Chapter 21: RLC Circuits. PHY2054: Chapter 21 1


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1 Chapter 21: RC Circuits PHY2054: Chapter 21 1
2 Voltage and Current in RC Circuits AC emf source: driving frequency f ε = ε sinωt ω = 2π f m If circuit contains only R + emf source, current is simple ε ε i = = Imsin t Im = R R m ( ω ) ( current amplitude) If and/or C present, current is not in phase with emf εm i = Imsin( ωt φ) Im = Z Z, φ shown later PHY2054: Chapter 21 2
3 AC Source and Resistor Only Driving voltage is ε = ε sinωt m i Relation of current and voltage i = ε / R ε i = Imsinωt Im = R m ε ~ R Current is in phase with voltage (φ = 0) PHY2054: Chapter 21 3
4 AC Source and Capacitor Only Voltage is vc = q m sin t C = ε ω Differentiate to find current q = Cεm sin i ωt i = dq/ dt =ωcv cosωt Rewrite using phase (check this!) C Relation of current and voltage εm i = Imsin( ωt+ 90 ) Im = X Capacitive reactance : Current leads voltage by 90 C ( ω ) i = ωcv sin t+ 90 X C C = 1/ ωc ε ~ ( X = 1/ ωc) C C PHY2054: Chapter 21 4
5 AC Source and Inductor Only Voltage is v = di / dt =ε sinωt m Integrate di/dt to find current: = ( εm ) ( ) di / dt / sinωt i = εm / ω cos ωt Rewrite using phase (check this!) i = ε / ω sin ωt 90 ( ) ( ) m ε ~ i Relation of current and voltage εm i = Imsin( ωt 90 ) Im = X ( X = ω) Inductive reactance : X Current lags voltage by 90 = ω PHY2054: Chapter 21 5
6 General Solution for RC Circuit We assume steady state solution of form I m is current amplitude φ is phase by which current lags the driving EMF Must determine I m and φ i = I t Plug in solution: differentiate & integrate sin(ωtφ) i = I t di dt m sin ( ω φ ) = ωi ω φ I m m cos ( t ) q= cos( ωt φ ) ω Substitute m sin ( ω φ ) di q Ri m sin t dt + + C = ε ω Im Imωcos( ωt φ) + ImRsin ( ωt φ) cos( ωt φ) = εmsinωt ωc PHY2054: Chapter 21 6
7 General Solution for RC Circuit (2) Im Imωcos( ωt φ) + ImRsin ( ωt φ) cos( ωt φ) = εmsinωt ωc Expand sin & cos expressions ( ) ( ) sin ωt φ = sinωt cosφ cosωt sinφ cos ωt φ = cosωt cosφ+ sinωt sinφ Collect sinωt &cosωt terms separately High school trig! ( ) ( ) ω 1/ ωc cosφ Rsinφ = 0 I ω 1/ ωc sinφ+ I Rcosφ = ε m m m cosωt terms sinωt terms These equations can be solved for I m and φ (next slide) PHY2054: Chapter 21 7
8 General Solution for RC Circuit (3) Solve for φ and I m ω 1/ ωc X XC ε tanφ = m I R R m = Z R, X, X C and Z have dimensions of resistance X XC = ω = 1/ ωc Inductive reactance Capacitive reactance 2 ( ) 2 Z = R + X X C Total impedance This is where φ, X, X C and Z come from! PHY2054: Chapter 21 8
9 I m = tanφ C ε m Z X AC Source and RC Circuits X C = Phase angle R = 1/ ωc ( 2 ) X = ω ω = π f X 2 ( ) 2 Maximum current Inductive reactance Capacitive reactance Z = R + X X Total impedance C φ= angle that current lags applied voltage PHY2054: Chapter 21 9
10 What is Reactance? Think of it as a frequencydependent resistance X C = 1 ωc Shrinks with increasing ω X = ω Grows with increasing ω (" X " = R ) R Independent of ω PHY2054: Chapter 21 10
11 Pictorial Understanding of Reactance 2 ( ) 2 Z = R + X X C X XC tanφ = R cosφ = R Z PHY2054: Chapter 21 11
12 Summary of Circuit Elements, Impedance, Phase Angles 2 ( ) 2 Z = R + X X C X XC tanφ = R PHY2054: Chapter 21 12
13 Quiz Three identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured as a function of frequency. Which one of the following curves corresponds to an inductive circuit? (1) a (2) b (3) c (4) Can t tell without more info I max a b c max max ( 2 ) X = ω ω = π f I = ε / X For inductor, higher frequency gives higher reactance, therefore lower current f PHY2054: Chapter 21 13
14 RC Example 1 Below are shown the driving emf and current vs time of an RC circuit. We can conclude the following Current leads the driving emf (φ<0) Circuit is capacitive (X C > X ) I ε t PHY2054: Chapter 21 14
15 RC Example 2 R = 200Ω, C = 15μF, = 230mH, ε max = 36v, f = 60 Hz X XC = 2π = 86.7Ω ( 6 π ) = 1/ = 177Ω X C > X Capacitive circuit 2 ( ) 2 Z = = 219Ω I = ε / Z = 36/ 219 = 0.164A max max φ = tan = Current leads emf (as expected) i ( ωt ) = 0.164sin PHY2054: Chapter 21 15
16 Resonance Consider impedance vs frequency ( ) ( ω 1/ ω ) C Z = R + X X = R + C Z is minimum when This is resonance! ω = 1/ ωc ω = ω0 = 1/ C At resonance Impedance = Z is minimum Current amplitude = I m is maximum PHY2054: Chapter 21 16
17 I max vs Frequency and Resonance Circuit parameters: C = 2.5μF, = 4mH, ε max = 10v f 0 = 1 / 2π(C) 1/2 = 1590 Hz Plot I max vs f 2 ( ω ω ) 2 Imax = 10 / R + 1/ C I max R = 5Ω R = 10Ω R = 20Ω Resonance f = f 0 f / f 0 PHY2054: Chapter 21 17
18 Power in AC Circuits Instantaneous power emitted by circuit: P = i 2 R 2 sin 2 ( ω φ ) P= I R t m d More useful to calculate power averaged over a cycle Use < > to indicate average over a cycle ( ω φ) m sin d 2 m P = I R t = I R Instantaneous power oscillates Define RMS quantities to avoid ½ factors in AC circuits I m ε I rms = ε rms = m 2 2 House current V rms = 110V V peak = 156V ave 2 rms P = I R PHY2054: Chapter 21 18
19 Power formula Rewrite using Power in AC Circuits I rms ave 2 rms P = I R = ε rms Z I rms = I max / 2 εrms Pave = IrmsR=εrmsIrms Z cos φ P ave = ε rms I rms cosφ cos R φ = Z φ Z R X X C cosφ is the power factor To maximize power delivered to circuit make φ close to zero Max power delivered to load happens at resonance E.g., too much inductive reactance (X ) can be cancelled by increasing X C (e.g., circuits with large motors) PHY2054: Chapter 21 19
20 Power Example 1 R = 200Ω, X C = 150Ω, X = 80Ω, ε rms = 120v, f = 60 Hz 2 ( ) 2 Z = = 211.9Ω I = ε / Z = 120/ = 0.566A rms rms φ = tan = cosφ = Current leads emf Capacitive circuit P = ε I cos φ = = 64.1W ave rms rms 2 2 Pave = IrmsR= = 64.1W Same PHY2054: Chapter 21 20
21 Power Example 1 (cont) R = 200Ω, X C = 150Ω, X = 80Ω, ε rms = 120v, f = 60 Hz How much capacitance must be added to maximize the power in the circuit (and thus bring it into resonance)? Want X C = X to minimize Z, so must decrease X C X = 150Ω= 1/ 2π fc C = 17.7μF C X = X = 80Ω C = 33.2μF Cnew new So we must add 15.5μF capacitance to maximize power PHY2054: Chapter 21 21
22 Power vs Frequency and Resonance Circuit parameters: C = 2.5μF, = 4mH, ε max = 10v f 0 = 1 / 2π(C) 1/2 = 1590 Hz Plot P ave vs f for different R values P ave Resonance f = f 0 R = 2Ω R = 5Ω R = 10Ω R = 20Ω f / f 0 PHY2054: Chapter 21 22
23 Resonance Tuner is Based on Resonance Vary C to set resonance frequency to (ugh!) Other radio stations. RC response is less Circuit response Q = 500 Tune for f = MHz PHY2054: Chapter 21 23
24 Quiz A generator produces current at a frequency of 60 Hz with peak voltage and current amplitudes of 100V and 10A, respectively. What is the average power produced if they are in phase? (1) 1000 W (2) 707 W (3) 1414 W (4) 500 W (5) 250 W 1 ave = ε 2 peak peak = εrms rms P I I PHY2054: Chapter 21 24
25 Quiz The figure shows the current and emf of a series RC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken? (1) Increase R (2) Decrease (3) Increase (4) Increase C X XC tanφ = R Current lags applied emf (φ > 0), thus circuit is inductive. Either (1) Reduce X by decreasing or (2) Cancel X by increasing X C (decrease C). PHY2054: Chapter 21 25
26 Example: R Circuit Variable frequency EMF source with ε m =6V connected to a resistor and inductor. R=80Ω and =40mH. At what frequency f does V R = V? X = ω= R ω = 2000 f = 2000 / 2π = 318Hz At that frequency, what is phase angle φ? tan φ = X / R= 1 φ = 45 What is the current amplitude and RMS value? I max max 2 2 = ε / = 6 /113 = 0.053A I rms I max / A = = i = 0.053sin ( ωt 45 ) PHY2054: Chapter 21 26
27 Transformers Purpose: change alternating (AC) voltage to a bigger (or smaller) value Input AC voltage in the primary turns produces a flux Changing flux in secondary turns induces an emf V p = N p ΔΦB Δt V s = N s ΔΦ Δt B N s Vs = Vp N p PHY2054: Chapter 21 27
28 Transformers Nothing comes for free, however! Increase in voltage comes at the cost of current. Output power cannot exceed input power! power in = power out (osses usually account for 1020%) iv = iv p p s s i V s p N = = i V N p p s s PHY2054: Chapter 21 28
29 Transformers: Sample Problem A transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120 V and the output current is 15.0 A. What is the output voltage and input current? Ns 1240 Vs = Vp = 120 = 451V N 330 p Stepup transformer iv = iv p p s s Vs 451 ip = is = 15 = 56.4A V 120 p PHY2054: Chapter 21 29
30 Transformers This is how first experiment by Faraday was done He only got a deflection of the galvanometer when the switch is opened or closed Steady current does not make induced emf. PHY2054: Chapter 21 30
31 Applications Microphone Tape recorder PHY2054: Chapter 21 31
32 ConcepTest: Power lines At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit (high V, low i) or (high i, low V)? (1) high V, low i (2) low V, high i (3) makes no difference Power loss is i 2 R PHY2054: Chapter 21 32
33 Electric Power Transmission i 2 R: 20x smaller current 400x smaller power loss PHY2054: Chapter 21 33
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