911_PartUniversity Physics Solution

911_PartUniversity Physics Solution - Alternating Current...

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Alternating Current 31-9 (b) The average power dissipated by the resistor is 2 2 rms 120 V (350 ) 7.32 W. 830 R PIR ⎛⎞ == Ω = ⎜⎟ Ω ⎝⎠ EVALUATE: Conservation of energy requires that the answers to parts (a) and (b) are equal. 31.27. IDENTIFY: The power factor is cos , φ where is the phase angle in Fig.31.13. The average power is given by Eq.(31.31). Use the result of part (a) to rewrite this expression. (a) SET UP: The phasor diagram is sketched in Figure 31.27. EXECUTE: From the diagram cos , R VI R R ZZ === as was to be shown. Figure 31.27 (b) rms av rms rms rms rms rms cos . Z RV PV I V I IR Z ⎛⎞⎛ ⎞ ⎜⎟⎜ ⎟ ⎝⎠⎝ ⎠ But 2 rms rms av rms , so . Z V I EVALUATE: In an L - R - C circuit, electrical energy is stored and released in the inductor and capacitor but none is dissipated in either of these circuit elements. The power delivered by the source equals the power dissipated in the resistor. 31.28. IDENTIFY and SET UP: av rms rms cos . I = rms rms . V I Z = cos . R Z = EXECUTE: rms 80.0 V 0.762 A. 105 I Ω 75.0 cos 0.714. 105 Ω Ω av (80.0 V)(0.762 A)(0.714) 43.5 W. P = = EVALUATE: Since the average power consumed by the inductor and by the capacitor is zero, we can also calculate the average power as 22 av rms (0.762 A) (75.0 ) 43.5 W. Ω = 31.29. IDENTIFY and SET UP: Use the equations of Section 31.3 to calculate rms , and . Z V The average power delivered by the source is given by Eq.(31.31) and the average power dissipated in the resistor is 2 rms EXECUTE: (a) 2 2 (400 Hz)(0.120 H) 301.6 L XLf L ω ππ = 6 11 1 54.51 2 2 (400 Hz)(7.3 10 Hz) C X Cf C ωπ π = = Ω × 301.6 54.41 tan , so 45.8 . 240 LC XX R φφ −Ω Ω = + ° Ω The power factor is cos 0.697. = + (b) 222 2 ( ) (240 ) (301.6 54.51 ) 344 ZRX X =+− = Ω + Ω Ω (c) rms rms (0.450 A)(344 ) 155 V Z Ω = (d) av rms rms cos (0.450 A)(155 V)(0.697) 48.6 W PI V = (e) av rms (0.450 A) (240 ) 48.6 W Ω = EVALUATE: The average electrical power delivered by the source equals the average electrical power consumed in the resistor. (f) All the energy stored in the capacitor during one cycle of the current is released back to the circuit in another part of the cycle. There is no net dissipation of energy in the capacitor. (g) The answer is the same as for the capacitor. Energy is repeatedly being stored and released in the inductor, but no net energy is dissipated there. 31.30. IDENTIFY: The angular frequency and the capacitance can be used to calculate the reactance C X of the capacitor. The angular frequency and the inductance can be used to calculate the reactance L X of the inductor. Calculate the phase angle and then the power factor is cos . Calculate the impedance of the circuit and then the rms current in the circuit. The average power is av rms rms cos . I = On the average no power is consumed in the capacitor or the inductor, it is all consumed in the resistor. SET UP: The source has rms voltage rms 45 V 31.8 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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911_PartUniversity Physics Solution - Alternating Current...

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