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webadm | 投稿日時: 2008-8-25 1:51 |
Webmaster 登録日: 2004-11-7 居住地: 投稿: 3068 |
【56】相互誘導回路(その21) 今度も二つぐらいひねった問題。AB端のインピーダンスを求めるのと、CD端を短絡してもインピーダンスが変わらない条件を導けというもの。
以下の式が成り立つ (R1+jωL1+R3)*I1+jωM*I2=E (R2+jωL2+R4)*I2+jωM*I1=E Z*(I1+I2)=E これをI1,I2,Zに関する3元連立方程式として解くと (%i60) e1:(R1+%i*o*L1+R3)*I1+%i*o*M*I2=E; (%o60) I1*(R3+R1+%i*o*L1)+%i*o*I2*M=E (%i61) e2:(R2+%i*o*L2+R4)*I2+%i*o*M*I1=E; (%o61) I2*(R4+R2+%i*o*L2)+%i*o*I1*M=E (%i62) e3:Z*(I1+I2)=E; (%o62) (I2+I1)*Z=E (%i63) solve([e1,e2,e3],[I1,I2,Z]); (%o63) [[I1=(E*R4+E*R2-%i*o*E*M+%i*o*E*L2)/((R3+R1+%i*o*L1)*R4+(R2+%i*o*L2)*R3+(R1+%i*o*L1)*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2),I2=( (E*R3^2+(2*E*R1-%i*o*E*M+2*%i*o*E*L1)*R3+E*R1^2+(2*%i*o*E*L1-%i*o*E*M)*R1+o^2*E*L1*M-o^2*E*L1^2)*R4^2+(E* R3^3+(2*E*R2+3*E*R1-3*%i*o*E*M+2*%i*o*E*L2+3*%i*o*E*L1)*R3^2+((4*E*R1-2*%i*o*E*M+4*%i*o*E*L1)*R2+3*E* R1^2+(-6*%i*o*E*M+4*%i*o*E*L2+6*%i*o*E*L1)*R1-o^2*E*M^2+(2*o^2*E*L2+6*o^2*E*L1)*M-4*o^2*E*L1*L2-3*o^2*E*L1^2)* R3+(2*E*R1^2+(4*%i*o*E*L1-2*%i*o*E*M)*R1+2*o^2*E*L1*M-2*o^2*E*L1^2)*R2+E*R1^3+ (-3*%i*o*E*M+2*%i*o*E*L2+3*%i*o*E*L1)*R1^2+(-o^2*E*M^2+(2*o^2*E*L2+6*o^2*E*L1)*M-4*o^2*E*L1*L2-3*o^2*E*L1^2)*R1 -%i*o^3*E*M^3-%i*o^3*E*L1*M^2+(2*%i*o^3*E*L1*L2+3*%i*o^3*E*L1^2)*M-2*%i*o^3*E*L1^2*L2-%i*o^3*E*L1^3)*R4+ (E*R2+%i*o*E*L2)*R3^3+(E*R2^2+(3*E*R1-3*%i*o*E*M+2*%i*o*E*L2+3*%i*o*E*L1)*R2+3*%i*o*E*L2*R1+o^2*E*M^2+3* o^2*E*L2*M-o^2*E*L2^2-3*o^2*E*L1*L2)*R3^2+((2*E*R1-%i*o*E*M+2*%i*o*E*L1)*R2^2+(3*E*R1^2+ (-6*%i*o*E*M+4*%i*o*E*L2+6*%i*o*E*L1)*R1-o^2*E*M^2+(2*o^2*E*L2+6*o^2*E*L1)*M-4*o^2*E*L1*L2-3*o^2*E*L1^2)*R2+ 3*%i*o*E*L2*R1^2+(2*o^2*E*M^2+6*o^2*E*L2*M-2*o^2*E*L2^2-6*o^2*E*L1*L2)*R1-3*%i*o^3*E*M^3+ (2*%i*o^3*E*L1-%i*o^3*E*L2)*M^2+(%i*o^3*E*L2^2+6*%i*o^3*E*L1*L2)*M-2*%i*o^3*E*L1*L2^2-3*%i*o^3*E*L1^2*L2)*R3+ (E*R1^2+(2*%i*o*E*L1-%i*o*E*M)*R1+o^2*E*L1*M-o^2*E*L1^2)*R2^2+(E*R1^3+(-3*%i*o*E*M+2*%i*o*E*L2+3*%i*o*E*L1)* R1^2+(-o^2*E*M^2+(2*o^2*E*L2+6*o^2*E*L1)*M-4*o^2*E*L1*L2-3*o^2*E*L1^2)*R1-%i*o^3*E*M^3-%i*o^3*E*L1*M^2+ (2*%i*o^3*E*L1*L2+3*%i*o^3*E*L1^2)*M-2*%i*o^3*E*L1^2*L2-%i*o^3*E*L1^3)*R2+%i*o*E*L2*R1^3+ (o^2*E*M^2+3*o^2*E*L2*M-o^2*E*L2^2-3*o^2*E*L1*L2)*R1^2+(-3*%i*o^3*E*M^3+(2*%i*o^3*E*L1-%i*o^3*E*L2)*M^2+ (%i*o^3*E*L2^2+6*%i*o^3*E*L1*L2)*M-2*%i*o^3*E*L1*L2^2-3*%i*o^3*E*L1^2*L2)*R1-2*o^4*E*M^4+(o^4*E*L2+3*o^4*E*L1)*M^3+ (o^4*E*L1*L2-o^4*E*L1^2)*M^2+(-o^4*E*L1*L2^2-3*o^4*E*L1^2*L2)*M+o^4*E*L1^2*L2^2+o^4*E*L1^3*L2)/( (R3^2+(2*R1+2*%i*o*L1)*R3+R1^2+2*%i*o*L1*R1-o^2*L1^2)*R4^3+(R3^3+ (3*R2+3*R1-2*%i*o*M+3*%i*o*L2+3*%i*o*L1)*R3^2+ ((6*R1+6*%i*o*L1)*R2+3*R1^2+(-4*%i*o*M+6*%i*o*L2+6*%i*o*L1)*R1+2*o^2*M^2+4*o^2*L1*M-6*o^2*L1*L2-3*o^2*L1^2)* R3+(3*R1^2+6*%i*o*L1*R1-3*o^2*L1^2)*R2+R1^3+(-2*%i*o*M+3*%i*o*L2+3*%i*o*L1)*R1^2+ (2*o^2*M^2+4*o^2*L1*M-6*o^2*L1*L2-3*o^2*L1^2)*R1+2*%i*o^3*L1*M^2+2*%i*o^3*L1^2*M-3*%i*o^3*L1^2*L2-%i*o^3*L1^3)*R4^2 +((2*R2+2*%i*o*L2)*R3^3+ (3*R2^2+(6*R1-4*%i*o*M+6*%i*o*L2+6*%i*o*L1)*R2+6*%i*o*L2*R1+2*o^2*M^2+4*o^2*L2*M-3*o^2*L2^2-6*o^2*L1*L2)*R3^2 +((6*R1+6*%i*o*L1)*R2^2+ (6*R1^2+(-8*%i*o*M+12*%i*o*L2+12*%i*o*L1)*R1+4*o^2*M^2+8*o^2*L1*M-12*o^2*L1*L2-6*o^2*L1^2)*R2+6*%i*o*L2*R1^2 +(4*o^2*M^2+8*o^2*L2*M-6*o^2*L2^2-12*o^2*L1*L2)*R1-4*%i*o^3*M^3+(4*%i*o^3*L2+4*%i*o^3*L1)*M^2+8*%i*o^3*L1*L2*M-6* %i*o^3*L1*L2^2-6*%i*o^3*L1^2*L2)*R3+(3*R1^2+6*%i*o*L1*R1-3*o^2*L1^2)*R2^2+(2*R1^3+ (-4*%i*o*M+6*%i*o*L2+6*%i*o*L1)*R1^2+(4*o^2*M^2+8*o^2*L1*M-12*o^2*L1*L2-6*o^2*L1^2)*R1+4*%i*o^3*L1*M^2+4*%i*o^3* L1^2*M-6*%i*o^3*L1^2*L2-2*%i*o^3*L1^3)*R2+2*%i*o*L2*R1^3+(2*o^2*M^2+4*o^2*L2*M-3*o^2*L2^2-6*o^2*L1*L2)*R1^2+ (-4*%i*o^3*M^3+(4*%i*o^3*L2+4*%i*o^3*L1)*M^2+8*%i*o^3*L1*L2*M-6*%i*o^3*L1*L2^2-6*%i*o^3*L1^2*L2)*R1+o^4*M^4+4*o^4* L1*M^3+(-4*o^4*L1*L2-2*o^4*L1^2)*M^2-4*o^4*L1^2*L2*M+3*o^4*L1^2*L2^2+2*o^4*L1^3*L2)*R4+ (R2^2+2*%i*o*L2*R2-o^2*L2^2)*R3^3+(R2^3+(3*R1-2*%i*o*M+3*%i*o*L2+3*%i*o*L1)*R2^2+ (6*%i*o*L2*R1+2*o^2*M^2+4*o^2*L2*M-3*o^2*L2^2-6*o^2*L1*L2)*R2-3*o^2*L2^2*R1+2*%i*o^3*L2*M^2+2*%i*o^3*L2^2*M-%i*o^3* L2^3-3*%i*o^3*L1*L2^2)*R3^2+((2*R1+2*%i*o*L1)*R2^3+ (3*R1^2+(-4*%i*o*M+6*%i*o*L2+6*%i*o*L1)*R1+2*o^2*M^2+4*o^2*L1*M-6*o^2*L1*L2-3*o^2*L1^2)*R2^2+(6*%i*o*L2*R1^2+ (4*o^2*M^2+8*o^2*L2*M-6*o^2*L2^2-12*o^2*L1*L2)*R1-4*%i*o^3*M^3+(4*%i*o^3*L2+4*%i*o^3*L1)*M^2+8*%i*o^3*L1*L2*M-6* %i*o^3*L1*L2^2-6*%i*o^3*L1^2*L2)*R2-3*o^2*L2^2*R1^2+(4*%i*o^3*L2*M^2+4*%i*o^3*L2^2*M-2*%i*o^3*L2^3-6*%i*o^3*L1*L2^2)* R1+o^4*M^4+4*o^4*L2*M^3+(-2*o^4*L2^2-4*o^4*L1*L2)*M^2-4*o^4*L1*L2^2*M+2*o^4*L1*L2^3+3*o^4*L1^2*L2^2)*R3+ (R1^2+2*%i*o*L1*R1-o^2*L1^2)*R2^3+(R1^3+(-2*%i*o*M+3*%i*o*L2+3*%i*o*L1)*R1^2+ (2*o^2*M^2+4*o^2*L1*M-6*o^2*L1*L2-3*o^2*L1^2)*R1+2*%i*o^3*L1*M^2+2*%i*o^3*L1^2*M-3*%i*o^3*L1^2*L2-%i*o^3*L1^3)*R2^2 +(2*%i*o*L2*R1^3+(2*o^2*M^2+4*o^2*L2*M-3*o^2*L2^2-6*o^2*L1*L2)*R1^2+ (-4*%i*o^3*M^3+(4*%i*o^3*L2+4*%i*o^3*L1)*M^2+8*%i*o^3*L1*L2*M-6*%i*o^3*L1*L2^2-6*%i*o^3*L1^2*L2)*R1+o^4*M^4+4*o^4* L1*M^3+(-4*o^4*L1*L2-2*o^4*L1^2)*M^2-4*o^4*L1^2*L2*M+3*o^4*L1^2*L2^2+2*o^4*L1^3*L2)*R2-o^2*L2^2*R1^3+ (2*%i*o^3*L2*M^2+2*%i*o^3*L2^2*M-%i*o^3*L2^3-3*%i*o^3*L1*L2^2)*R1^2+ (o^4*M^4+4*o^4*L2*M^3+(-2*o^4*L2^2-4*o^4*L1*L2)*M^2-4*o^4*L1*L2^2*M+2*o^4*L1*L2^3+3*o^4*L1^2*L2^2)*R1-2*%i*o^5*M^5+ (%i*o^5*L2+%i*o^5*L1)*M^4+4*%i*o^5*L1*L2*M^3+(-2*%i*o^5*L1*L2^2-2*%i*o^5*L1^2*L2)*M^2-2*%i*o^5*L1^2*L2^2*M+%i*o^5* L1^2*L2^3+%i*o^5*L1^3*L2^2),Z=((%i*R3+%i*R1-o*L1)*R4+(%i*R2-o*L2)*R3+(%i*R1-o*L1)*R2-o*L2*R1+%i*o^2*M^2-%i*o^2*L1*L2)/(%i*R4+%i*R3+%i*R2+%i*R1+2*o*M-o*L2-o*L1) ]] 異常にI2の式が長いのだが大丈夫かMaxima? Zに関して直交形式に整理すると (%i68) factor(Z=((%i*R3+%i*R1-o*L1)*R4+(%i*R2-o*L2)*R3+(%i*R1-o*L1)*R2-o*L2*R1+%i*o^2*M^2 -%i*o^2*L1*L2)/(%i*R4+%i*R3+%i*R2+%i*R1+2*o*M-o*L2-o*L1)); (%o68) Z=(%i*R3*R4+%i*R1*R4-o*L1*R4+%i*R2*R3-o*L2*R3+%i*R1*R2-o*L1*R2-o*L2*R1+%i*o^2*M^2-%i*o^2*L1*L2)/(%i*R4+%i*R3+%i*R2+%i*R1+2*o*M-o*L2-o*L1) (%i69) rectform(%); (%o69) Z= ((2*o*M-o*L2-o*L1)*(-o*L1*R4-o*L2*R3-o*L1*R2-o*L2*R1)-(-R4-R3-R2-R1)*(R3*R4+R1*R4+R2*R3+R1*R2+o^2*M^2-o^2*L1*L2))/((R4+R3+R2+R1)^2+(2*o*M-o*L2-o*L1)^2) +(%i*((2*o*M-o*L2-o*L1)*(R3*R4+R1*R4+R2*R3+R1*R2+o^2*M^2-o^2*L1*L2)+(-R4-R3-R2-R1)* (-o*L1*R4-o*L2*R3-o*L1*R2-o*L2*R1)))/((R4+R3+R2+R1)^2+(2*o*M-o*L2-o*L1)^2) Zについて直交形式に整理すると Z=((2*ω*M-ω*L2-ω*L1)*(-ω*L1*R4-ω*L2*R3-ω*L1*R2-ω*L2*R1)-(-R4-R3-R2-R1)*(R3*R4+R1*R4+R2*R3+R1*R2+ω^2*M^2-ω^2*L1*L2))/((R4+R3+R2+R1)^2+(2*ω*M-ω*L2-ω*L1)^2)+(j*((2*ω*M-ω*L2-ω*L1)*(R3*R4+R1*R4+R2*R3+R1*R2+ω^2*M^2-ω^2*L1*L2)+(-R4-R3-R2-R1)*(-ω*L1*R4-ω*L2*R3-ω*L1*R2-ω*L2*R1)))/((R4+R3+R2+R1)^2+(2*ω*M-ω*L2-ω*L1)^2) =(ω^2*(L1+L2-2*M)*(L1*(R2+R4)+L2*(R1+R3))+(R1+R2+R3+R4)*((R1+R3)*(R2+R4)+ω^2*(M^2-L1*L2))/((R1+R2+R3+R4)^2+ω^2*(L1+L2-2*M))+j*ω*((R1+R2+R3+R4)*(L1*R4+L2*R3+L1*R2+L2*R1)-(L1+L2-2*M)*((R1+R3)*(R2+R4)+ω^2*(M^2-L1*L2)))/(R1+R2+R3+R4)^2+ω^2*(L1+L2-2*M)^2) ということになる。 同じ式でI1,I2だけ解くと (%i131) solve([e1,e2],[I1,I2]); (%o131) [[I1=-(E*(-R4-R2-%i*o*L2)+%i*o*E*M)/(L1*(%i*o*(R4+R2)-o^2*L2)+R3*(R4+R2+%i*o*L2)+R1*(R4+R2+%i*o*L2)+o^2*M^2),I2= (E*R3+E*R1-%i*o*E*M+%i*o*E*L1)/(L1*(%i*o*(R4+R2)-o^2*L2)+R3*(R4+R2+%i*o*L2)+R1*(R4+R2+%i*o*L2)+o^2*M^2)]] (%i132) factor(%); (%o132) [[I1=(E*(R4+R2-%i*o*M+%i*o*L2))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2),I2= (E*(R3+R1-%i*o*M+%i*o*L1))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2)]] 今度は至ってシンプルな式が出る。 ここでAC間の電圧をEac,AD間の電圧をEadとするとCD間の電圧Ecdは Eac=(R1+jωL1)*I1+jωM*I2 Ead=R2*I2 Ecd=Eac-Ead =(R1+jωL1)*I1+jωM*I2-R2*I2 =(R1+jωL1)*I1+(jωM-R2)*I2 これに先のI1,I2の式を代入すると (%i149) subst((E*(R4+R2-%i*o*M+%i*o*L2))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2 +%i*o*L2*R1+o^2*M^2-o^2*L1*L2), I1, Ecd=I2*(%i*o*M-R2)+I1*(R1+%i*o*L1)); (%o149) Ecd=(E*(R1+%i*o*L1)*(R4+R2-%i*o*M+%i*o*L2))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2)+I2* (%i*o*M-R2) (%i150) subst((E*(R3+R1-%i*o*M+%i*o*L1))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2 +%i*o*L2*R1+o^2*M^2-o^2*L1*L2), I2, Ecd=(E*(R1+%i*o*L1)*(R4+R2-%i*o*M+%i*o*L2))/(R3*R4 +R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2) +I2*(%i*o*M-R2)); (%o150) Ecd=(E*(R1+%i*o*L1)*(R4+R2-%i*o*M+%i*o*L2))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2)+ (E*(%i*o*M-R2)*(R3+R1-%i*o*M+%i*o*L1))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2) (%i151) factor(%); (%o151) Ecd=(E*(R1*R4+%i*o*L1*R4-R2*R3+%i*o*M*R3+%i*o*M*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2))/(R3*R4+R1*R4+%i*o*L1*R4+R2*R3+%i*o*L2*R3+R1*R2+%i*o*L1*R2+%i*o*L2*R1+o^2*M^2-o^2*L1*L2) 従ってCD間の電圧Ecdが0となるためには分子が0となる条件 R1*R4+j*ω*L1*R4-R2*R3+j*ω*M*R3+j*ω*M*R2+j*ω*L2*R1+ω^2*M^2-ω^2*L1*L2=0 が成り立つ必要がある。このためには左辺の実数部と虚数部が共に0でなければならないので直交形式に直すと R1*R4-R2*R3+ω^2*M^2-ω^2*L1*L2+j*ω*(L1*R4+M*R3+M*R2+L2*R1) =R1*R4-R2*R3+ω^2*(M^2-L1*L2)+j*ω*(L1*R4+L2*R1+M*(R2+R3)) =0 従って実数部が R1*R4-R2*R3+ω^2*(M^2-L1*L2)=0 虚数部が L1*R4+L2*R1+M*(R2+R3)=0 でなければならない。 実数部の式より ∴R1*R4-R2*R3=ω^2*(L1*L2-M^2) 虚数部の式より ∴M=-(L1*R4+L2*R1)/(R2+R3) ということになる。 当初式をたてる際にはMは正だという前提だったが、実際には負でないといけないということになる。それと特定のωに限定されるので、CD間の電圧を測定すればその周波数の時に0となることがわかる。この回路はある種のブリッジ回路である。 ちなみに著者のZの式には誤植がある。虚数部の分母の式で+ω^2とあるべきところが-ω^2となってしまっている。 |
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