Question

Two RL circuits are connected by the mutual inductance L_12. Find the currents I_1 and I_2 flowing in the left and the right circuits if the left circuit is driven by a time dependent emf V = V_0 e^I omega t. Set both currents as I_1(t) = i_1 e^i(omega t - phi_1) and I_2(t_ = i_2 e^(I omega t - phi_2) and find the amplitude ratio i_1/i_2 and the phase difference phi_1 - phi_2

Answer 1

impedence of circuit 1

Z = R +i $\omega$ L₁₁

current I₁(t) = V/z = V_o exp(i $\omega$ t) / ( R + i $\omega$ L₁₁ )

                                 = V_o/ ( R² + $\omega$ ² L²₁₁ ) exp(i $\omega$ t) *( R - i $\omega$ L₁₁ )

                                 = V_o/ ( R² + $\omega$ ² L²₁₁ ) exp(i $\omega$ t) *exp(- i $\phi$ ₁   ) , where tan $\phi$ ₁ = $\omega$ L₁₁ /R ,

                                 = i₁ exp i( $\omega$ t - $\phi$ ₁ ) , i₁ = V_o/ ( R² + $\omega$ ² L²₁₁ )

mutual inductance = L₁₂

emf in circuit 2

V₂(t) =L₁₂ d I₁(t) /dt = L₁₂ i $\omega$ i₁exp i( $\omega$ t - $\phi$ ₁ )

current in circuit 2:

I₂(t) = V2/ (R+i L₂₂) = L₁₂ i $\omega$ i₁exp i( $\omega$ t - $\phi$ ₁ ) (R - i L₂₂) /(R²+ $\omega$ L²₂₂)

                               = L₁₂ $\omega$ i₁exp i( $\omega$ t - $\phi$ ₁ ) ( L₂₂ - i R ) /(R²+ L²₂₂)

                              = L₁₂ $\omega$ i₁exp i( $\omega$ t - $\phi$ ₁ ) exp(- i $\phi$ ₂ )/(R²+ $\omega$ ²L²₂₂) , where tan $\phi$ ₂ = R/ $\omega$ L₂₂

                             = i₂ exp i( $\omega$ t - $\phi$ ₁ - $\phi$ ₂ ) where i₂ = L₁₂ $\omega$ i₁/(R²+ $\omega$ ²L²₂₂)

i₁/i₂ = (R²+ $\omega$ ²L²₂₂)/L₁₂    

phase difference

( $\phi$ ₁ - $\phi$ ₂) = arcTan( $\omega$ L₁₁ /R ) - arcTan(R/ $\omega$ L₂₂ )