AC i-V relationship for R, L, and C

Resistive Load Source vS(t) Asint

tR

A

R

vRi

tAtvv

R

SR

sin

sin

VR and iR are in phase

Phasor representation: vS(t) =Asint = Acos(t-90°)= A -90°=VS(j)

IS(jw) =(A / R)-90°

Impendence: complex number of resistance Z=VS(j)/ IS(j)=R

Generalized Ohm’s law VS(j) = Z IS(j)

Everything we learnt before applies for phasors with generalized ohm’s law

Capacitor Load

CjCj

jj

C

j

XjI

jVZ o

CC

CC

1

90

tCAdt

dqi

Cvq

tAv

CC

CC

C

cos

sin

90sin1

tC

AiC

ICE

VC(j)= A -90°

Notice the impedance of a capacitance decreases with increasing frequency

o

cC X

AjI 0

Inductive Load

tL

Adt

L

Ai

dt

diLv

tAv

L

LL

L

cossin

sin

90sin tL

AiL

Phasor: VL(j-90°IL(j)=(A/L) -180°

ZL=jL

ELI

Opposite to ZC, ZL increases with frequency

AC circuit analysis

• Effective impedance: example

• Procedure to solve a problem– Identify the sinusoidal and note the excitation frequency.

– Covert the source(s) to phasor form

– Represent each circuit element by its impedance

– Solve the resulting phasor circuit using previous learnt analysis tools

– Convert the (phasor form) answer to its time domain equivalent.

Ex. 4.16, p180

Ex. 4.21 P188

R1=100 R2=75 C= 1F, L=0.5 H, vS(t)=15cos(1500t) V.Determine i1(t) and i2(t).

Step 1: vS(t)=15cos(1500t), =1500 rad/s.Step 2: VS(j)=15 0Step 3: ZR1=R1, ZR2=R2, ZC=1/jC, ZL=jLStep 4: mesh equation