26 - LCR Circuits

26 - LCR Circuits - Physics 212 LCR Circuits 10/28/2009...

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Unformatted text preview: Physics 212 LCR Circuits 10/28/2009 Fall 2009 Summary AC Voltage RMS (Root-Mean-Square) VRMS, IRMS V = V0 sin(ωt ) ω = 2πf € f= 1 T ω= € IRMS = 2π T IPVP = ISVS Magnetic flux change is the same due to iron core N VS = S V p NP IS = IP NS>NP - step up; NS<NP - step down € € I0 2 keep power constant € € 12 PAVG = I0 R 2 Average Power VP N = IP P VS NS Step up the voltage and step down the current Transformer The ac adapter for a laptop computer contains a transformer. The input of the adapter is the 120 volts from the ac wall outlet. The output from the transformer is 20 volts. What is the turns ratio of the transformer? a) 0.17 b) 6 c) 100 d) This cannot be determined without knowing how many turns one of the coils in the transformer has. What is a Phasor? Vector that rotates about the origin - represents oscillating quantities Angular speed - rotation speed is equal to oscillation angular frequency ω Length - size of the vector represents the amplitude of the alternating quantity Projection - projection onto vertical axis is the instantaneous value at time (t) Rotation Angle - Phase of the quantity at time (t) Resistive Load Peak Current and Reactance Voltage across R I0 I0 = v R = V0 sin(ωt ) V0 V0 R ⇒ V0 = I0 R Current € € I0 v V iR = R = 0 sin(ωt ) RR € Time varying vR or iR can be depicted by projection onto yaxis of rotating phasor V0 Current and voltage are “in phase” Peak current is not frequency dependent Capacitative Reactance Voltage across C vC = V0 sin(ωt ) I0 V0 Peak Current and Reactance I0 = CV0ω ⇒ V0 = Charge and Current φ = 90° = π 2 rad € € XC = q = CvC = CV0 sin(ωt ) iC = CV0ω sin(ωt + 90°) 1 ωC V0 = I0 X C dq d sin(ωt ) i= = CV0 = CV0ω cos(ωt ) dt dt € I0 ωC € € I0 V0 € Time varying vC or iC can be depicted by projection onto yaxis of rotating phasor Current “leads” the voltage Peak current is frequency dependent Circuit “reacts” differently for low and high f Example Consider a circuit with a purely capacitative load where C=15.0 μF and the sinusoidal emf device operates at amplitude of V0=36.0 V and f=60.0 Hz, a) what is the potential difference across C and what is the current in the circuit? Reactance Angular Frequency 1 1 XC = = ω = 2πf = 2π (60 s−1 ) = 120π ωC 2πfC Voltage vC ( t ) = ( 36.0 V ) sin(120πt ) XC = Units € I0 = Current V0 36 V = = 0.203 A X C 177 Ω iC ( t ) = (0.203 A) sin(120πt + π 2) € € € 1 = 177 Ω 2π (60 s )(15 E − 6 F ) XC = −1 1 1 V V = = = =Ω 1C C ( s )( F ) A s sV −1 Inductive Reactance Voltage across L Peak Current and Reactance v L = V0 sin(ωt ) Current φ = −90° = −π 2 rad V0 I0 diL = V0 sin(ωt ) dt 1 V iL = ∫ V0 sin(ωt ) dt = − 0 cos(ωt ) L ωL V iL = 0 sin(ωt − 90°) ωL € € V0 € I0 I0 = Current “lags” the voltage ⇒ V0 = I0ωL X L = ωL L Time varying vL or iL can be depicted by projection onto yaxis of rotating phasor V0 ωL V0 = I0 X L € Peak current is frequency dependent Circuit “reacts” differently for low and high f Example Consider a circuit with a purely inductive load where L=230 mH and the sinusoidal emf device operates at amplitude of V0=36.0 V and f=60.0 Hz, a) what is the potential difference across L and what is the current in the circuit? Reactance Angular Frequency X L = ωL = 2πfL ω = 2πf = 2π (60 s−1 ) = 120π Voltage v L ( t ) = ( 36.0 V ) sin(120πt ) X L = 2π (60 s−1 )(230 E − 3 H ) = 86.7 Ω Units 1 Tm 2 1 N m 2 Nm J 1 V X L = ( € )( H ) = s = = = = =Ω sA s C m A CA C A A s −1 € Current I0 = V0 36. V = = 0.415 A X L 86.7 Ω € iL ( t ) = (0.415 A) sin(120πt − π 2) € € RCL Series Circuit V = εm sin(ωt ) Applied emf i = I0 sin(ωt + φ ) Solve for amplitude I0 and phase ϕ Use phasors to simplify the analysis € AC source creates oscillating current Individual voltage drops across components Use vector addition to find magnitude of applied EMF Inductive Reactance Four circuits are set up so that each contains an ac source and an inductor. Circuit A has a 50-Hz source and a 15-H inductor. Circuit B has a 60-Hz source and a 10-H inductor. Circuit C has a 100-Hz source and a 2-H inductor. Circuit D has a 50-Hz source and a 4-H inductor. Rank the inductive reactance for these circuits in order, from largest to smallest. a) A > B > C > D b) A > B > C = D c) D > C > A > B d) B > A > C > D Next Time LCR Oscillations ...
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This note was uploaded on 10/03/2011 for the course PHYS 212 taught by Professor Tedeschi during the Spring '08 term at South Carolina.

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