YF_ISM_31 - ALTERNATING CURRENT 31 31.1 IDENTIFY i = I cos...

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31-1 A LTERNATING C URRENT 31.1. I DENTIFY : cos i I t ω = and rms / 2. I I = S ET U P : The specified value is the root-mean-square current; rms 0.34 A. I = E XECUTE : (a) rms 0.34 A I = (b) rms 2 2(0.34 A) 0.48 A. I I = = = (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 2 2 2 rms (0.34 A) 0.12 A . I = = E VALUATE : The current amplitude is larger than its rms value. 31.2. I DENTIFY and S ET U P : Apply Eqs.(31.3) and (31.4) E XECUTE : (a) rms 2 2(2.10 A) 2.97 A. I I = = = (b) rav 2 2 (2.97 A) 1.89 A. I I π π = = = E VALUATE : (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. I DENTIFY and S ET U P : Apply Eq.(31.5). E XECUTE : (a) rms 45.0 V 31.8 V. 2 2 V V = = = (b) Since the voltage is sinusoidal, the average is zero. E VALUATE : The voltage amplitude is larger than rms . V 31.4. I DENTIFY : C V IX = with 1 C X C ω = . S ET U P : ω is the angular frequency, in rad/s. E XECUTE : (a) C I V IX C ω = = so 6 (60.0 V)(100 rad s)(2.20 10 F) 0.0132 A. I V C ω = = × = (b) 6 (60.0 V)(1000 rad s)(2.20 10 F) 0.132 A. I V C ω = = × = (c) 6 (60.0 V)(10,000 rad s)(2.20 10 F) 1.32 A. I V C ω = = × = (d) The plot of log I versus log ω is given in Figure 31.4. E VALUATE : I VC ω = so log log( ) log . I VC ω = + A graph of log I versus log ω 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. I DENTIFY : L V IX = with L X L ω = . S ET U P : ω is the angular frequency, in rad/s. E XECUTE : (a) L V IX I L ω = = and 60.0 V 0.120 A. (100 rad s)(5.00 H) V I L ω = = = (b) 60.0 V 0.0120 A (1000 rad s)(5.00 H) V I L ω = = = . (c) 60.0 V 0.00120 A (10,000 rad s)(5.00 H) V I L ω = = = . (d) The plot of log I versus log ω is given in Figure 31.5. E VALUATE : V I L ω = so log log( / ) log I V L ω = . A graph of log I versus log ω should be a straight line with slope 1 , and that is what Figure 31.5 shows. Figure 31.5 31.6. I DENTIFY : The reactance of capacitors and inductors depends on the angular frequency at which they are operated, as well as their capacitance or inductance. S ET U P : The reactances are 1/ C X C ω = and L X L ω = . E XECUTE : (a) Equating the reactances gives 1 1 L C LC ω ω ω = = (b) Using the numerical values we get 1 1 (5.00 mH)(3.50 F) LC μ ω = = = 7560 rad/s X C = X L = ω L = (7560 rad/s)(5.00 mH) = 37.8 E VALUATE : At other angular frequencies, the two reactances could be very different. 31.7. I DENTIFY and S ET U P : For a resistor . R v iR = For an inductor, cos( 90 ). L v V t ω = + ° For a capacitor, cos( 90 ). C v V t ω = ° E XECUTE : The graphs are sketched in Figures 31.7a-c. The phasor diagrams are given in Figure 31.7d. E VALUATE : 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 °.
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