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Unformatted text preview: ALTERNATINGCURRENT 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 = = 1 2 (50 Hz) 314 s .  = Then the voltage expressed in the form of Equation 28.3 is 1 ( ) (325 V)sin[(314 s ) ]. V t t = 15. I NTERPRET Were 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 (rootmeansquare) 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  = = (Equation 28.2). 17. I NTERPRET Were 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 = + 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) .  = 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  = = = A SSESS The phase I is zero in this problem. Note that since the rms (rootmeansquare) 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 = Since sin( ) sin( ), V V = one must also consider the slope of the sinusoidal signal function at 0. t = In addition, the conventional range for V usually runs from 180 to 180 , + or . V  Thus, 1 sin [ (0)/ ] V p V V  = when ( / ) 0, dV dt but V = 1 sin [ (0)/ ] p V V  when ( / ) dV dt according as (0) 0. V < (Here, we are taking 1 sin [ (0)/ ] /2 p V V  or 90 , as common on most electronic calculators, since the sine function is onetoone 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 and ) and the slope at zero is positive, so 1 sin (1/ 2) /4 or 45 . a  = = (This signal is sin( ) sin( /4), p a p V t V t + = + which leads a signal with zero phase constant by 45 .) For the other signals, (b) (0) 0, V = ( / ) 0, dV dt so 0; b = (c) (0) , ( / ) 0, p V V dV dt = = so 1 1 sin (1) sin (1) /2 c  = =  + = or 90 ; (d) (0) 0, V = ( / ) 0, dV dt < so d = or 180 ; and (e) (0) , ( / ) 0, p V...
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics, Current

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