X X L C = 2π fl = 1/2π fc 2 AC Source and RLC Circuits ( ) 2 Inductive reactance Capacitive reactance Z = R + X X Total impedance L C εmax Imax = Z XL XC tanφ = R Maximum current Phase angle PHY2054: Chapter 21 1
Pictorial Understanding of Reactance 2 ( ) 2 Z = R + X X L C XL XC tanφ = R cosφ = R Z PHY2054: Chapter 21 2
Summary of Circuit Elements, Impedance, Phase Angles 2 ( ) 2 Z = R + X X L C XL XC tanφ = R PHY2054: Chapter 21 3
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 X I L max = 2π fl = ε max / X L For inductor, higher frequency gives higher reactance, therefore lower current f PHY2054: Chapter 21 4
RLC Example 1 Below are shown the driving emf and current vs time of an RLC circuit. We can conclude the following Current leads the driving emf (φ<0) Circuit is capacitive (X C > X L ) I ε t PHY2054: Chapter 21 5
RLC Example 2 R = 200Ω, C = 15μF, L = 230mH, ε max = 36v, f = 60 Hz Resonant frequency ( )( 6 ) f0 = 1/ 2π 0.230 15 10 = 85.6Hz X L XC = 2π 60 0.23 = 86.7Ω ( 6 π ) = 1/ 2 60 15 10 = 177Ω X C > X L Capacitive circuit 2 ( ) 2 Z = 200 + 86.7 177 = 219Ω I = ε / Z = 36/ 219 = 0.164A max max 1 86.7 177 φ = tan = 24.3 200 Current leads emf (as expected) PHY2054: Chapter 21 6
I max vs Frequency and Resonance Circuit parameters: C = 2.5μF, L = 4mH, ε max = 10v f 0 = 1 / 2π(LC) 1/2 = 1590 Hz Plot I max vs f 2 ( π π ) 2 Imax = 10 / R + 2 fl 1/ 2 fc I max R = 5Ω R = 10Ω R = 20Ω Resonance f = f 0 f / f 0 PHY2054: Chapter 21 7
Power in AC Circuits Recall power formula ave 2 rms P = I R I rms = I max / 2 Rewrite using I rms = ε rms Z εrms Pave = IrmsR=εrmsIrms Z cos φ P ave = ε rms I rms cosφ cos R φ = Z 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 L ) can be cancelled by increasing X C (e.g., circuits with large motors) PHY2054: Chapter 21 8
Power Example 1 R = 200Ω, X C = 150Ω, X L = 80Ω, ε rms = 120v, f = 60 Hz 2 ( ) 2 Z = 200 + 80 150 = 211.9Ω I = ε / Z = 120/ 211.9 = 0.566A rms rms 1 80 150 φ = tan = 19.3 200 cosφ = 0.944 Current leads emf Capacitive circuit P = ε I cos φ = 120 0.566 0.944 = 64.1W ave rms rms 2 2 Pave = IrmsR= 0.566 200 = 64.1W Same PHY2054: Chapter 21 9
Power Example 1 (cont) R = 200Ω, X C = 150Ω, X L = 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 L to minimize Z, so must decrease X C X = 150Ω= 1/ 2π fc C = 17.7μF C X = X = 80Ω C = 33.2μF Cnew L new So we must add 15.5μF capacitance to maximize power PHY2054: Chapter 21 10
Power vs Frequency and Resonance Circuit parameters: C = 2.5μF, L = 4mH, ε max = 10v f 0 = 1 / 2π(LC) 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 11
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 12
Quiz The figure shows the current and emf of a series RLC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken? (1) Increase R (2) Decrease L (3) Increase L (4) Increase C XL XC tanφ = R Current lags applied emf (φ > 0), thus circuit is inductive. Either (1) Reduce X L by decreasing L or (2) Cancel X L by increasing X C (decrease C). PHY2054: Chapter 21 13
Example: LR Circuit Variable frequency EMF source with ε m =6V connected to a resistor and inductor. R=80Ω and L=40mH. At what frequency f does V R = V L? X = 2π fl= R f = 318Hz L At that frequency, what is phase angle φ? tan φ = X / R= 1 φ = 45 L What is the current amplitude? I max max 2 2 = ε / 80 + 80 = 6 /113 = 0.053A What is the rms current? I rms = I max / 2 = 0.037 A PHY2054: Chapter 21 14
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 15
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 (Losses usually account for 10-20%) iv = iv p p s s i V s p N = = i V N p p s s PHY2054: Chapter 21 16
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 Step-up transformer iv = iv p p s s Vs 451 ip = is = 15 = 56.4A V 120 p PHY2054: Chapter 21 17
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 18
Applications Microphone Tape recorder PHY2054: Chapter 21 19
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 20
Electric Power Transmission i 2 R: 20x smaller current 400x smaller power loss PHY2054: Chapter 21 21