921_PartUniversity Physics Solution

# 921_PartUniversity Physics Solution - Alternating Current...

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Alternating Current 31-19 (d) 6 11 1000 rad sec (2.0 H)(0.50 10 F) LC ω == = × and 159 Hz. f = The phasor diagram is sketched in Figure 31.57b. (e) 22 . VV IV C R L ⎛⎞ ⎛ =+ ⎜⎟ ⎝⎠ 2 2 16 1 100 V 100 V (100 V)(1000 s )(0.50 10 F) 0.50 A 200 (1000 s )(2.0 H) I −− ⎛⎞ × = Ω (f) At resonance (100 V)(1000 s )(0.50 10 F) 0.0500 A LC IIV C = × = and 100 V 0.50 A. 200 R V I R = Ω EVALUATE: At resonance 0 CL ii at all times and the current through the source equals the current through the resistor. Figure 31.57 31.58. IDENTIFY: The average power depends on the phase angle φ . SET UP: The average power is P av = V rms I rms cos , and the impedance is 2 2 1 . ZR L C =+− EXECUTE: (a) P av = V rms I rms cos = 1 2 ( V rms I rms ), which gives cos = 1 2 , so = π /3 = 60&. tan = ( X L & X C )/ R , which gives tan 60& = ( L & 1/ C )/ R . Using R = 75.0 , L = 5.00 mH, and C = 2.50 ± F and solving for we get = 28760 rad/s = 28,800 rad/s. (b) () , ZRX X where X L = L = (28,760 rad/s)(5.00 mH) = 144 and X C = 1/ C = 1/[(28,760 rad/s)(2.50 ± F)] = 13.9 , giving (75 ) (144 13.9 ) Z = Ω+ Ω Ω = 150 ; I = V/Z = (15.0 V)/(150 ) = 0.100 A and P av = 1 2 VI cos = 1 2 (15.0 V)(0.100 A)(1/2) = 0.375 W. EVALUATE: All this power is dissipated in the resistor because the average power delivered to the inductor and capacitor is zero. 31.59. IDENTIFY: We know R , and C X so Eq.(31.24) tells us . L X Use 2 av rms PIR = from Exercise 31.27 to calculate rms . I Then calculate Z and use Eq.(31.26) to calculate rms V for the source. SET UP: Source voltage lags current so av 54.0 . 350 , 180 , 140 W C XR P = −° = Ω = Ω = EXECUTE: (a) tan L C XX R = tan (180 )tan( 54.0 ) 350 248 350 102 X = Ω ° + Ω = Ω + Ω = Ω (b) 2 av rms rms rms cos PV I IR (Exercise 31.27). av rms 140 W 0.882 A 180 P I R = Ω (c) 222 2 ( ) (180 ) (102 350 ) 306 X = Ω Ω rms rms (0.882 A)(306 ) 270 V. VI Z Ω = EVALUATE: We could also use Eq.(31.31): av rms rms cos I = av rms rms 140 W 270 V, cos (0.882 A)cos( 54.0 ) P V I = −° which agrees. The source voltage lags the current when , > and this agrees with what we found.

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31-20 Chapter 31 31.60. IDENTIFY and SET UP: Calculate Z and / . IVZ = EXECUTE: (a) For 800 rad s: ω = 22 2 72 ( 1 ) (500 ) ((800 rad/s)(2.0 H) 1 ((800 rad/s)(5.0 10 F))) . ZR L C ωω =+− = Ω + × 1030 Z . 100 V 0.0971A. 1030 V I Z == = Ω (0.0971A)(500 ) 48.6 V, R VI R Ω = 7 1 0.0971A 243 V (800 rad s)(5.0 10 F) C V C = × and (0.0971A)(800 rad s)(2.00 H) 155 V. L L = 1( ) arctan 60.9 . LC R φ ⎛⎞ = =− ° ⎜⎟ ⎝⎠ The graph of each voltage versus time is given in Figure 31.60a. (b) Repeating exactly the same calculations as above for 1000 rad/s: = C Z 500 ; 0; 0.200 A; 100 V; 400 V.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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921_PartUniversity Physics Solution - Alternating Current...

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