0
$\begingroup$

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

enter image description here

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – The Photon Aug 3 at 18:04
0
$\begingroup$

Your solution looks good to me, you are unlikely to find a formal solution much shorter than yours. It may or may not make the solution slightly easier to use the equivalent circuit for two mutual inductors, involving an ideal transformer.

Mutual inductors equivalent circuit

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.