20. Alternating Currents
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1 University of hode sland PHY 204: Elementary Physics Physics Course Materials lternating Currents Gerhard Müller University of hode sland, Creative Commons License This work is licensed under a Creative Commons ttribution-noncommercial-share like 4.0 License. Follow this and additional works at: bstract Part twenty of course materials for Elementary Physics (PHY 204), taught by Gerhard Müller at the University of hode sland. Documents will be updated periodically as more entries become presentable. Some of the slides contain figures from the textbook, Paul. Tipler and Gene Mosca. Physics for Scientists and Engineers, 5 th /6 th editions. The copyright to these figures is owned by W.H. Freeman. We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook figures are not to be used or copied for any purpose outside this class without direct permission from W.H. Freeman. ecommended Citation Müller, Gerhard, "20. lternating Currents" (2015). PHY 204: Elementary Physics. Paper 4. This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@U. t has been accepted for inclusion in PHY 204: Elementary Physics by an authorized administrator of DigitalCommons@U. For more information, please contact digitalcommons@etal.uri.edu.
2 lternating Current Generator Coil of N turns and cross-sectional area rotating with angular frequency ω in uniform magnetic field B. ngle between area vector and magnetic field vector: θ = ωt. Flux through coil: Φ B = NB cos(ωt). nduced EMF: E = dφ B = E max sin(ωt) with amplitude E max = NBω. dt U.S. household outlet values: E max = f = 60Hz, ω = 2πf 377rad/s. 4/11/2015 [tsl287 1/34]
3 Single Device in C Circuit: esistor oltage of ac source : E = E max cos ωt Current through device: = max cos(ωt δ) esistor = = E max cos ωt = E max amplitude: max = E max impedance: X E max = max cos ωt, phase angle: δ = 0 (resistance) ωt 4/11/2015 [tsl288 2/34]
4 Single Device in C Circuit: nductor oltage of ac source : E = E max cos ωt Current through device: = max cos(ωt δ) nductor L = L d dt = E max cos ωt = E max ωl sin(ωt) amplitude: max = E max ωl, phase angle: δ = π 2 impedance: X L E max = ωl (inductive reactance) max L ωt 4/11/2015 [tsl289 3/34]
5 Single Device in C Circuit: Capacitor oltage of ac source : E = E max cos ωt Current through device: = max cos(ωt δ) Capacitor C = Q C = E max cos ωt = dq dt = ωce max sin(ωt) amplitude: max = ωce max, phase angle: δ = π 2 impedance: X C E max max = 1 ωc (capacitive reactance) C ωt 4/11/2015 [tsl290 4/34]
6 Single Device in C Circuit: pplication (1) The ac voltage source E = E max sin ωt has an amplitude of E max = 24 and an angular frequency of ω = 10rad/s. n each of the three circuits, find (a) the current amplitude max, (b) the current at time t = 1s. (1) (2) (3) 8 25mF 0.6H 4/11/2015 [tsl291 5/34]
7 Single Device in C Circuit: pplication (2) Consider an ac generator E(t) = E max cos(ωt), E max = 25, ω = 377rad/s connected to an inductor with inductance L = 12.7H. (a) Find the maximum value of the current. (b) Find the current when the emf is zero and decreasing. (c) Find the current when the emf is 12.5 and decreasing. (d) Find the power supplied by the generator at the instant described in (c). 4/11/2015 [tsl292 6/34]
8 Single Device in C Circuit: pplication (3) Consider an ac generator E(t) = E max cos(ωt), E max = 25, ω = 377rad/s connected to a capacitor with capacitance C = 4.15µF. (a) Find the maximum value of the current. (b) Find the current when the emf is zero and decreasing. (c) Find the current when the emf is 12.5 and increasing. (d) Find the power supplied by the generator at the instant described in (c). 4/11/2015 [tsl293 7/34]
9 LC Series Circuit (1) pplied alternating voltage: E = E max cos ωt esulting alternating current: = max cos(ωt δ) Goals: Find max, δ for given E max, ω. Find voltages, L, C across devices. Loop rule: E C L = 0 C L ε Note: ll voltages are time-dependent. n general, all voltages have a different phase. C L has the same phase as. 4/11/2015 [tsl301 8/34]
10 LC Series Circuit (2) Phasor diagram (for ωt = δ): oltage amplitudes:,max = max X = max L L C ε L,max = max X L = max ωl C,max = max X C = max ωc elation between E max and max from geometry: C δ Emax 2 =,max 2 + ( L,max C,max ) 2 " = max ωl 1 «# 2 ωc 4/11/2015 [tsl302 9/34]
11 LC Series Circuit (3) mpedance: Z E max max = s 2 + Current amplitude and phase angle: max = E max Z = E max q 2 + `ωl 1 tan δ = L,max C,max,max oltages across devices: = ωl 1 «2 ωc ωc 2 ωl 1/ωC L L C C δ ε = = max cos(ωt δ) =,max cos(ωt δ) L = L d dt = ωl max sin(ωt δ) = L,max cos ωt δ + π 2 C = 1 Z dt = max C ωc sin(ωt δ) = C,max cos ωt δ π 2 4/11/2015 [tsl303 10/34]
12 C Circuit pplication (1) n this LC circuit, the voltage amplitude is E max = 100. Find the impedance Z, the current amplitude max, and the voltage amplitudes, C, L, LC (a) for angular frequency is ω = 1000rad/s, (b) for angular frequency is ω = 500rad/s µF 2H C LC L 4/11/2015 [tsl304 11/34]
13 C Circuit pplication (2) n this LC circuit, we know the voltage amplitudes, C, L across each device, the current amplitude max = 5, and the angular frequency ω = 2rad/s. Find the device properties, C, L and the voltage amplitude E max of the ac source. ε max C L /11/2015 [tsl305 12/34]
14 mpedances: LC in Series (1) Z = s 2 + ωl 1 «2 ωc resonance at ω 0 = 1 LC limit 0 Z = ωl 1 ωc L Z 1 C L Z 1 C 0 0 4/11/2015 [tsl467 13/34]
15 mpedances: LC in Series (2) limit C limit L 0 Z = q 2 + (ωl) 2 Z = s (ωc) 2 Z L Z 1 C 4/11/2015 [tsl502 14/34]
16 Filters 4/11/2015 [tsl468 15/34]
17 LC Series esonance (1) impedance current phase angle Z = s 2 + ωl 1 «2 ωc max = max q 2 + `ωl 1 2 ωc δ = ωl 1/ωC L max max Π 2 Z 1 C Π 2 resonance angular frequency: ω 0 = 1 LC 4/11/2015 [tsl501 16/34]
18 LC Series esonance (2) resistor inductor capacitor max = max max L = max ωl max C = max /ωc max L max 0 max C max 0 max max max max relaxation times: τ C = C, τ L = L/ angular frequencies: ω L = ω 0 p 1 (ω0 τ C ) 2 /2, ω C = ω 0 q1 (ω 0 τ C ) 2 /2 voltages: 0 max = max ω 0 τ L, max (ω L ) = C max (ω C ) = L max 0 p 1 (ω0 τ C ) 2 /4 4/11/2015 [tsl503 17/34]
19 LC Parallel Circuit (1) pplied alternating voltage: E = E max cos ωt esulting alternating current: = max cos(ωt δ) Goals: Find max, δ for given E max, ω. Find currents, L, C through devices. ε Junction rule: = + L + C L L Note: ll currents are time-dependent. n general, each current has a different phase C C has the same phase as E. 4/11/2015 [tsl306 18/34]
20 LC Parallel Circuit (2) Phasor diagram (for ωt = δ): Current amplitudes:,max = E max X L,max = E max X L C,max = E max X C = E max = E max ωl = E max ωc L C C L δ ε elation between E max and max from geometry: max 2 =,max 2 + ( L,max C,max ) 2 " «# = Emax ωl ωc 4/11/2015 [tsl307 19/34]
21 LC Parallel Circuit (3) mpedance: 1 Z max E max = s Current amplitude and phase angle: max = E max Z = E max s tan δ = L,max C,max,max = «1 2 ωl ωc «1 2 ωl ωc 1/ωL ωc 1/ L C C δ ε Currents through devices: L = E = E max cos(ωt) =,max cos(ωt) L = 1 Z Edt = E max L ωl sin(ωt) = L,max cos C = C de dt = ωce max sin(ωt) = C,max cos ωt π 2 ωt + π 2 4/11/2015 [tsl308 20/34]
22 C Circuit pplication (3) Find the current amplitudes 1, 2, 3 (a) for angular frequency ω = 2rad/s, (b) for angular frequency ω = 4rad/s. ε max =50 =10 3 L=2.5H 2 C=0.1F 1 4/11/2015 [tsl309 21/34]
23 C Circuit pplication (4) Given the current amplitudes 1, 2, 3 through the three branches of this LC circuit, and given the amplitude E max = 100 and angular frequency ω = 500rad/s of the ac source, find the device properties, L, C. ε 3 =213.6m L 2 =75m C 1 =25m 4/11/2015 [tsl310 22/34]
24 mpedances: LC in Parallel (1) 1 Z = s ωc 1 «2 ωl resonance at ω 0 = 1 LC limit 1 Z = ωc 1 ωl C C 1 Z 1 Z 0 1 L L 4/11/2015 [tsl504 23/34]
25 mpedances: LC in Parallel (2) 1 Z = limit C 0 s (ωl) 2 1 Z = limit L r (ωc)2 1 Z 1 1 L 1 Z C 1 4/11/2015 [tsl505 24/34]
26 LC Parallel esonance (1) impedance current phase angle 1 Z = s ωc 1 «2 ωl max = max Z δ = 1/ωL ωc 1/ max Π 2 C 1 Z L 1 max 0 Π 2 resonance angular frequency: ω 0 = 1 LC 4/11/2015 [tsl506 25/34]
27 LC Parallel esonance (2) resistor inductor capacitor max = max / max L = max /ωl max C = max ωc max L max C max max 0 max 0 max currents at resonance: max = max, max L = max C = max 0 = max r C L. 4/11/2015 [tsl507 26/34]
28 Power in C Circuits oltage of ac source: E = E max cos ωt Current through circuit: = max cos(ωt δ) nstantaneous power supplied: P(t) = E(t)(t) = [E max cos ωt][ max cos(ωt δ)] Use cos(ωt δ) = cos ωt cos δ + sin ωtsin δ P(t) = E max max [cos 2 ωtcos δ + cos ωtsin ωtsin δ] Time averages: [cos 2 ωt] = 1 2, [cos ωtsin ωt] = 0 verage power supplied by source: P = 1 2 E max max cos δ = E rms rms cos δ Power factor: cos δ ε Z, δ L C 4/11/2015 [tsl311 27/34]
29 Transformer Primary winding: N 1 turns 1 (t) = (rms) 1 cos(ωt), 1 (t) = (rms) 1 cos(ωt δ 1 ) Secondary winding: N 2 turns 2 (t) = (rms) 2 cos(ωt), 2 (t) = (rms) 2 cos(ωt δ 2 ) oltage amplitude ratio: (rms) 1 (rms) 2 = N 1 N 2 Power transfer: (rms) 1 (rms) 1 cos δ 1 = (rms) 2 (rms) 2 cos δ 2 4/11/2015 [tsl418 28/34]
30 C Circuit pplication (5) Find the current amplitudes 1, 2, 3, 4 in the four LC circuits shown. εmax =1 ω =1rad/s 1 1H 1F 1 εmax =1 1 1H ω =1rad/s 2 1F εmax =1 1 ω =1rad/s 1H 1F 3 εmax =1 ω =1rad/s 1 1H 1F 4 4/11/2015 [tsl312 29/34]
31 C Circuit pplication (6) Consider an LC series circuit with inductance L = 88mH, capacitance C = 0.94µF, and unknown resistance. The ac generator E = E max sin(ωt) has amplitude E max = 24 and frequency f = 930Hz. The phase angle is δ = 75. (a) Find the resistance. (b) Find the current amplitude max. (c) Find the maximum energy UL max (d) Find the maximum energy UC max stored in the inductor. stored in the capacitor. (e) Find the time t 1 at which the current has its maximum value max. (f) Find the time t 2 at which the charge on the capacitor has its maximum value Q max. 4/11/2015 [tsl313 30/34]
32 C Circuit pplication (7) Consider the two ac circuits shown. (a) n the circuit on the left, determine the current amplitude 1 and the voltage amplitudes 1 and 2. (b) n the circuit on the right, determine the current amplitudes 2, 3, and 4. ω = 5rad/s ε max = 50 = 3 C = 0.1F 1 ω = 5rad/s C=0.1F ε max = L=2H /11/2015 [tsl384 31/34]
33 C Circuit pplication (8) Consider the two ac circuits shown. (a) n the circuit on the left, determine the maximum value of current 1 and the maximum value of voltages 1 and 2. (b) n the circuit on the right, determine the maximum value of currents 2, 3, and 4. ω = 5rad/s L=2H ε max = 50 C = 0.1F 1 ω = 5rad/s C=0.1F ε max = = /11/2015 [tsl385 32/34]
34 C Circuit pplication (9) n the two ac circuits shown the ammeter and voltmeter readings are rms values. (a) n the circuit on the left, find the resistance of the resistor, the capacitance C of the capacitor, the impedance Z of the two devices combined, and the voltage E rms of the power source. (b) n the circuit on the right, find the capacitance C of the capacitor, the inductance L of the inductor, the impedance Z of the two devices combined, and the rms value of the current 4. ω = 5rad/s ε rms C 1 = 9 ω = 5rad/s C 3 ε rms = 5 = 20 4 L 2 = 10 1 = 30 2 = 20 4/11/2015 [tsl386 33/34]
35 C Circuit pplication (10) n the two ac circuits shown the ammeter and voltmeter readings are rms values. (a) n the circuit on the left, find the capacitance C of the capacitor, the inductance L of the inductor, the impedance Z of the two devices combined, and the voltage E rms of the power source. (b) n the circuit on the right, find the capacitance C of the capacitor, the resistance of the resistor, the impedance Z of the two devices combined, and the rms value of the current 4. ε rms ω = 5rad/s 1 = 2 L C ω = 5rad/s C 3 ε rms = 5 = = 2 1 = 40 2 = 25 4/11/2015 [tsl387 34/34]
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