First-order RL switching transient: 22-Elec-A1 Circuits, May 2014, Q2

Question 2: First-order RL switching transient [3 + 2 + 5 + 7 + 3]

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.

Element	Value
Source	20 V (DC)
Series resistor (source side)	4 Ω
Series resistor (to node B)	6 Ω
Inductor	$L=3$ H
Shunt resistor at node B	10 Ω
Switch	node A $\to$ ground, closes at $t=0$

Find. $i_L(0^{+})$, $i_L(\infty)$, $\tau$, and $i_L(t)$.

Figure 2 — RL network. Before $t=0$ the switch is open; at $t=0$ it closes, grounding node A and splitting the source branch off from the inductor mesh.

Approach. One inductor gives a first-order response $i_L(t)=i_L(\infty)+[\,i_L(0^{+})-i_L(\infty)\,]e^{-t/\tau}$. Find the two steady states (inductor $\to$ short) and $\tau=L/R_{th}$ from the resistance seen at the inductor terminals after the switch closes.

Initial value (switch open, steady state). At DC the inductor is a short, which shorts out the 10 Ω; the 20 V source drives the series $4+6$ Ω: $$i_L(0^{-})=\frac{20}{4+6}=\boxed{2\ \text{A}}.$$ Inductor current is continuous, so $i_L(0^{+})=2$ A.
Final value (switch closed, steady state). The closed switch grounds node A, and the inductor (short at DC) pulls node B to ground as well, so no voltage remains to drive it: $$i_L(\infty)=\boxed{0\ \text{A}}.$$
Time constant. Deactivate the 20 V source (short it); with the switch closed, node A is grounded, so from the inductor’s terminals the 6 Ω and 10 Ω appear in parallel: $$R_{th}=6\,\|\,10=\frac{6\cdot 10}{16}=3.75\ \Omega,\qquad \tau=\frac{L}{R_{th}}=\frac{3}{3.75}=\boxed{0.8\ \text{s}}.$$
Assemble the response. With $i_L(\infty)=0$: $$i_L(t)=2\,e^{-t/0.8}=\boxed{2\,e^{-1.25t}\ \text{A}},\qquad t\ge 0.$$
Sketch. A single decaying exponential from 2 A toward 0, reaching $2/e\approx 0.74$ A at one time constant ($t=0.8$ s) and essentially zero after $\sim 5\tau=4$ s.

Figure 2(b) — Sketch of $i_L(t)=2e^{-1.25t}$ A: a first-order decay with $\tau=0.8$ s.

Quantity	Result
$i_L(0^{+})$	$\boxed{2\ \text{A}}$
$i_L(\infty)$	$\boxed{0\ \text{A}}$
$R_{th}$ / time constant	$3.75\ \Omega$ / $\tau=0.8\ \text{s}$
$i_L(t)$	$\boxed{2\,e^{-1.25t}\ \text{A}},\ t\ge 0$

Concept & Theory

Topic. Source-free / step response of a first-order RL circuit by the initial/final-value method.

Key relations. $x(t)=x(\infty)+[x(0^{+})-x(\infty)]e^{-t/\tau}$; $\tau=L/R_{th}$; inductor current is continuous, $i_L(0^{+})=i_L(0^{-})$.

Why this works. After the switch settles, the inductor sees a Thévenin resistance $R_{th}$; the loop obeys $L\,\dot i_L+R_{th}i_L=$ const, whose solution is one exponential from the pre-switch current to the new steady value. Because only $L$ stores energy, $i_L$ cannot jump at $t=0$.

Common pitfall. Using the pre-switch resistance for $\tau$, or forgetting that closing the switch removes the source from the inductor’s mesh (node A is grounded), which is why $i_L(\infty)=0$.

Source. Sadiku §7.5–7.6 (source-free and step response of RL circuits).

Question 2 of 6: First-order RL switching transient

Question 2: First-order RL switching transient [3 + 2 + 5 + 7 + 3]