I would like to set up differential equations (please, no phasors in answers) for the circuit below:


So far I have two equations:



where I assume perfect magnetic coupling (so I treat M as a known constant).

Initial conditions are that at time=0, both i1 and i2 are zero.

The system is underdetermined and I have to put one more equation. What equation is it?

My intention is to solve for i2. Again, please no suggestions with phasors. I am looking into a more general case where v1 isn't necessarily a perfect sinusoidal function.


You have two unknowns, \$i_1, i_2\$ and two equations, so your equations are solvable.

Solving these equations are generally done using the Laplace transform.

$$\left\{ \begin{align} i_1R_1 + L_1\frac{di_1}{dt} - M\frac{di_2}{dt} &= v_1(t)\\ -M\frac{di_1}{dt} + i_2R_2 + L_2\frac{di_2}{dt} &= 0 \end{align}\right.$$

Leads to

$$\left\{ \begin{align} (R_1+L_1s)&\cdot I_1 &- Ms\cdot I_2 & = V_1(s)\\ -Ms&\cdot I_1 &+ (R_2+L_2s)\cdot I_2 & = 0 \end{align}\right.$$

$$I_2 = \frac{\left|\begin{matrix} R_1+L_1s & V_1(s) \\ -Ms & 0 \end{matrix}\right| }{\left|\begin{matrix} R_1+L_1s & -Ms \\ -Ms & R_2+L_2s \end{matrix}\right| }=\frac{Ms\cdot V_1(s)}{(R_1+L_1s)(R_2+L_2s)-M^2s^2}$$

If you prefer differential equations you can always go back using:

$$\begin{align} \left[(R_1+L_1s)(R_2+L_2s)-M^2s^2\right]\cdot I_2(s) &= Ms\cdot V_1(s)\\ &\Downarrow\\ \left[R_1R_2 + (R_1L_2 + R_2L_1)s + (L_1L_2-M^2)s^2\right]\cdot I_2(s) &= Ms\cdot V_1(s)\\ &\Downarrow \mathcal{L}^{-1}\\ R_1R_2\cdot i_2(t)+(R_1L_2+R_2L_1)\frac{di_2}{dt}+(L_1L_2-M^2)\frac{d^2i_2}{dt^2} &= M\frac{dv_1}{dt} \end{align}$$

| improve this answer | |
  • \$\begingroup\$ Thank you! I just solved the last equation numerically for i2 using a sine wave for v1 and it gives the solution that I see in LTSpice. But this wouldn't work with a step function, would it? Correct me if I am wrong. Say, I want to solve this for source v1 that outputs a square wave with a very sharp rise time. \$\endgroup\$ – space bobcat Oct 22 '18 at 9:53
  • \$\begingroup\$ You can solve it for a step function, but I believe it will be easier using the Laplace transform. If you want to use the time-domain equation, you'll have to work with initial conditions (situation before \$t=0\$, then solve \$t>0\$). \$\endgroup\$ – Sven B Oct 22 '18 at 9:55

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.