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Lesson 19 Impedance Learning Objectives For purely resistive, inductive and capacitive elements define the voltage and current phase differences. Define inductive reactance. Understand the variation of inductive reactance as a function of frequency. Define capacitive reactance. Understand the variation of capacitive reactance as a function of frequency. Define impedance. Graph impedances of purely resistive, inductive and capacitive elements as a function of phase. Review R, L and C circuits with Sinusoidal Excitation R, L, C have very different voltage-current relationships vR iR R dvC iC C dt diL vL L dt (Ohm's law) (capacitor current relationship) (inductor voltage relationship) Sinusoidal (ac) sources are a special case The Impedance Concept Impedance (Z) is the opposition that a circuit element presents to current in the phasor domain. It is defined V V Z = = Ðq = ZÐq I I Ohm’s law for ac circuits V IZ Impedance Impedance is a complex quantity that can be made up of resistance (real part) and reactance (imaginary part). Unit of impedance is ohms (). Z X q R Resistance and Sinusoidal AC For a purely resistive circuit, current and voltage are in phase. Resistors For resistors, voltage and current are in phase. VR VR q VR ZR 0 R0 R I I q I Z R R0 Example Problem 1 Two resistors R1=10 kΩ and R2=12.5 kΩ are in series. If i(t) = 14.7 sin (ωt + 39˚) mA a) b) c) d) Compute VR1 and VR2 Compute VT=VR1 + VR2 Calculate ZT Compare VT to the results of VT=IZT Inductance and Sinusoidal AC Voltage-Current relationship for an inductor diL d L I m sin t dt dt LI m cos t LI m sin t 90 vL L vL LI m sin t 90 ZL iL I m si n t LI m 90 2 Im 0 2 L90 ( ) It should be noted that for a purely inductive circuit voltage leads current by 90º. Inductive Impedance Impedance can be written as a complex number (in rectangular or polar form): Z L L90 j L ( ) Since an ideal inductor has no real resistive component, this means the reactance of an inductor is the pure imaginary part: X L L Inductance and Sinusoidal AC Voltage leads current by 90˚ Inductance For inductors, voltage leads current by 90º. VL VL 90 VL ZL 90 L90 j L I I 0 I Z L jX L X L90 X L L 2 fL Impedance and AC Circuits Solution technique 1. Transform time domain currents and voltages into phasors Calculate impedances for circuit elements Perform all calculations using complex math Transform resulting phasors back to time domain (if reqd) 2. 3. 4. Example Problem 2 For the inductive circuit: vL = 40 sin (ωt + 30˚) V f = 26.53 kHz L = 2 mH Determine VL and IL Graph vL and iL Example Problem 2 solution vL = 40 sin (ωt + 30˚) V iL = 120 sin (ωt - 60˚) mA iL Notice 90°phase difference! vL Example Problem 3 For the inductive circuit: vL = 40 sin (ωt + Ө) V iL = 250 sin (ωt + 40˚) μA f = 500 kHz What is L and Ө? Capacitance and Sinusoidal AC Current-voltage relationship for an capacitor dvC d iC C C Vm sin t dt dt CVm cos t CVm sin t 90 Zc vc Vm sin t ic CVm sin t 90 Vm 0 1 2 90 ( ) Vm C 90 C 2 It should be noted that, for a purely capacitive circuit current leads voltage by 90º. Capacitive Impedance Impedance can be written as a complex number (in rectangular or polar form): 1 1 Zc 90 j ( ) C C Since a capacitor has no real resistive component, this means the reactance of a capacitor is the pure imaginary part: 1 Xc C Capacitance and Sinusoidal AC Capacitance For capacitors, voltage lags current by 90º. VC VC 0 VC 1 1 ZC 90 90 j I I 90 I C C ZC jX C X C 90 1 1 XC C 2 fC Example Problem 4 For the capacitive circuit: vC = 3.6 sin (ωt-50°) V f = 12 kHz C=1.29 uF Determine VC and IC Example Problem 5 For the capacitive circuit: vC = 362 sin (ωt - 33˚) V iC = 94 sin (ωt + 57˚) mA C = 2.2 μF Determine the frequency ELI the ICE man E leads I When voltage is applied to an inductor, it resists the change of current. The current builds up more slowly, lagging in time and phase. I leads E Since the voltage on a capacitor is directly proportional to the charge on it, the current must lead the voltage in time and phase to conduct charge to the capacitor plate and raise the voltage