Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8EE201 Lecture 32 P. 1Phasor Impedance / AdmittanceDefinitions:Impedance AdmittanceZR(j) = R Resistors YR(j) = 1/RZL(j) = jL Inductors YL(j) = 1/(jL)ZC(j) = 1/(jC) Capacitors YC(j) = jCAdmittance = 1Impedance(mho) = 1/ ohm I =Y(j)VEE201 Lecture 32 P. 2For any 2 terminal device with input voltage and current (Vin and I in , respectively),V inI inZin (j)=I inV inYin (j)=Impedance and admittance are complex quantities that can be separated into real and imaginary parts.Z(j)Re[Z(j)] – ResistanceIm [Z(j)] – ReactanceY(j)Re[Y(j)] – ConductanceIm [Y(j)] – SusceptanceVi (j) = Vin = VinEE201 Lecture 32 P. 3Equivalent impedance for 2-terminal devices in series.Z1(j)Z2(jI inI 1I 2+V1-+V2-Vin+_From KVL: Vin =V1 + V2 V1 = Z1(j) I1 V2 = Z2(j) I2 Iin = I1 = I2Vin = (Z1 (j)+ Z2 (j)) IinZin (j) = Z1 (j) + Z2(j) Impedances add in series.Voltage division:Zi (j) Zi (j) Z1 (j)+ Z2 (j) Zeq(j) Z in(j)EE201 Lecture 32 P. 4 Y2(j)Equivalent admittance for 2 terminal devices in parallel.Y1(j)I in+V in-Yin(j)+V 1-+V 2-I 1I 2Vin = V1 = V2Iin = I1 + I2 Iin I1 + I2 I1 I2 Vin Vin V1 V2Yin(j) = = = + Yin(j) = Y1 (j)+ Y2(j)For parallel configuration of elements admittances add.EE201 Lecture 32 P. 5Current division formulaIi= IinIi= IinYi(j)Yeq(j)Yi(j) Yk(j)k•Impedances are manipulated in same way as resistances.•Admittances are manipulated in same way as conductances.EE201 Lecture 32 P. 6Example: Compute Zeq for the series circuit shown below. = 200 rad/sec4mH 1.2 2mFZ1(j) Z2(j) Z3(j)Z1(j) = jL = j(200*0.004) = j0.8 Z2(j) = R = 1.2 Z3(j) = 1/(jC) = -j/(c) = - j2.5 Zeq(j) = Z1 + Z2 + Z3Zeq(j) = (1.2 - j1.7) EE201 Lecture 32 P. 7Example: Find value of L to yield Zin(j) = (0.25 + j0.25) when = 4rad/secLStep 1: Find Z11 for parallel elements 1 1 1 (1/R)+ jC (1/0.5)+j2 2+j2 2-j2 80.5 0.5FZin(j)Z11 = = =Z11 =EE201 Lecture 32 P. 8Step 2: Add ZL and solve for LZin(j) = ZL(j) + Z11(j)Zin(j) = jL + 0.25 - j0.25Grouping j termsj(4L - 0.25) = j 0.25 L = 0.125
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