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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 t, show explicitly that Vrms Vp= 2....
View Full Document

This note was uploaded on 03/28/2010 for the course PHYSICS 101 taught by Professor Gfos during the Spring '08 term at Adams State University.

Page1 / 12

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online