まず複数の素子を直列に接続した場合の合成複素インピーダンスとアドミッタンスを求めてみよう

最初に抵抗の直列接続



Z=R1+R2
Y=1/(R1+R2)
|Z|=sqrt((R1+R2)^2)=R1+R2
φ=0

次ぎにインダクタンスの直列接続



Z=jωL1+jωL2=j(ωL1+ωL2)=jω(L1+L2)=jωL=jXL
Y=1/j(ωL1+ωL2)=-j/(ωL1+ωL2)=-j/ω(L1+L2)=-j/ωL=-j/XL
|Z|=sqrt((ωL1+ωL2)^2)=ωL1+ωL2=ω(L1+L2)=ωL=XL
φ=atan(XL/R)=atan(∞)=π/2
L=L1+L2
XL=ωL

今度はキャパシタンスの直列接続



Z=1/jωC1+1/jωC2=-(j/ω)*(1/C1+1/C2)=-(j/ω)*((C1+C2)/(C1*C2))=-j/ωC=-jXC
Y=1/(1/jωC1+1/jωC2)=jω/(1/C1+1/C2)=jω(C1*C2/(C1+C2))=jωC=j/XC
|Z|=sqrt((-1/ω)*(C1+C2)/(C1*C2))^2)=sqrt((1/ω^2)*(C1+C2)^2/(C1*C2)^2)=(1/ω)*(C1+C2)/(C1*C2)=1/ωC=XC
φ=atan(-XC/R)=atan(-∞)=-π/2
C=(C1*C2)/(C1+C2)
XC=1/ωC

次ぎはRL直列回路



Z=R+jωL
Y=1/(R+jωL)=R-jωL/(R^2+(ωL)^2)=R-jXL/(R^2+XL^2)
|Z|=sqrt(R^2+(ωL)^2)=sqrt(R^2+XL^2)
φ=atan(XL/R)=atan(ωL/R)

続いてRC直列回路



Z=R-j(1/ωC)=R-jXC
Y=1/(R-j(1/ωC))=(R+j(1/ωC))/((R+j(1/ωC))*(R-j(1/ωC)))
=(R+j(1/ωC))/(R^2+(1/ωC)^2)
=(R+j(1/ωC))/(((ωC)^2*R^2+1)/(ωC)^2)
=((ωC)^2*R+jωC)/((ωC)^2*R^2+1)
|Z|=sqrt(R^2+(-1/ωC)^2)=sqrt(R^2+1/(ωC)^2)
=sqrt(((ωC)^2*R^2+1)/(ωC)^2)
φ=atan(-XC/R)=-atan(1/ωCR)

今度はLC直列回路



Z=j(ωL-1/ωC)=j(XL-XC)
Y=1/j(ωL-1/ωC)=-j/(ωL-1/ωC)=-jωC/(ω^2*C*L-1)
|Z|=sqrt((ωL-1/ωC)^2)=|ωL-1/ωC|=|XL-XC|
φ=atan((ωL-1/ωC)/R)=atan(±∞)=±π/2

最後RLC直列回路



Z=R+j(ωL-1/ωC)=R+j(XL-XC)
Y=1/(R+j(ωL-1/ωC))=(R-j(ωL-1/ωC))/((R-j(ωL-1/ωC))*(R+j(ωL-1/ωC)))
=(R-j(ωL-1/ωC))/(R^2+(ωL-1/ωC)^2)
=(R-j(XL-XC))/(R^2+(XL-XC)^2)
|Z|=sqrt(R^2+(ωL-1/ωC)^2)=sqrt(R^2+(XL-XC)^2)
φ=atan((ωL-1/ωC)/R)=atan((XL-XC)/R)

並列回路はまた今度