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:
- $$\epsilon=I_1R+LdI_1/dt+MdI_2/dt $$
- $$I_2R+LdI_2/dt+MdI_1/dt=0 $$ 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!