906_PartUniversity Physics Solution

906_PartUniversity Physics Solution - 31-4 Chapter 31...

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31-4 Chapter 31 31.12. IDENTIFY: Compare C v that is given in the problem to the general form sin C I vt C ω = and determine . SET UP: 1 C X C = . R vi R = and cos . iI = EXECUTE: (a) 6 11 1736 (120 rad s)(4.80 10 F) C X C == = Ω × (b) 3 7.60 V 4.378 10 A 1736 C C V I X = × Ω and 3 cos (4.378 10 A)cos[(120 rad/s) ]. t t × Then 3 (4.38 10 A)(250 )cos((120 rad s) ) (1.10 V)cos((120 rad s) ). R R t t × Ω = EVALUATE: The voltage across the resistor has a different phase than the voltage across the capacitor. 31.13. IDENTIFY and SET UP: The voltage and current for a resistor are related by . R R = Deduce the frequency of the voltage and use this in Eq.(31.12) to calculate the inductive reactance. Eq.(31.10) gives the voltage across the inductor. EXECUTE: (a) (3.80 V)cos[(720 rad/s) ] R = 3.80 V , so cos[(720 rad/s) ] (0.0253 A)cos[(720 rad/s) ] 150 R R v R i t t R ⎛⎞ = = ⎜⎟ Ω ⎝⎠ (b) L XL = 720 rad/s, 0.250 H, so (720 rad/s)(0.250 H) 180 L LX L = = = Ω (c) If cos t = then cos( 90 ) LL vV t =+ ° (from Eq.31.10). (0.02533 A)(180 ) 4.56 V VI L I X = Ω = (4.56 V)cos[(720 rad/s) 90 ] L ° But cos( 90 ) sin aa = (Appendix B), so (4.56 V)sin[(720 rad/s) ]. L = − EVALUATE: The current is the same in the resistor and inductor and the voltages are 90 ° out of phase, with the voltage across the inductor leading. 31.14. IDENTIFY: Calculate the reactance of the inductor and of the capacitor. Calculate the impedance and use that result to calculate the current amplitude. SET UP: With no capacitor, 22 L Z RX and tan . L X R φ = . L = . V I Z = L L X = and . R R = For an inductor, the voltage leads the current. EXECUTE: (a) (250 rad/s)(0.400 H) 100 . L = Ω (200 ) (100 ) 224 . Z = Ω+ Ω= Ω (b) 30.0 V 0.134 A 224 V I Z = Ω (c) (0.134 A)(200 ) 26.8 V. R R Ω = (0.134 A)(100 ) 13.4 V. X = = (d) 100 tan 200 L X R Ω Ω and 26.6 °. Since is positive, the source voltage leads the current. (e) The phasor diagram is sketched in Figure 31.14. EVALUATE: Note that R L VV + is greater than V . The loop rule is satisfied at each instance of time but the voltages across R and L reach their maxima at different times. Figure 31.14 31.15. IDENTIFY: () R is given by Eq.(31.8). ( ) L is given by Eq.(31.10). SET UP: From Exercise 31.14, 30.0 V V = , 26.8 V R V = , 13.4 V L V = and 26.6 = ° . EXECUTE: (a) The graph is given in Figure 31.15. (b) The different voltages are (30.0 V)cos(250 26.6 ), (26.8 V)cos(250 ), R v t ° = (13.4 V)cos(250 90 ). L ° At 20 ms: t = 20.5 V, 7.60 V, 12.85 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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906_PartUniversity Physics Solution - 31-4 Chapter 31...

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