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Last Time: LRC Circuits with Phasors… Last time: LRC Circuits with phasors… I X where . R L εm XLL ≡ ω C L ⇒ φ IR 1 ∼ X C ≡ I XC ωC ε The phasor diagram gives us graphical solutions for φ and I: 2 2 2 2 ε=IRXXm ( +( LC − ) ) I XL I XC εm ⇓ φ 2 2 XX− I=ε m R + X()LC − X = IZ IR tanφ = LC R 2 2 ZRXX≡ +()LC − LRC series circuit; Summary of instantaneous Current and voltages VR = IR VL = IXL VC = IXC i t()= Icos (ω ) t I XL vR t()= IRcos (ω ) t ε 1 m v t()= IXcos (ω− t ) 90 = cos I ()ωt − 90 C C ωC v tL ()= IXcosL (ω+ t ) 90ω = cos I L( ω+ ) 90 t IR tε()= v t() = ω IZcos (+ φ ) t = εcos (t ω+ ) φ ad m I XC VVLC− ωLC−1 / ω 2 2 tanφ = = ZXXX=(()R + LC - ) VR R Lagging & Leading The phase φ between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances. XL≡ ω ε XXLC− L m tan φ = I = 1 Z R X ≡ C ω C XL Z XL XL φ Z R φ R R Z XC XC XC XL > XC XL < XC XL = XC φ > 0 φ < 0 φ = 0 current current current LAGS LEADS IN PHASE WITH applied voltage applied voltage applied voltage Lecture 19, Act 2 2A A series RC circuit is driven by emf ε. Which of the following could be an appropriate phasor ~ diagram? V ε L εm m VC VR VR VR V C εm VC (a) (b) (c) 2B For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b) The drive voltage lags the current. (c) The drive voltage leads the current. Lecture 19, Act 2 2A A series RC circuit is driven by emf ε. Which of the following could be an appropriate phasor ~ diagram? V ε L εm m VC VR VR VR V C εm VC (a) (b) (c) • The phasor diagram for the driven series RLC circuit always has the voltage across the capacitor lagging the current by 90°. The vector sum of the VC and VR phasors must equal the generator emf phasor εm. Lecture 19, Act 2 ~ 2B For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b) The drive voltage lags the current. (c) The drive voltage leads the current. First, remember that the current phasor I is always in the same orientation as the resistor voltage phasor VR (since the current and voltage VR , I are always in phase). From the diagram, we see φ that the drive phasor εm is lagging (clockwise) I. Just as V lags I by 90°, in an AC driven RC ε C m circuit, the drive voltage will also lag I by some angle less than 90°. The precise phase lag φ depends on the values of R, C and ω. Impedance, Z • From the phasor diagram we found that the current amplitude I was related to the drive voltage amplitude εm by ε = I Z mmm • Z is known as the “impedance”, and is basically the frequency dependent equivalent resistance of the series LRC circuit, given by: Z “ Impedance IZ IXX− XXLC− Triangle” LC | |φ | |φ IR R εm 2 2 R ZRXX≡= +()LC − or Z = Im cos(φ ) • Note that Z achieves its minimum value (R) when φ = 0. Under this condition the maximum current flows in the circuit. Resonance • For fixed R, C, L the current I will be a maximum at the resonant frequency ω which makes the impedance Z purely resistive (Z = R). i.e., εm εm Im= = Z 2 2 RXX+()LC − reaches a maximum when: X LC= X This condition is obtained when: 1 1 ωL = ⇒ ω = ωC LC • Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! • At this frequency, the current and the driving voltage are in phase: XX− tan φ = LC = 0 R Resonance Plot the current versus ω, the frequency of the voltage source: εm R •For ω very large, XL >> XC, φ Æ 90°, Im Æ 0 •For ω very small, XC >> XL, φ Æ -90°, Im Æ 0 Im 0 0 2ω ω Example: vary R V=100 v ω=1000 rad/s R=200, 500, 2000 ohm L=2 H C=0.5 μC Consider a general AC circuit containing a resistor, capacitor, and inductor, driven by an AC generator. 1) As the frequency of the circuit is either raised above or lowered below the resonant frequency, the impedance of the circuit ________________. a) always increases b) only increases for lowering the frequency below resonance c) only increases for raising the frequency above resonance 2) At the resonant frequency, which of the following is true? a) The current leads the voltage across the generator. b) The current lags the voltage across the generator. c) The current is in phase with the voltage across the generator. 2 2 • Impedance = Z = sqrt( R + (XL-XC) ) • At resonance, (XL-XC) = 0, and the impedance has its minimum value: Z = R • As frequency is changed from resonance, either up or down, (XL-XC) no longer is zero and Z must therefore increase. Changing the frequency away from the resonant frequency will change both the reductive and capacitive reactance such that XL -XC is no longer 0. This, when squared, gives a positive term to the impedance, increasing its value. By definition, at the resonance frequency, Imax is at its greatest and the phase angle is 0, so the current is in phase with the voltage across the generator. 4) Fill in the blank. This circuit is being driven __________ its resonance frequency. a) above b) below c) exactly at 5) The generator voltage ________________ the current. a) leads b) lags c) is in phase with Power in LRC circuit i t() I=cos t () (ω t ; ) vad (= ) t V(ω cos+ φ) t The instantaneous power delivered to L-R-C is: P t() i= t () vad () t = cos V(ω + φ) tcos(ω I) t We can use trig identities to expand the above to, cosP() t= V[] cos (ω ) t (φ )− sin(ωt) (φ) sin I(ω ) t cos VI=cos2ω)() t cossin φ (− VIω () cos )t ω () t sin( φ 1 1 P= P() t(cos = VI 2()ωt + 1 )()φ − cosVIsin()ω 2t () sin φ ave 2 2 ⎛ 1 ⎞ = VI⎜ ⎟cos()φ ⎝ 2 ⎠ 1 VI Pave= P() t = cos VI()φ = cos()φ 2 2 2 =RMSVI RMScos()φ ueder in LRC circuit, continPow 1 P= P() t = cos VI()φ =VI cos()φ ave 2 RMS RMS General result. VRMS is voltage across element, I RMS is current gh element, and throuϕ is phase angle btween them.e Example; 100Watt light bulb plugged into 120V house outlet, Pure resistive load (no L and no C), ϕ = 0. V 2 PIV= = rms rms rms R V 2 1202 =R rms = =144 Ω Pave 100 Pave 100 I rms= =0 . = 83A Vrms 120 Note; 120V house voltage is rms and has peak voltage of 120 √2 = 170V Question: What is PAVE for an inductor or capacitor? Summary • Power “power factor” 1 1 { ε rms ≡ ε m I ≡ I 2 rms 2 m 〈〉=Pt()εφrms I rms cos 2 2 2 XX− =()IRrms ZRXX≡ +() − tanφ = LC LC R • Driven Series LRC Circuit: 1 • Resonance condition ω = » Resonant frequency LC.
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