Presentation on theme: "Lecture 211 Phasor Diagrams and Impedance. Lecture 212 Set Phasors on Stun 1.Sinusoids-amplitude, frequency and phase (Section 8.1) 2.Phasors-amplitude."— Presentation transcript:
Lecture 211 Phasor Diagrams and Impedance
Lecture 212 Set Phasors on Stun 1.Sinusoids-amplitude, frequency and phase (Section 8.1) 2.Phasors-amplitude and phase (Section 8.3 [sort of]) 2a. Complex numbers (Appendix B).
Lecture 213 Set Phasors on Kill 3.Complex exponentials-amplitude and phase 4.Relationship between phasors, complex exponentials, and sinusoids 5.Phasor relationships for circuit elements (Section 8.4) 5a. Arithmetic with complex numbers (Appendix B).
Lecture 214 Set Phasors on Vaporize 6.Fundamentals of impedance and admittance (some of Section 8.5) 7.Phasor diagrams (some of Section 8.6)
Lecture 215 Phasor Diagrams A phasor diagram is just a graph of several phasors on the complex plane (using real and imaginary axes). A phasor diagram helps to visualize the relationships between currents and voltages.
Lecture 216 An Example - 1F1F VCVC + - 2mA 40 1k VRVR V
Lecture 217 An Example (cont.) I = 2mA 40 V R = 2V 40 V C = 5.31V -50 V = 5.67V
Lecture 218 Phasor Diagram Real Axis Imaginary Axis VRVR VCVC V
Lecture 219 Impedance AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law: V = I Z Z is called impedance.
Lecture 2110 Impedance Resistor: –The impedance is R Inductor: –The impedance is j L
Lecture 2111 Impedance Capacitor: –The impedance is 1/j L
Lecture 2112 Some Thoughts on Impedance Impedance depends on the frequency . Impedance is (often) a complex number. Impedance is not a phasor (why?). Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.
Lecture 2113 Impedance Example: Single Loop Circuit 20k + - 1F1F10V 0 VCVC + - = 377 Find V C
Lecture 2114 Impedance Example How do we find V C ? First compute impedances for resistor and capacitor: Z R = 20k = 20k 0 Z C = 1/j (377 1 F) = 2.65k -90
Lecture 2115 Impedance Example 20k 0 k -90 10V 0 VCVC + -
Lecture 2116 Impedance Example Now use the voltage divider to find V C :
Lecture 2117 What happens when changes? 20k + - 1F1F10V 0 VCVC + - = 10 Find V C
Lecture 2118 Low Pass Filter: A Single Node-pair Circuit Find v(t) for =2 k 0.1 F 5mA 0 + - V