Modeling — Question 5

Question 5 – Solution

Strategy. Invert the inductance matrix, not just its diagonal; the sum and difference currents see different inductances.

Step 1: Form the coupled voltage balances. Voltage drops give $2i_1'+i_2'=3-i_1$ and $i_1'+2i_2'=-i_2$. The inductance determinant is $4-1=3$ H$^2>0$, so $$\boxed{i_1'=2-\tfrac 23i_1+\tfrac 13i_2,\qquad i_2'=-1+\tfrac 13i_1-\tfrac 23i_2.}$$ Each right side has units A/s. Ignoring the mutual terms would predict $i_2\equiv 0$ and would fail the second voltage balance during the transient.

Step 2: Solve the sum and difference. Let $s=i_1+i_2$, $d=i_1-i_2$. Adding and subtracting the voltage equations gives $3s'=3-s$ and $d'=3-d$, with $s(0)=d(0)=0$. Thus $s=3(1-e^{-t/3})$, $d=3(1-e^{-t})$, and $$\boxed{i_1=3-\tfrac 32(e^{-t/3}+e^{-t}),\qquad i_2=\tfrac 32(e^{-t}-e^{-t/3}).}$$ They tend to $(3,0)$ A, the resistive steady state. Their initial slopes are $(2,-1)$ A/s, agreeing with the inverted model.

Step 3: Locate the induced-current extremum. For $t>0$, $e^{-t}<e^{-t/3}$, so $i_2<0$. Its derivative vanishes when $e^{-t/3}/3=e^{-t}$, at $t_*=(3/2)\log 3$ s. The derivative changes from negative to positive, giving $\boxed{i_{2,\min}=-1/\sqrt 3\text{ A}}$. The sign means actual current flows opposite its chosen reference direction; it is not a negative stored energy or a violation of the circuit model.

Step 4: Verify the power identity. Writing $L=\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$, $W'=i^TLi'=i^T(V-i)$ gives $\boxed{W'=3i_1-i_1^2-i_2^2}$ in watts. With $V=0$ instead, $W'=-i_1^2-i_2^2\le 0$. Since $L$ has positive eigenvalues $1,3$, $W>0$ for nonzero currents. The driven circuit can gain magnetic energy from its source; the source-free circuit can only dissipate it.