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28_InstSolManual_PC

# 28_InstSolManual_PC - ALTERNATING-CURRENT CIRCUITS 28...

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ALTERNATING-CURRENT CIRCUITS 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 ϖ π = = 1 2 (50 Hz) 314 s . π - = Then the voltage expressed in the form of Equation 28.3 is 1 ( ) (325 V)sin[(314 s ) ]. V t t - = 15. I NTERPRET We’re given the rms voltage and asked to find the peak voltage. D EVELOP The rms voltage rms V and the peak voltage 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) 3 1 2 (400 Hz) 2.51 10 s ϖ π - = = × (Equation 28.2). 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 ϖ φ = + where p I is the peak current amplitude, ϖ 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) . ϖ - = 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 ϖ 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 0, t = i.e., (0) sin( ). p V V V φ = Since sin( ) sin( ), V V φ φ π = - ± one must also consider the slope of the sinusoidal signal function at 0. t = In addition, the conventional range for V φ usually runs from 180 - ° to 180 , + ° or . V π φ π - Thus, 1 sin [ (0)/ ] V p V V φ - = when 0 ( / ) 0, dV dt but V φ = 1 sin [ (0)/ ] p V V π - - ± when 0 ( / ) 0 dV dt according as (0) 0. V < (Here, we are taking 1 |sin [ (0)/ ]| /2 p V V 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 V (since that curve next crosses zero about halfway between /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 ϖ φ ϖ π + = + which leads a signal with zero phase constant by 45 .) ° For the other signals, (b) (0) 0, V = 0 ( / ) 0, dV dt so 0; b φ = (c) 0 (0) , ( / ) 0, p V V dV dt = = so 1 1 sin (1) sin (1) /2 c φ π π - - = = - + = or 90 ; ° (d) (0) 0, V = 0 ( / ) 0, dV dt < so d φ π = ± or 180 ; ± ° and (e) 0 (0) , ( / ) 0, p V 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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