33 CHAPTER ALTERNATINGCURRENT CIRCUITS ActivPhysics can help with these problems:
Activities 14.2, 14.3 Section 331: 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 0. Alternating Current Solution Use of
Equations 331 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 333 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 331 gives Vp 2 (6.3 V) 8.91 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 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 331), and (b) 2 (400 Hz) 2.51 103 s 1 (Equation 332). 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 333 shows that its amplitude and angular
frequency are I p 495 mA and 9.43 (ms)1. Application of Equations 331 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. 3327? FIGURE 3327 Problem 6. Solution The phase constant is a
solution of Equation 333 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 onetoone only in such a restricted range.) For signal (a)
in Fig. 3327, 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 156), 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. 1616. Problem 8. How are the rms and peak voltages related for the square wave shown in Fig.
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 Spring '08
 GFOS
 Current, Power, Alternating Current

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