Given. $V_s$ in parallel with $R_1$ feeds $R_2$ in series to the switch; after the switch, $R_3\parallel R_4$ (with $v_4(t)$ measured across $R_4$) is followed by $R_5$ in series, reaching a second parallel pair $L\parallel R_6$. The switch opens at $t=0$, disconnecting the source side from the $R_3$-$R_4$-$R_5$-$L$-$R_6$ group.
Given data
Quantity
Value
$R_1,R_2,R_3$
$3,\ 3,\ 6\ \Omega$
$R_4,R_5,R_6$
$4,\ 4,\ 8\ \Omega$
$L$
$20\text{ mH}$
$V_s$
$12\text{ V}$
Find. $v_4$ and $I_L$ in steady state (switch closed); the inductor's stored energy; the post-switch time constant; and a plot of $I_L(t)$.
[Figure not reproduced: Figure 3: RL transient network for Question 3 -- switch S opens at t=0, isolating R3-R4-R5-L-R6 from the source. See the official exam paper.]
Approach. In DC steady state an inductor behaves as a short circuit, so reduce the closed-switch network to series/parallel resistors first; after the switch opens, find the Thevenin resistance the inductor sees and use the standard first-order decay $i_L(t)=i_L(0)e^{-t/\tau}$.
a) Steady state, switch closed. $R_1$ is directly across the ideal source, so it carries current but does not change the source node voltage ($V_s=12\text{ V}$ at that node). With the inductor shorted, $R_6$ (in parallel with a short) carries no current, so the network reduces to $R_2$ feeding $(R_3\parallel R_4)\parallel R_5$: $$R_{34} = \dfrac{R_3R_4}{R_3+R_4} = \dfrac{(6)(4)}{10} = 2.4\,\Omega, \qquad R_{eq} = \dfrac{1}{\tfrac{1}{R_{34}}+\tfrac{1}{R_5}} = \dfrac{1}{\tfrac{1}{2.4}+\tfrac{1}{4}} = 1.5\,\Omega$$ $$V_B = V_s\cdot\dfrac{R_{eq}}{R_2+R_{eq}} = 12\cdot\dfrac{1.5}{3+1.5} = 4.0\text{ V} \;\Rightarrow\; \boxed{v_4 = V_B = 4.0\text{ V}}$$ since $R_3\parallel R_4$ spans the same two nodes as $v_4$. All of the $R_5$-branch current continues through the shorted inductor (since $R_6$, in parallel with the short, carries zero current): $$\boxed{I_L(0^-) = \dfrac{V_B}{R_5} = \dfrac{4.0}{4} = 1.0\text{ A}}$$
b) Stored energy. $$W_L = \tfrac12 L I_L(0^-)^2 = \tfrac12(0.020)(1.0)^2 = \boxed{0.010\text{ J} = 10\text{ mJ}}$$
c) Time constant, switch open. Opening the switch disconnects $R_1,R_2,V_s$; the inductor now sees $R_6$ in parallel with the series path $R_5+(R_3\parallel R_4)$ back to the same two nodes: $$R_{eq,open} = R_6 \parallel \big(R_5+R_{34}\big) = 8\parallel(4+2.4) = \dfrac{(8)(6.4)}{8+6.4} = 3.556\,\Omega$$ $$\tau = \dfrac{L}{R_{eq,open}} = \dfrac{0.020}{3.556} = \boxed{5.625\text{ ms}}$$
d) $I_L(t)$. Inductor current cannot jump, so $I_L(0^+)=I_L(0^-)=1.0\text{ A}$, and with no source remaining in the loop after $t=0$, the current decays exponentially toward zero: $$I_L(t) = \begin{cases} 1.0\text{ A}, & t\lt 0 \\ 1.0\,e^{-t/0.005625}\text{ A}, & t\ge 0\end{cases}$$ plotted below from $-5\text{ ms}$ to $25\text{ ms}$.
Figure 3b: I_L(t), -5 ms to 25 ms -- 1.000 A steady state for t<0, decaying with τ=5.625 ms after S opens at t=0
