Two inductively coupled loops (self-inductances L_1,L_2…
Two inductively coupled loops (self-inductances $L_1,L_2$, mutual inductance $M$, resistances $R_1,R_2$) are connected to a constant EMF $E$, with zero initial currents. Find $i_1(t)$ and $i_2(t)$: $$ L_1\frac{di_1}{dt} + M\frac{di_2}{dt} + R_1 i_1 = E,\qquad M\frac{di_1}{dt} + L_2\frac{di_2}{dt} + R_2 i_2 = 0 $$
Explanation of Method
Writing Kirchhoff's voltage law for both loops in operator form and eliminating $i_2$ by Cramer's rule reduces the problem to a single second-order linear ODE for $i_1$ with constant forcing; the physical coupling inequality $L_1L_2>M^2$ guarantees the characteristic roots are always real, distinct, and negative, so $i_2$ can be reconstructed mode-by-mode and both currents settle to a physically sensible steady state.
- Write Kirchhoff's equations for both loops in operator notation.
- Eliminate $i_2$ using Cramer's rule to get a single ODE for $i_1$.
- Show the characteristic roots are always real, distinct, and negative.
- Solve for $i_1$, then reconstruct $i_2$ mode-by-mode from the second equation.
- Apply the zero initial conditions to fix both constants.
Step 1 · Operator form
With $D=d/dt$, the two loop equations become
$$ (L_1D+R_1)i_1+MD\,i_2=E,\qquad MD\,i_1+(L_2D+R_2)i_2=0. $$
Step 2 · Eliminate $i_2$
Applying Cramer's rule to the operator system, the determinant $(L_1D+R_1)(L_2D+R_2)-M^2D^2$ acting on $i_1$ equals $(L_2D+R_2)$ acting on $E$, giving
$$ (L_1L_2-M^2)\,\ddot i_1+(L_1R_2+L_2R_1)\,\dot i_1+R_1R_2\,i_1=R_2E. $$
Writing $L=L_1L_2-M^2$, $a=(L_1R_2+L_2R_1)/L$, $b=R_1R_2/L$, $f=R_2E/L$, this is $\ddot i_1+a\dot i_1+bi_1=f$.
Step 3 · Particular (steady-state) solution
$$ i_1^{(p)}=\frac{f}{b}=\frac{E}{R_1}. $$
Step 4 · Nature of the characteristic roots
The discriminant of $r^2+ar+b=0$ works out to
$$ a^2-4b=\frac{(L_1R_2-L_2R_1)^2+4M^2R_1R_2}{L^2}>0, $$
a sum of a square and a positive term — always positive. So the roots $r_1,r_2$ are always real and distinct, and both negative (since $a>0$, $b>0$):
$$ r_{1,2}=\frac{-(L_1R_2+L_2R_1)\pm\sqrt{(L_1R_2-L_2R_1)^2+4M^2R_1R_2}}{2(L_1L_2-M^2)}. $$
Step 5 · General solution for $i_1$, then $i_2$
$i_1(t)=\dfrac{E}{R_1}+C_1e^{r_1t}+C_2e^{r_2t}$. Since $i_2$ must share the same exponential modes, writing $\mu_j=L_2r_j+R_2$, the second equation gives
$$ i_2(t)=-\frac{Mr_1C_1}{\mu_1}e^{r_1t}-\frac{Mr_2C_2}{\mu_2}e^{r_2t}. $$
Step 6 · Apply zero initial conditions
From $i_1(0)=0$ and $i_2(0)=0$, solving the resulting $2\times2$ linear system gives
$$ C_1=\frac{Er_2\mu_1}{R_1R_2(r_1-r_2)},\qquad C_2=-\frac{Er_1\mu_2}{R_1R_2(r_1-r_2)}. $$
Step 7 · Simplify $i_2$ using $r_1r_2=R_1R_2/L$
Substituting $C_1,C_2$ into $i_2(t)$, the $\mu_1,\mu_2$ factors cancel and, using $r_1r_2=b=R_1R_2/L$,
$$ i_2(t)=\frac{ME}{L_1L_2-M^2}\cdot\frac{e^{r_2t}-e^{r_1t}}{r_1-r_2}. $$
Step 8 · Verification and steady state
Both $i_1(0)=E/R_1+C_1+C_2=0$ and $i_2(0)=0$ check out directly from the constants above, and as $t\to\infty$ (both roots negative) the currents settle to $i_1\to E/R_1$, $i_2\to0$ — exactly the DC steady state predicted by setting all derivatives to zero in the original equations. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced ODE for $i_1$ | $(L_1L_2-M^2)\ddot i_1+(L_1R_2+L_2R_1)\dot i_1+R_1R_2i_1=R_2E$ |
| Discriminant | $(L_1R_2-L_2R_1)^2+4M^2R_1R_2 > 0$ always |
| Steady state | $i_1\to E/R_1,\ i_2\to0$ as $t\to\infty$ |
Additional Notes
- The physical coupling inequality $M^2<L_1L_2$ (mutual inductance cannot exceed the geometric mean of the self-inductances) is exactly what keeps $L=L_1L_2-M^2>0$ and the whole reduction well-defined.
- Because the discriminant is a sum of a perfect square and a strictly positive term, this circuit can never exhibit oscillatory (underdamped) behaviour — the response is always overdamped.
- The steady-state currents match simple DC intuition: the second loop, having no independent EMF, carries zero current once all transients die out.
Result
With $r_{1,2}$ the roots above and $\mu_j=L_2r_j+R_2$, the currents are: