Phasor Relationships for Circuit Elements (8.4); Impedance and Admittance (8.5)Dr. HolbertFebruary 1, 2006

ECE201 Lect-5

Phasor Relationships for Circuit Elements Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor.A complex exponential is the mathematical tool needed to obtain this relationship.

ECE201 Lect-5

I-V Relationship for a ResistorSuppose that i(t) is a sinusoid:i(t) = IM ej(wt+q)Find v(t)Rv(t)+i(t)

ECE201 Lect-5

Computing the Voltage

ECE201 Lect-5

Class ExampleLearning Extension E8.5

ECE201 Lect-5

I-V Relationship for a CapacitorSuppose that v(t) is a sinusoid:v(t) = VM ej(wt+q)Find i(t)

ECE201 Lect-5

Computing the Current

ECE201 Lect-5

Phasor RelationshipRepresent v(t) and i(t) as phasors:V = VM qI = jwC VThe derivative in the relationship between v(t) and i(t) becomes a multiplication by jw in the relationship between V and I.

ECE201 Lect-5

Examplev(t) = 120V cos(377t + 30) C = 2mF

What is V?What is I?What is i(t)?

ECE201 Lect-5

Class ExampleLearning Extension E8.7

ECE201 Lect-5

I-V Relationship for an InductorV = jwL ILv(t)+i(t)

ECE201 Lect-5

Examplei(t) = 1mA cos(2p 9.15107t + 30) L = 1mH

What is I?What is V?What is v(t)?

ECE201 Lect-5

Class ExampleLearning Extension E8.6

ECE201 Lect-5

Circuit Element Phasor Relations(ELI and ICE man)

ECE201 Lect-5

Element

V/I Relation

Phasor Relation

Phase

Capacitor

I = C dV/dt

I = j C V

= CV (90

I leads V

by 90

Inductor

V = L dI/dt

V = j L I

= LI (90

V leads I

by 90

Resistor

V = I R

V = R I

= R I (0

In-phase

ImpedanceAC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohms law:V = I ZZ is called impedance (units of ohms, W)

ECE201 Lect-5

ImpedanceResistor:V = I R

The impedance is ZR = R

Inductor:V = I jL

The impedance is ZL = jwL

ECE201 Lect-5

ImpedanceCapacitor:

The impedance is ZC = 1/jwC

ECE201 Lect-5

Some Thoughts on ImpedanceImpedance depends on the frequency, w=2fImpedance 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.

ECE201 Lect-5

Impedance Example:Single Loop Circuitw = 377Find VC20kW+1mF10V 0VC+

ECE201 Lect-5

Impedance ExampleHow do we find VC?First compute impedances for resistor and capacitor:ZR = 20kW= 20kW 0 ZC = 1/j (3771mF) = 2.65kW -90

ECE201 Lect-5

Impedance Example20kW 0+2.65kW -9010V 0VC+

ECE201 Lect-5

Impedance ExampleNow use the voltage divider to find VC:

ECE201 Lect-5

Low Pass Filter:A Single Node-pair CircuitFind v(t) for w=2p 30001kW0.1mF5mA 0+V

ECE201 Lect-5

Find Impedances1kW-j530W5mA 0+V

ECE201 Lect-5

Find the Equivalent Impedance5mA 0+VZeq

ECE201 Lect-5

Parallel Impedances

ECE201 Lect-5

Computing V

ECE201 Lect-5

Impedance Summary

ECE201 Lect-5

Element

Impedance

Capacitor

ZC = 1 / j(C = -1/(C ( 90(

Inductor

ZL = j(L = (L ( 90(

Resistor

ZR = R = R ( 0(

Class ExamplesLearning Extension E8.8Learning Extension E8.9

ECE201 Lect-5