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# chap28 - ALTERNATING-CURRENT CIRCUITS 28 EXERCISES Section...

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28.1 ALTERNATING-CURRENT CIRCUITS 28 E XERCISES Section 28.1 Alternating Current 14. Use of Equations 28.1 and 28.2 allows us to write rms 2 2(230 V) 325 V, p = = = V V and 2 f ω π = = 1 2 (50 Hz) 314 s . π = Then the voltage expressed in the form of Equation 28.3 is V t 1 ( ) (325 V)sin[(314 s ) ]. t = 15. I NTERPRET We’re given the rms voltage and asked to find the peak voltage. D EVELOP The rms voltage V and the peak voltage rms p V are related by Equation 28.1, rms / 2. p V V = E VALUATE Equation 28.1 gives rms 2 2(6.3 V) 8.91 V. p V V = = = A SSESS The rms (root-mean-square) voltage is obtained by squaring the voltage and taking its time average, and then taking the square root. Therefore, it is smaller than the peak voltage by a factor of 2. 16. (a) 2(208 V) 294 V p V (Equation 28.1), and (b) (Equation 28.2). = = 3 1 2 (400 Hz) 2.51 10 s ω π = = × 17. I NTERPRET We’re given the AC current in terms of sinusoidal function, and asked to deduce the rms current and the frequency of the current. D EVELOP As shown in Equation 28.3, the AC current can be written as p sin( ) I I I t ω φ = + where p I is the peak current amplitude, ω is the angular frequency, and I φ is the phase constant. Comparison of the current with Equation 28.3 shows that its amplitude and angular frequency are p 495 mA I = and 1 9.43 (ms) . ω = E VALUATE (a) Applying Equation 28.1 gives rms p / 2 495 mA/ 2 350 mA. I I = = = (b) Similarly, using Equation 28.2 we have 1 /2 9.43 ms /2 1.50 kHz. f ω π π = = = A SSESS The phase I φ is zero in this problem. Note that since the rms (root-mean-square) current is obtained by squaring the current and taking its time average, and then taking the square root, it is smaller than the peak current by a factor of 2. 18. The phase constant is a solution of Equation 28.3 for t 0, = i.e., V V (0) sin( ). p V φ = Since sin( ) sin( ), V V φ φ π = ± one must also consider the slope of the sinusoidal signal function at t 0. = In addition, the conventional range for V φ usually runs from to or 180 ° 180 , + ° . V π φ π Thus, when but 1 sin [ (0)/ ] V p V V φ = 0 ( / ) 0, dV dt V φ = 1 sin [ (0)/ ] p V V π ± when according as V (Here, we are taking |s 0 ( / ) 0 dV dt (0) 0. > < in [ (0)/ ]| /2 p V V 1 π or 90 as common on most electronic calculators, since the sine function is one-to-one only in such a restricted range.) For signal (a) in Figure 28.25, we guess that , ° (0) / 2 p V (since that curve next crosses zero about halfway between V /2 π and ) π and the slope at zero is positive, so 1 sin (1/ 2) /4 or 45 . a φ π = = ° (This signal is sin( ) sin( /4), p a p V t V t ω φ ω + = + π = which leads a signal with zero phase constant by For the other signals, (b) V so 45 .) ° (0) 0, 0 ( / ) 0, dV dt > 0; b φ = (c) V 0 (0) , ( / ) 0, p V dV dt = = so 1 1 sin (1) sin (1) /2 c φ π π = = − + = or (d) V so 90 ; ° (0) 0, = 0 ( / ) 0, dV dt < d φ π = ± or 180 ; ± ° and (e) V 0 (0) , ( / ) 0, p V dV dt = − = = so 1 sin ( 1) e φ = 1 sin ( 1) /2 π π = − or 90 . ° Section 28.2 Circuit Elements in AC Circuits 19. I NTERPRET In this problem we want to find the rms current in a capacitor connected to an AC power source.

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