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Unformatted text preview: CHAPTER 33—ALTERNATINGCURRENT CIRCUITS ActivPhysics can help with these problems: Activities 14.2, 14.3 Section 331:—Alternating Current Problem 1. Much of Europe uses AC power at 230 V rms and 50 Hz. Express this AC voltage in the form of Equation 333, taking φ = 0. Solution Use of Equations 331 and 2 allows us to write V V p = = = 2 2 230 325 V V rms ( ) , and ϖ π π = = = 2 2 50 f ( ) Hz 314 1 s . Then the voltage expressed in the form of Equation 333 is V t t ( ) ( ) sin[( ) ]. = 325 314 1 V s Problem 2. An rms voltmeter connected across the filament of a TV picture tube reads 6.3 V. What is the peak voltage across the filament? Solution Equation 331 gives V p = = 2 6 3 8 91 ( . ) . . V V Problem 3. An oscilloscope displays a sinusoidal signal whose peaktopeak voltage (see Fig. 331) is 28 V. What is the rms voltage? Solution As shown in Fig. 331, the peaktopeak voltage is twice the peak voltage, so Equation 331 gives V V p rms = = = 2 V p p V V = = 2 2 28 2 2 9 90 = = . . Problem 4. An industrial electric motor runs at 208 V rms and 400 Hz. What are (a) the peak voltage and (b) the angular frequency? Solution (a) V p = = 2 208 294 ( ) V V (Equation 331), and (b) ϖ π = = × 2 400 2 51 10 3 1 ( ) . Hz s (Equation 332). Problem 5. An AC current is given by I t = 495 9 sin( .43 ), with I in milliamperes and t in milliseconds. Find (a) the rms current and (b) the frequency in Hz. Solution Comparison of the current with Equation 333 shows that its amplitude and angular frequency are I p = 495 mA and ϖ = 9 1 .43 ( ) . ms Application of Equations 331 and 2 give (a) I rms mA mA, = = 495 2 350 = and (b) f = = 9 2 .43 ( ) = π ms 150 . . kHz Problem 6. What are the phase constants for each of the signals shown in Fig. 3327? 2 CHAPTER 33 FIGURE 3327 Problem 6. Solution The phase constant is a solution of Equation 333 for t = 0, i.e., V V p V ( ) sin( ). = φ Since sin( ) sin( ), φ φ π V V = ± one must also consider the slope of the sinusoidal signal function at t = 0. In addition, the conventional range for φ V usually runs from  ° 180 to + ° 180 , or  ≤ ≤ π φ π V . Thus, φ V p V V = sin [ ( ) ] 1 = when ( ) , dV dt = ≥ but φ V = ± sin [ ( ) ] 1 V V p = π when ( ) dV dt = ≤ according as V ( ) . ? (Here, we are taking sin [ ( ) ] ≤ 1 2 V V p = = π or 90 ° , as common on most electronic calculators, since the sine function is onetoone only in such a restricted range.) For signal (a) in Fig. 3327, we guess that V V p ( ) 2 ¼ = (since that curve next crosses zero about halfway between π = 2 and π ) and the slope at zero is positive, so φ π a = = ° sin ( ) . 1 1 2 4 45 = = or (This signal is V t V t p a p sin( ) sin( ), ϖ φ ϖ π + = + = 4 which leads a signal with zero phase constant by 45 ° .) For the other signals, (b) V dV dt ( ) , ( ) , = = so φ b = 0; (c) V V dV dt p ( ) , ( ) , = = = so φ π π c = =  + = sin ( ) sin ( ) 1 1 1 1 2 = or 90 ° ; (d) V dV dt ( ) , ( ) , =...
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This note was uploaded on 05/20/2009 for the course PHYS 662895 taught by Professor Staff during the Spring '09 term at UCSD.
 Spring '09
 staff
 Current, Power

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