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33 CHAPTER ALTERNATING - 33 CHAPTER ALTERNATING-CURRENT...

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33 CHAPTER ALTERNATING-CURRENT CIRCUITS ActivPhysics can help with these problems: Activities 14.2, 14.3 Section 33-1: 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 0. Alternating Current Solution Use of Equations 33-1 and 2 allows us to write Vp 1 2 Vrms 2 (230 V) 325 V, and 2 f 2 (50 Hz) 314 s . Then the voltage expressed in the form of Equation 33-3 is V (t ) (325 V) sin[(314 s 1 )t ]. 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 Vp 2 (6.3 V) 8.91 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 Vrms Vp= 2 Vp - p=2 2 28 V=2 2 9.90 V. 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) Vp 2 (208 V) 294 V (Equation 33-1), and (b) 2 (400 Hz) 2.51 103 s 1 (Equation 33-2). Problem 5. An AC current is given by I 495 sin(9.43t), 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 9.43 (ms)1. Application of Equations 33-1 and 2 give (a) Irms 495 mA= 2 350 mA, and (b) f 9.43=2 ( ms) 1.50 kHz. CHAPTER 33 777 Problem 6. What are the phase constants for each of the signals shown in Fig. 33-27? FIGURE 33-27 Problem 6. Solution The phase constant is a solution of Equation 33-3 for t 0, i.e., V (0) Vp sin( V ). Since sin( V ) sin( V ), one must also consider the slope of the sinusoidal signal function at t 0. In addition, the conventional range for V usually runs from 180 to 180, or V . Thus, V sin 1[V(0)= p ] when ( dV=dt ) 0 0, but V V V sin 1[V(0)= p ] when ( dV=dt ) 0 0 according as V(0) ? 0. (Here, we are taking sin 1[V (0)= p ] =2 or 90, as V 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 (0) Vp= 2 (since that curve next crosses zero about halfway between =2 and ) and the slope at zero is positive, so a sin 1 (1= 2 ) =4 or 45. (This signal is Vp sin( t a ) Vp sin( t =4), which leads a signal with zero phase constant by 45.) For the other signals, (b) V (0) 0, ( dV=dt ) 0 0, so b 0; (c) V (0) Vp , (dV=dt )0 0, so c sin 1 (1) sin 1 (1) =2 or 90 ; (d) V (0) 0, ( dV=dt ) 0 0, so d or 180 ; and (e) V (0) Vp , (dV= )0 0, so e sin 1 (1) sin 1 (1) =2 or 90. dt Problem 7. The rms amplitude is defined as the square root of the average of the square of the signal. For a periodic function, the time average is the integral over one period, divided by the period. For a sinusoidal voltage given by V Vp sin t, show explicitly that Vrms Vp= 2. Solution The period of a sinusoidal signal is T 2= (Equation 15-6), so 2 V rms V 2 1 T z T 0 2 (V p sin 2 t )dt V2 p 2T z T 0 (1 cos 2 t ) dt 2 Vp L 1 sin 2 t T M 2T M 2 N 1 2 T 0 OV . P2 P Q 2 p (We used an identity from Appendix A.) A graphical argument that sin 2 t is suggested in Fig. 16-16. Problem 8. How are the rms and peak voltages related for the square wave shown in Fig.
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