First-Order RL Transient — Steady State, Stored Energy, Time Constant, Response Plot: 04-BS-4, May 2018, Q3

Question 3: First-Order RL Transient — Steady State, Stored Energy, Time Constant, Response Plot (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Figure 3: $V_s$ is wired directly in parallel with $R_1$ (a decoy pair); $R_2$ then carries the current onward to a node where $R_3$ shunts to ground and, through the closed switch $S$, continues to a node where $R_4$ shunts to ground ($v_4(t)$ measured here); $R_5$ then leads to the final node where $L$ and $R_6$ both shunt to ground.

Given data
$R_1$	$R_2$	$R_3$	$R_4$	$R_5$	$R_6$	$L$	$V_s$
3 Ω	3 Ω	6 Ω	4 Ω	4 Ω	8 Ω	20 mH	12 V

Find. $v_4$ and $I_L$ in steady state (S closed); the energy stored in $L$; the time constant after S opens; and a plot of $I_L(t)$ from $-5$ ms to $25$ ms.

Figure 3: RL transient network -- switch opens at t=0 (isolates R4-R5-L-R6 group)

Approach. $V_s\parallel R_1$ is a decoy (the ideal source pins that node regardless of $R_1$'s current); in steady state the inductor is a short, so with $S$ closed the network reduces to a single divider, and with $S$ open the right-hand group becomes an isolated decaying loop with no source at all.

a) Steady state with S closed. $L$ is a short at DC, so the far node is at $0$ V, which puts $R_6$ across $0$ V too (it carries no current in steady state). With $S$ closed, $R_3$, $R_4$, and $R_5$ (the last because it now dead-ends into the $L$-short) are all in parallel between the switch node and ground; that combination forms a divider with $R_2$ off the fixed $V_s$ node. $$R_3\parallel R_4\parallel R_5=\left(\frac{1}{6}+\frac{1}{4}+\frac{1}{4}\right)^{-1}=1.500\,\Omega\ ,\qquad V_{node}=V_s\cdot\frac{R_3\parallel R_4\parallel R_5}{R_2+R_3\parallel R_4\parallel R_5}=12\cdot\frac{1.5}{3+1.5}=\boxed{4.000\text{ V}}$$ This node voltage IS $v_4(t)$ (measured directly across $R_4$ to ground), so $v_4=\boxed{4.000\text{ V}}$. All of the current into $R_5$ continues through the shorted inductor (since $R_6$ carries none): $$I_L(0^-)=\frac{V_{node}}{R_5}=\frac{4.000}{4}=\boxed{1.000\text{ A}}$$
b) Stored energy. $$W_L=\tfrac{1}{2}LI_L(0^-)^2=\tfrac{1}{2}(0.020)(1.000)^2=\boxed{10.0\text{ mJ}}$$
c) Time constant after S opens. Opening $S$ disconnects $R_2$-$R_3$ (and the $V_s$-$R_1$ decoy) from the rest of the network entirely, isolating $R_4$, $R_5$, $L$, $R_6$ with no source. Seen from $L$'s terminals, $R_6$ is directly in parallel with the series pair $R_4+R_5$ (reached via the now-dangling $R_4$ node). $$R_{eq}=R_6\parallel(R_4+R_5)=8\parallel(4+4)=\frac{8\times8}{16}=\boxed{4.000\,\Omega}\ ,\qquad \tau=\frac{L}{R_{eq}}=\frac{0.020}{4.000}=\boxed{5.00\text{ ms}}$$
d) Plot of $I_L(t)$. For $t\lt0$ the switch has been closed a long time, so $I_L$ sits at its steady value; at $t=0$ the switch opens and $I_L$ decays exponentially toward zero with the time constant found above (no source remains to sustain it). $$I_L(t)=\begin{cases}1.000\text{ A}, & t\lt0\\[2pt] 1.000\,e^{-t/0.005}\text{ A}, & t\ge0\end{cases}$$

Figure 3b: I_L(t) -- 1 A steady state, decaying with τ=5 ms after S opens at t=0

Final results — Question 3
Quantity	Value
$v_4$ (steady state)	4.000 V
$I_L(0^-)$	1.000 A
$W_L$	10.0 mJ
$R_{eq}$ (S open)	4.000 Ω
$\tau$	5.00 ms

Topic: First-order RL transients — steady-state analysis and natural (source-free) decay.

Key relations: inductor = short at DC steady state; $W_L=\tfrac12LI^2$; $\tau=L/R_{eq}$ where $R_{eq}$ is the Thévenin resistance the inductor sees with all sources deactivated; $I_L(t)=I_L(0)e^{-t/\tau}$ for $t\ge0$.

Why this works: a switch opening at $t=0$ can completely sever a sourced sub-network from a source-free one; recognizing that the post-switch network has literally no driving source (rather than a reduced one) is what confirms the response is a pure exponential decay to zero, not to some new nonzero steady state.

Common pitfall: forgetting that $R_6$ carries zero current (and hence has zero effect on $v_4$/$I_L$) specifically because the inductor short pins its far node to $0$ V — that fact only holds while S is closed, and $R_6$ becomes essential (part of $R_{eq}$) the instant S opens.

Source: Alexander & Sadiku, Fundamentals of Electric Circuits, Ch. 7 (first-order RL/RC circuits).

Question 3 of 7: First-Order RL Transient — Steady State, Stored Energy, Time Constant, Response Plot

Question 3: First-Order RL Transient — Steady State, Stored Energy, Time Constant, Response Plot (20 marks)