926_PartUniversity Physics Solution

# 926_PartUniversity Physics Solution - 31-24 Chapter 31...

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31-24 Chapter 31 (e) 2 2 2 1 . 2 9 2 4 L LV UL I L R C == + EVALUATE: For 0 , ω > L C VV > and the maximum energy stored in the inductor is greater than the maximum energy stored in the capacitor. 31.71. IDENTIFY and SET UP: Assume the angular frequency of the source and the resistance R of the resistor are known. EXECUTE: Connect the source, capacitor, resistor, and inductor in series. Measure R V and . L V L R VI L L RR and L can be calculated. EVALUATE: There are a number of other approaches. The frequency could be varied until , CL = and then this frequency is equal to 1/ . LC If C is known, then L can be calculated. 31.72. IDENTIFY: av rms rms cos PV I φ = and rms rms . V I Z = Calculate Z . cos . RZ = SET UP: 50.0 Hz f = and 2 . f π = The power factor is cos . EXECUTE: (a) 2 rms av cos . V P Z = 22 rms av cos (120 V) (0.560) 36.7 . (220 W) V Z P = Ω cos (36.7 )(0.560) 20.6 . Ω = Ω (b) 2 2 2 2 (36.7 ) (20.6 ) 30.4 . LL ZR X X Z R =+⋅ =− = Ω Ω But 0 = is at resonance, so the inductive and capacitive reactances equal each other. Therefore we need to add 30.4 . C X = Ω 1 C X C = therefore gives 4 11 1 1.05 10 F. 2 2 (50.0 Hz)(30.4 ) CC C Xf X ωπ = = × Ω (c) At resonance, av (120 V) 699 W. 20.6 V P R = Ω EVALUATE: 2 av rms PIR = and rms I is maximum at resonance, so the power drawn from the line is maximum at resonance. 31.73. IDENTIFY: 2 . R pi R = . L di L dt = . C q p i C = SET UP: cos iI t = EXECUTE: (a) 2 2 1 cos ( ) cos ( ) (1 cos(2 )). 2 R p iR I tR VI t t ωω = = + 1 av 0 2 00 1 () ( 1c o s ( 2 ) ) [ ] . TT T PR pd t td t t V I T + = = ∫∫ (b) 2 1 2 cos( )sin( ) sin(2 ). di p Li LI t t V I t dt = But av 0 sin(2 ) 0 ( ) 0. T t P L = ⇒= (c) 1 2 sin( )cos( ) sin(2 ). C C q p iv iV I t t V I t C === = But av 0 sin(2 ) 0 ( ) 0. T t P C = (d) 2 cos ( ) sin(2 ) sin(2 ) RLcR L C pp p pV I I I t =++= + and cos( )( cos( ) sin( ) sin( )). RLC p It V t Vt V t + But cos R V V = and sin , L C V = so cos( )(cos cos( ) sin sin( )), p VI t t t φω at any instant of time. EVALUATE: At an instant of time the energy stored in the capacitor and inductor can be changing, but there is no net consumption of electrical energy in these components. 31.74. IDENTIFY: . L L X = 0 L dV d = at the where L V is a maximum. . X = 0 C dV d = at the where C V is a maximum. SET UP: Problem 31.51 shows that . (1 / ) V I R LC = +− EXECUTE: (a) R V = maximum when 0 1 . LC =⇒==

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Alternating Current 31-25 (b) L V = maximum when 0. L dV d ω = Therefore: 22 0. (1) L dV d V L dd RL C ωω ⎛⎞ ⎜⎟ == +− ⎝⎠ 2 3 2 (1 ) ) ((1) ) VL V L L C L C RLC RL C −+ =− 2 2 4 2 ( 1 ) . C L C ωωω = 2 12 1 . L R CC C = 2 1 2 R C LC and 1 . 2 LC R C = (c) C V = maximum when 0. C dV d = Therefore: C dV d V CR L C 2 232 2 ) ) ) VV L C L C L C L C 2 2 4 2 ( 1 ) . R LC L C = 2 2 2 2 L L C = and 2 2 1 .
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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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926_PartUniversity Physics Solution - 31-24 Chapter 31...

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