# Mutual inductance circuit analysis

given the following circuit, I need to find $$I_1(t)$$ and $$I_2(t)$$ (forgive my microsoft paint skills ;) )

I've come up with a solution, and I wish to check with you if it is current or, better yet, if there are easier ways to solve this problem:

step 1: derive the differential equations:

1. $$\epsilon=I_1R+LdI_1/dt+MdI_2/dt$$
2. $$I_2R+LdI_2/dt+MdI_1/dt=0$$ step 2: I now use $$I'$$ and $$I''$$ as followed $$I' = I_1+I_2$$ $$I''=I_1-I_2$$ this leads me to solve the following equations: $$dI'/dt+R/(L+M) I'=\epsilon/(L+M)$$ $$dI''/dt+R/(L-M) I'=\epsilon/(L-M)$$ ill spare the solving part (solving a Non-homogonous linear eq' and an homogonous linear eq).

step 3: the solution is: $$I_1+I_2=I'=\epsilon/R (1-exp(-Rt/(L+M)))$$ $$I_1-I_2=I''=\epsilon/R (1-exp(-Rt/(L-M)))$$

therefore the final solution for $$I_1$$ and $$I_2$$ I received is $$I_1=\epsilon/2R \cdot (2-exp(-Rt/(L+M)-exp(-Rt/(L-M))$$ $$I_2=\epsilon/2R \cdot (exp(-Rt/(L+M)-exp(-Rt/(L-M))$$

would appreciate your response!

• Your diagram shows mutual inductance, not conductance. Can you correct it to make your intention clear? Also, you should include a dot on each inductor to show the sign of the coupling. – The Photon Aug 3 at 18:04 In general, you use $$L_{l1} = \left(1 - \frac{M}{\sqrt{L_1L_2}} \right)L_1$$ $$L_{l2} = \left(1 - \frac{M}{\sqrt{L_1L_2}} \right)L_2$$ $$L_{Mag} = \frac{M}{\sqrt{L_1L_2}}L_1$$ $$N_1 : N_2 = L_1^2:L_2^2.$$
The transformer is nice because it lets you refer any impedance on the secondary to the primary. In some cases, especially when $$M = \sqrt{L_1 L_2}$$, this equivalent circuit is often more convenient to work with.