(31.3) and (31.4) EXECUTE: (a) rms ( 2 . 1 0 A ) 2 . 9 7 A . = (b) rav (2.97 A) 1.89 A. ππ = EVALUATE: (c) The root-mean-square voltage is always greater than the rectified average, because squaring the current before averaging, and then taking the square root to get the root-mean-square value will always give a larger value than just averaging. 31.3. IDENTIFY and SET UP: Apply Eq.(31.5). EXECUTE: (a) rms 45.0 V 31.8 V. V V = (b) Since the voltage is sinusoidal, the average is zero. EVALUATE: The voltage amplitude is larger than rms . V 31.4. IDENTIFY: C VI X = with 1 C X C = . SET UP: is the angular frequency, in rad/s. EXECUTE: (a) C I X C so 6 (60.0 V)(100 rad s)(2.20 10 F) 0.0132 A. IVC × = (b) 6 (60.0 V)(1000 rad s)(2.20 10 F) 0.132 A. × = (c) 6 (60.0 V)(10,000 rad s)(2.20 10 F) 1.32 A. × = (d) The plot of log I versus log is given in Figure 31.4. EVALUATE: IV C = so log log( ) log . C =+ A graph of log I versus log should be a straight line with slope +1, and that is what Figure 31.4 shows. Figure 31.4 31
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31-2 Chapter 31 31.5. IDENTIFY: L VI X = with L XL ω = . SET UP: is the angular frequency, in rad/s. EXECUTE: (a) L X IL == and 60.0 V 0.120 A. (100 rad s)(5.00 H) V I L = (b) 60.0 V 0.0120 A (1000 rad s)(5.00 H) V I L = . (c) 60.0 V 0.00120 A (10,000 rad s)(5.00 H) V I L = . (d) The plot of log I versus log is given in Figure 31.5. EVALUATE: V I L = so log log( / ) log IV L =− . A graph of log I versus log should be a straight line with slope 1 , and that is what Figure 31.5 shows. Figure 31.5 31.6. IDENTIFY: The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance. SET UP: The reactances are 1/ C XC = and L = . EXECUTE: (a) Equating the reactances gives 11 L C LC =⇒ = (b) Using the numerical values we get (5.00 mH)(3.50 F) L C& = 7560 rad/s X C = X L = L = (7560 rad/s)(5.00 mH) = 37.8 EVALUATE: At other angular frequencies, the two reactances could be very different. 31.7. IDENTIFY and SET UP: For a resistor . R vi R = For an inductor, cos( 90 ). L vV t =+ ° For a capacitor, cos( 90 ). C t ° EXECUTE: The graphs are sketched in Figures 31.7a-c. The phasor diagrams are given in Figure 31.7d. EVALUATE: For a resistor only in the circuit, the current and voltage in phase. For an inductor only, the voltage leads the current by 90 °. For a capacitor only, the voltage lags the current by 90 °. Figure 31.7a and b
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Alternating Current 31-3 Figure 31.7c Figure 31.7d 31.8. IDENTIFY: The reactance of an inductor is 2 L XLf L ω π == . The reactance of a capacitor is 11 2 C X Cf C ωπ . SET UP: The frequency f is in Hz.
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This homework help was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona- Tucson.

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31_InstructorSolutions - ALTERNATING CURRENT 31 31.1....

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