CH33 - CHAPTER 33—ALTERNATING-CURRENT CIRCUITS...

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Unformatted text preview: CHAPTER 33—ALTERNATING-CURRENT CIRCUITS ActivPhysics can help with these problems: Activities 14.2, 14.3 Section 33-1:—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 33-3, taking φ = 0. Solution Use of Equations 33-1 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 33-3 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 33-1 gives V p = = 2 6 3 8 91 ( . ) . . V V Problem 3. An oscilloscope displays a sinusoidal signal whose peak-to-peak voltage (see Fig. 33-1) is 28 V. What is the rms voltage? Solution As shown in Fig. 33-1, the peak-to-peak voltage is twice the peak voltage, so Equation 33-1 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 33-1), and (b) ϖ π = = ×- 2 400 2 51 10 3 1 ( ) . Hz s (Equation 33-2). 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 33-3 shows that its amplitude and angular frequency are I p = 495 mA and ϖ =- 9 1 .43 ( ) . ms Application of Equations 33-1 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. 33-27? 2 CHAPTER 33 FIGURE 33-27 Problem 6. Solution The phase constant is a solution of Equation 33-3 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 one-to-one only in such a restricted range.) For signal (a) in Fig. 33-27, 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.

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CH33 - CHAPTER 33—ALTERNATING-CURRENT CIRCUITS...

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