AC Circuit Application (1)
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Alternating Current Generator Coil of N turns and cross-sectional area A rotating with angular frequency ω ~ in uniform magnetic field B. • Angle between area vector and magnetic field vector: θ = ωt. • Flux through coil: ΦB = N BA cos(ωt). dΦB = Emax sin(ωt) with amplitude Emax = N BAω. dt • U.S. household outlet values: √ ◦ Emax = 120V 2 ≃ 170V ◦ f = 60Hz, ω = 2πf ≃ 377rad/s. • Induced EMF: E = − 4/11/2015 [tsl287 – 1/34] Single Device in AC Circuit: Resistor Voltage of ac source : E = Emax cos ωt Current through device: I = Imax cos(ωt − δ) Resistor VR = RI = Emax cos ωt ⇒ I = Emax cos ωt R Emax , phase angle: δ = 0 R Emax ≡ = R (resistance) Imax amplitude: Imax = impedance: XR I VR ωt 4/11/2015 [tsl288 – 2/34] Single Device in AC Circuit: Inductor Voltage of ac source : E = Emax cos ωt Current through device: I = Imax cos(ωt − δ) Inductor Emax dI = Emax cos ωt ⇒ I = sin(ωt) dt ωL Emax π amplitude: Imax = , phase angle: δ = ωL 2 Emax impedance: XL ≡ = ωL (inductive reactance) Imax VL = L VL ωt I 4/11/2015 [tsl289 – 3/34] Single Device in AC Circuit: Capacitor Voltage of ac source : E = Emax cos ωt Current through device: I = Imax cos(ωt − δ) Capacitor dQ Q = Emax cos ωt ⇒ I = = −ωCEmax sin(ωt) C dt π amplitude: Imax = ωCEmax , phase angle: δ = − 2 1 Emax (capacitive reactance) = impedance: XC ≡ Imax ωC VC = I VC ωt 4/11/2015 [tsl290 – 4/34] Single Device in AC Circuit: Application (1) The ac voltage source E = Emax sin ωt has an amplitude of Emax = 24V and an angular frequency of ω = 10rad/s. In each of the three circuits, find (a) the current amplitude Imax , (b) the current I at time t = 1s. (1) ~ (2) ~ (3) ~ 8Ω 25mF 0.6H 4/11/2015 [tsl291 – 5/34] Single Device in AC Circuit: Application (2) Consider an ac generator E(t) = Emax cos(ωt), Emax = 25V, 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 −12.5V and decreasing. (d) Find the power supplied by the generator at the instant described in (c). 4/11/2015 [tsl292 – 6/34] Single Device in AC Circuit: Application (3) Consider an ac generator E(t) = Emax cos(ωt), Emax = 25V, 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 −12.5V and increasing. (d) Find the power supplied by the generator at the instant described in (c). 4/11/2015 [tsl293 – 7/34] RLC Series Circuit (1) Applied alternating voltage: E = Emax cos ωt Resulting alternating current: I = Imax cos(ωt − δ) Goals: • Find Imax , δ for given Emax , ω. • Find voltages VR , VL , VC across devices. ~ ε A Loop rule: E − VR − VC − VL = 0 Note: • All voltages are time-dependent. R C L V V V VR VC VL I • In general, all voltages have a different phase. • VR has the same phase as I. 4/11/2015 [tsl301 – 8/34] RLC Series Circuit (2) Phasor diagram (for ωt = δ): VL Voltage amplitudes: • VR,max = Imax XR = Imax R ε VL−VC • VL,max = Imax XL = Imax ωL • VC,max = Imax XC δ Imax = ωC VC I VR Relation between Emax and Imax from geometry: 2 Emax = = 2 + (VL,max − VC,max )2 VR,max " «2 # „ 1 2 R2 + ωL − Imax ωC 4/11/2015 [tsl302 – 9/34] RLC Series Circuit (3) Emax Impedance: Z ≡ = Imax s «2 „ 1 R2 + ωL − ωC VL ε Current amplitude and phase angle: • Imax Emax Emax = q = ` Z 2 R + ωL − • tan δ = VL−VC ´ 1 2 ωC δ VL,max − VC,max ωL − 1/ωC = VR,max R VC I VR Voltages across devices: • VR = RI = RImax cos(ωt − δ) = VR,max cos(ωt − δ) “ dI = −ωLImax sin(ωt − δ) = VL,max cos ωt − δ + • VL = L dt Z “ Imax 1 sin(ωt − δ) = VC,max cos ωt − δ − Idt = • VC = C ωC π” 2 π” 2 4/11/2015 [tsl303 – 10/34] AC Circuit Application (1) In this RLC circuit, the voltage amplitude is Emax = 100V. Find the impedance Z, the current amplitude Imax , and the voltage amplitudes VR , VC , VL , VLC (a) for angular frequency is ω = 1000rad/s, (b) for angular frequency is ω = 500rad/s. ~ 100V A 500 Ω 0.5µF 2H V V V VR VC VLC VL V 4/11/2015 [tsl304 – 11/34] AC Circuit Application (2) In this RLC circuit, we know the voltage amplitudes VR , VC , VL across each device, the current amplitude Imax = 5A, and the angular frequency ω = 2rad/s. • Find the device properties R, C, L and the voltage amplitude Emax of the ac source. ~ ε max A R C L V V V 50V 25V 25V 4/11/2015 [tsl305 – 12/34] Impedances: RLC in Series (1) Z= s R2 + „ 1 ωL − ωC resonance at ω0 = √ Ω0 «2 limit R → 0 ˛ ˛ ˛ 1 ˛˛ Z = ˛˛ωL − ωC ˛ 1 LC ΩL ΩL Z Z 1ΩC 1ΩC R Ω Ω0 Ω 4/11/2015 [tsl467 – 13/34] Impedances: RLC in Series (2) limit L → 0 limit C → ∞ q Z = R2 + (ωL)2 Z= s R2 + 1 (ωC)2 Z ΩL Z R R Ω 1ΩC Ω 4/11/2015 [tsl502 – 14/34] Filters 4/11/2015 [tsl468 – 15/34] RLC Series Resonance (1) impedance Z= s R2 + „ current 1 ωL − ωC «2 Vmax Imax = q ` R2 + ωL − Imax phase angle ´ 1 2 ωC δ= ωL − 1/ωC R ∆ Π2 VmaxR ΩL Z Ω Ω0 1ΩC Ω0 R Ω Ω0 Ω -Π2 resonance angular frequency: ω0 = √ 1 LC 4/11/2015 [tsl501 – 16/34] RLC Series Resonance (2) resistor inductor capacitor VRmax = Imax R VLmax = Imax ωL VCmax = Imax /ωC VRmax VLmax VC max V0max V0max Vmax Vmax Vmax Ω0 Ω • relaxation times: τRC = RC, • angular frequencies: ωL = p • voltages: V0max Ω Ω0 Ω Ω0 τRL = L/R ω0 1 − (ω0 τRC = Vmax ω0 τRL , )2 /2 VLmax (ωL ) q = ω0 1 − (ω0 τRC )2 /2 , ωC = VCmax (ωC ) V0max = p 1 − (ω0 τRC )2 /4 4/11/2015 [tsl503 – 17/34] RLC Parallel Circuit (1) Applied alternating voltage: E = Emax cos ωt Resulting alternating current: I = Imax cos(ωt − δ) ~ Goals: • Find Imax , δ for given Emax , ω. R L Junction rule: I = IR + IL + IC • All currents are time-dependent. IR A I A • Find currents IR , IL , IC through devices. Note: ε IL A C IC A • In general, each current has a different phase • IR has the same phase as E. 4/11/2015 [tsl306 – 18/34] RLC Parallel Circuit (2) Phasor diagram (for ωt = δ): IC Current amplitudes: • IR,max Emax Emax = = XR R • IL,max Emax Emax = = XL ωL • IC,max = IR ε δ I L − IC I Emax = Emax ωC XC IL Relation between Emax and Imax from geometry: 2 Imax = = 2 IR,max + (IL,max − IC,max )2 " «2 # „ 1 1 2 − ωC + Emax 2 R ωL 4/11/2015 [tsl307 – 19/34] RLC Parallel Circuit (3) 1 Imax Impedance: ≡ = Z Emax s 1 + R2 „ 1 − ωC ωL «2 Current amplitude and phase angle: s «2 „ 1 1 Emax = Emax − ωC + • Imax = Z R2 ωL • tan δ = IL,max − IC,max 1/ωL − ωC = IR,max 1/R Currents through devices: IC IR ε δ I L − IC I IL E Emax = cos(ωt) = IR,max cos(ωt) R R Z “ 1 Emax π” • IL = Edt = sin(ωt) = IL,max cos ωt − L ωL 2 “ π” dE = −ωCEmax sin(ωt) = IC,max cos ωt + • IC = C dt 2 • IR = 4/11/2015 [tsl308 – 20/34] AC Circuit Application (3) Find the current amplitudes I1 , I2 , I3 (a) for angular frequency ω = 2rad/s, (b) for angular frequency ω = 4rad/s. ~ εmax =50V R=10Ω L=2.5H C=0.1F A I3 A I2 A I1 4/11/2015 [tsl309 – 21/34] AC Circuit Application (4) Given the current amplitudes I1 , I2 , I3 through the three branches of this RLC circuit, and given the amplitude Emax = 100V and angular frequency ω = 500rad/s of the ac source, find the device properties R, L, C. ε ~ R L C A I 3 =213.6mA A I 2 =75mA A I 1 =25mA 4/11/2015 [tsl310 – 22/34] Impedances: RLC in Parallel (1) 1 = Z s 1 + R2 „ « 1 2 ωC − ωL resonance at ω0 = √ Ω0 limit R → ∞ ˛ ˛ ˛ 1 ˛˛ 1 = ˛˛ωC − Z ωL ˛ 1 LC ΩC ΩC 1Z 1Z 1ΩL 1ΩL Ω0 1R Ω Ω 4/11/2015 [tsl504 – 23/34] Impedances: RLC in Parallel (2) limit C → 0 1 = Z s limit L → ∞ 1 1 + R2 (ωL)2 1 = Z r 1 + (ωC)2 2 R 1Z ΩC 1Z 1R 1R 1ΩL Ω Ω 4/11/2015 [tsl505 – 24/34] RLC Parallel Resonance (1) current impedance 1 = Z s 1 + R2 „ 1 ωC − ωL «2 Imax = phase angle Vmax Z δ= 1/ωL − ωC 1/R ∆ Imax Π2 ΩC 1Z Ω0 1ΩL Ω0 Ω VmaxR 1R Ω Ω0 Ω -Π2 resonance angular frequency: ω0 = √ 1 LC 4/11/2015 [tsl506 – 25/34] RLC Parallel Resonance (2) resistor inductor capacitor max IR = Vmax /R max IL = Vmax /ωL max IC = Vmax ωC ILmax IRmax IC max I0max I0max VmaxR Ω Ω0 Ω0 Ω Ω Ω0 currents at resonance: max IR Vmax = , R max max IL = IC = I0max = Vmax r C . L 4/11/2015 [tsl507 – 26/34] Power in AC Circuits Voltage of ac source: E = Emax cos ωt Current through circuit: I = Imax cos(ωt − δ) Instantaneous power supplied: P (t) = E(t)I(t) = [Emax cos ωt][Imax cos(ωt − δ)] Use cos(ωt − δ) = cos ωt cos δ + sin ωt sin δ ⇒ P (t) = Emax Imax [cos2 ωt cos δ + cos ωt sin ωt sin δ] Time averages: [cos2 ωt]AV = 1 , 2 [cos ωt sin ωt]AV = 0 Average power supplied by source: PAV = Power factor: cos δ 1 Emax Imax cos δ = Erms Irms cos δ 2 ε ~ ~ A Z,δ R L C 4/11/2015 [tsl311 – 27/34] Transformer • Primary winding: N1 turns (rms) cos(ωt − δ1 ) • Secondary winding: N2 turns (rms) cos(ωt − δ2 ) (rms) cos(ωt), I1 (t) = I1 V1 (t) = V1 (rms) cos(ωt), I2 (t) = I2 V2 (t) = V2 (rms) • Voltage amplitude ratio: V1 (rms) V2 (rms) (rms) I1 • Power transfer: V1 = N1 N2 (rms) (rms) I2 cos δ1 = V2 cos δ2 4/11/2015 [tsl418 – 28/34] AC Circuit Application (5) Find the current amplitudes I1 , I2 , I3 , I4 in the four RLC circuits shown. εmax =1V ~ I1 1Ω εmax =1V ω =1rad/s ~ 1Ω A 1F 1H ω =1rad/s I2 A 1H 1F εmax =1V ~ 1H 1Ω εmax =1V ω =1rad/s 1F I3 ~ 1Ω A 1H ω =1rad/s I4 A 1F 4/11/2015 [tsl312 – 29/34] AC Circuit Application (6) Consider an RLC series circuit with inductance L = 88mH, capacitance C = 0.94µF, and unknown resistance R. The ac generator E = Emax sin(ωt) has amplitude Emax = 24V and frequency f = 930Hz. The phase angle is δ = 75◦ . (a) Find the resistance R. (b) Find the current amplitude Imax . max stored in the inductor. (c) Find the maximum energy UL max stored in the capacitor. (d) Find the maximum energy UC (e) Find the time t1 at which the current has its maximum value Imax . (f) Find the time t2 at which the charge on the capacitor has its maximum value Qmax . 4/11/2015 [tsl313 – 30/34] AC Circuit Application (7) Consider the two ac circuits shown. (a) In the circuit on the left, determine the current amplitude I1 and the voltage amplitudes V1 and V2 . (b) In the circuit on the right, determine the current amplitudes I2 , I3 , and I4 . ω = 5rad/s R = 3Ω ~ ε max = 50V A I1 ω = 5rad/s ~ C=0.1F C = 0.1F V V1 V2 I3 A I4 A I2 A L=2H V εmax = 50V 4/11/2015 [tsl384 – 31/34] AC Circuit Application (8) Consider the two ac circuits shown. (a) In the circuit on the left, determine the maximum value of current I1 and the maximum value of voltages V1 and V2 . (b) In the circuit on the right, determine the maximum value of currents I2 , I3 , and I4 . ω = 5rad/s ~ ε max = 50V A L=2H C = 0.1F I1 ω = 5rad/s ~ C=0.1F εmax = 50V I3 V V1 V2 I4 A I2 A R = 3Ω V A 4/11/2015 [tsl385 – 32/34] AC Circuit Application (9) In the two ac circuits shown the ammeter and voltmeter readings are rms values. (a) In the circuit on the left, find the resistance R of the resistor, the capacitance C of the capacitor, the impedance Z of the two devices combined, and the voltage Erms of the power source. (b) In 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 I4 . ω = 5rad/s ~ ε rms A R C I1= 9A ω = 5rad/s ~ C I 3 = 5A V1 = 30V A I4 A I 2 = 10A A L V εrms = 20V V V2 = 20V 4/11/2015 [tsl386 – 33/34] AC Circuit Application (10) In the two ac circuits shown the ammeter and voltmeter readings are rms values. (a) In 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 Erms of the power source. (b) In the circuit on the right, find the capacitance C of the capacitor, the resistance R of the resistor, the impedance Z of the two devices combined, and the rms value of the current I4 . ω = 5rad/s L ~ ε rms I 1 = 2A C ω = 5rad/s ~ A C I 3 = 5A V1 = 40V A I4 A I 2 = 2A A R V εrms = 50V V V2 = 25V 4/11/2015 [tsl387 – 34/34]
