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

circuit

So far I have two equations:

eq1

eq2

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.

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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:

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

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  • \$\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

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