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

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28.1 ALTERNATING-CURRENT CIRCUITS 28 EXERCISES 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 == = VV and 2 f ω π 1 2 (50 Hz) 314 s . = Then the voltage expressed in the form of Equation 28.3 is Vt 1 ( ) (325 V)sin[(314 s ) ]. t = 15. INTERPRET We’re given the rms voltage and asked to find the peak voltage. DEVELOP The rms voltage V and the peak voltage rms p V are related by Equation 28.1, rms /2 . p = EVALUATE Equation 28.1 gives rms 2 2(6.3 V) 8.91 V. p = ASSESS 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). 31 2( 4 0 0 H z ) 2 . 5 11 0 s ωπ × 17. INTERPRET We’re given the AC current in terms of sinusoidal function, and asked to deduce the rms current and the frequency of the current. DEVELOP As shown in Equation 28.3, the AC current can be written as p sin( ) I II t φ = + where p I is the peak current amplitude, is the angular frequency, and I 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) . = EVALUATE (a) Applying Equation 28.1 gives rms p /2 4 9 5 mA /2 3 5 0 . = (b) Similarly, using Equation 28.2 we have 1 /2 9.43 ms /2 1.50 kHz. f = ASSESS The phase I 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 18. The phase constant is a solution of Equation 28.3 for t 0, = i.e., (0) sin( ). pV = Since sin( ) sin( ), φπ =− ± one must also consider the slope of the sinusoidal signal function at t 0. = In addition, the conventional range for V usually runs from to or 180 −° 180 , . V ≤≤ Thus, when but 1 sin [ (0)/ ] Vp = 0 (/) 0 , dV dt V = 1 sin [ (0)/ ] p −± when according as V (Here, we are taking|s 0 dV dt (0) 0. > < in [ (0)/ ]| /2 p 1 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 and ) and the slope at zero is positive, so 1 sin (1/ 2) /4 or 45 . a = (This signal is sin( ) sin( /4), pa p φω += + = which leads a signal with zero phase constant by For the other signals, (b) V so 45 .) ° (0) 0, 0 , dV dt > 0; b = (c) V 0 (0) , ( / ) 0, p V dV dt = = so sin (1) sin (1) /2 c ππ −− −+ = or (d) V so 90 ; ° (0) 0, = 0 , dV dt < d or 180 ; ± ° and (e) V 0 (0) , ( / ) 0, p V dV dt = −= = so 1 sin ( 1) e 1 sin ( 1) /2 = or 90 . Section 28.2 Circuit Elements in AC Circuits 19. INTERPRET In this problem we want to find the rms current in a capacitor connected to an AC power source.
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28.2 Chapter 28 DEVELOP The amplitude of the current in a capacitor is given by Equation 28.5: pp 1/ C VV I CV XC ω == = Using Equation 28.1, the corresponding rms current is rms rms . I CV = EVALUATE Substituting the values given in the problem statement, we find the rms current to be rms rms (2 60 Hz)(1 F)(120 V) 45.2 mA IC V πμ × = ASSESS
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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

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