First-order $RL$ transient (switch closes): 22-Elec-A1 Circuits, May 2017, Q3

Question 3: First-order $RL$ transient (switch closes) [8 + 12]

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. $18\text{ V}$ dc source, $3\,\Omega$ in series to node $A$; a $6\,\Omega$ from $A$ to ground; from $A$ a series $2\,\Omega+5\,\Omega$ then the inductor $L=0.5\text{ H}$ back to ground. The switch, when closed, connects $A$ directly to the node just before the inductor — i.e. it short-circuits the series $2\,\Omega+5\,\Omega$.

Find. $i_L(0^+)$ and the full expression $i_L(t)$, $t>0$.

[Figure not reproduced: Figure 3 — redrawn. The upper branch is the switch; closing it ties node $A$ to the inductor node, bypassing the $2\,\Omega+5\,\Omega$. The inductor current $i_L$ cannot change instantaneously, which sets $i_L(0^+)$. See the official exam paper.]

Approach. Use continuity of inductor current: find the dc steady state before the switch closes (inductor = short) for $i_L(0^+)$; after the switch closes find the new steady value and the Thévenin resistance seen by $L$, then write the standard first-order response.

(a) Pre-switch steady state ($t<0$). The inductor behaves as a short, so from $A$ two paths reach ground: the $6\,\Omega$ and the $2+5=7\,\Omega$ inductor branch, in parallel. $$6\,\|\,7=\frac{42}{13}=3.231\,\Omega,\quad I_{src}=\frac{18}{3+3.231}=2.889\text{ A},\quad V_A=2.889(3.231)=9.333\text{ V}.$$
(a) Inductor current, carried across the switch. $$i_L(0^-)=\frac{V_A}{7}=\frac{9.333}{7}=1.333\text{ A}\ \Rightarrow\ \boxed{i_L(0^+)=1.333\text{ A}}$$ (continuity of $i_L$).
(b) Final value ($t\to\infty$, switch closed). Now $A$ connects straight to the inductor; at steady state $L$ is again a short, pulling node $A$ to $0\text{ V}$, so the $6\,\Omega$ carries nothing and all of the $3\,\Omega$ current flows in $L$: $$i_L(\infty)=\frac{18}{3}=6\text{ A}.$$
(b) Thévenin resistance and time constant. Kill the source (short it); the inductor sees $3\,\Omega\,\|\,6\,\Omega$ (the shorted $2+5$ branch is now a dead loop at node $A$): $$R_{th}=3\,\|\,6=2\,\Omega,\qquad \tau=\frac{L}{R_{th}}=\frac{0.5}{2}=0.25\text{ s}.$$
(b) Assemble the first-order response. $i_L(t)=i_L(\infty)+[\,i_L(0^+)-i_L(\infty)\,]e^{-t/\tau}$: $$\boxed{i_L(t)=6-4.667\,e^{-4t}\ \text{A},\qquad t>0.}$$ Check: $i_L(0^+)=6-4.667=1.333\text{ A}$ and $i_L(\infty)=6\text{ A}$. ✓

Quantity	Value
$i_L(0^+)$	$1.333\text{ A}$
Final value $i_L(\infty)$	$6\text{ A}$
Time constant $\tau$	$0.25\text{ s}$
$i_L(t),\ t>0$	$6-4.667\,e^{-4t}\text{ A}$

Topic: Source-free / step response of a first-order $RL$ circuit; inductor-current continuity.

Key relations: $i_L(0^+)=i_L(0^-)$; $x(t)=x(\infty)+[x(0^+)-x(\infty)]e^{-t/\tau}$; $\tau=L/R_{th}$ with $R_{th}$ seen from the inductor terminals with sources killed.

Why this works: The inductor stores energy in its current, which cannot jump; that single conserved quantity carries the circuit through the switching instant and supplies the initial condition. For $t>0$ the circuit is linear with one energy-storage element, so every current and voltage relaxes exponentially toward its new dc value with the one time constant $L/R_{th}$. Only the endpoints (initial, final) and $R_{th}$ are needed — no differential equation must be integrated by hand.

Common pitfall: computing $R_{th}$ from the pre-switch circuit, or forgetting that closing the switch shorts the $2\,\Omega+5\,\Omega$ so they drop out of both the final value and $R_{th}$. The $6\,\Omega$ still matters for $R_{th}$ but carries zero current at $t\to\infty$.

Source: Sadiku, Fundamentals of Electric Circuits, §7.3–7.5 (source-free and step response of $RL$ circuits).

Question 3 of 6: First-order $RL$ transient (switch closes)

Question 3: First-order $RL$ transient (switch closes) [8 + 12]