First-Order RL Switching Transient: 04-BS-4, May 2015, Q3

Question 3: First-Order RL Switching Transient (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. A DC ladder with a switch S; the inductor L sits at the far node with R6 across it. For t < 0 (S closed, long time) the circuit is in DC steady state (L is a short). At t = 0 the switch opens, disconnecting the source side.

Given data
R1	R2	R3	R4	R5	R6	L	Vs
3 Ω	3 Ω	6 Ω	4 Ω	4 Ω	8 Ω	20 mH	12 V

Find. (a) v_R4 and I_L in the closed steady state; (b) stored inductor energy; (c) open-switch time constant; (d) the plot of I_L(t).

Figure 3 — With S closed the source drives the ladder; nodes 2 and 3 merge through the closed switch. When S opens, the source side (Vs, R1, R2, R3) is cut off and the inductor discharges into R6 and the R5–R4 path.

Approach. (i) Closed steady state: replace L by a short, solve the resistive node; (ii) energy from $\tfrac12 L I_L^2$; (iii) open-switch $\tau=L/R_{Th}$ with R_Th seen by the inductor; (iv) assemble $I_L(t)$ as a flat value then an exponential decay.

a) Closed steady state. With L shorted, the far node sits at 0 V, so R6 carries no current. The closed switch merges nodes 2 and 3 to a single voltage $V_m$; nodal balance gives $$V_m\!\left(\tfrac1{R_2}+\tfrac1{R_3}+\tfrac1{R_4}+\tfrac1{R_5}\right)=\tfrac{V_s}{R_2}\;\Rightarrow\;V_m = 4\text{ V}.$$ The voltage across R4 is this node voltage, $v_{R4}=V_m=\boxed{4\text{ V}}$. The R5 current all flows into the shorted inductor, so $I_L=\dfrac{V_m}{R_5}=\dfrac{4}{4}=1\text{ A}$.
b) Stored energy. $W=\tfrac12 L I_L^2 = \tfrac12(0.02)(1)^2$: $$\boxed{W = 0.01\text{ J} = 10\text{ mJ}}$$
c) Open-switch time constant. With S open the source side is disconnected. The inductor sees R6 in parallel with the series path R5 + R4: $$R_{Th}=R_6\,\|\,(R_5+R_4)=8\,\|\,(4+4)=8\,\|\,8 = 4\ \Omega,\qquad \tau=\frac{L}{R_{Th}}=\frac{0.02}{4}.$$ $$\boxed{\tau = 5\text{ ms}}$$
d) The current waveform. $I_L$ is continuous: it holds at 1 A for $t<0$ and decays with no source after the switch opens, so $$I_L(t)=\begin{cases}1\text{ A}, & t<0\\[2pt] 1\cdot e^{-t/5\text{ms}}\text{ A}, & t\ge 0\end{cases}$$ At $t=5$ ms it is $0.368$ A (one $\tau$); by $t=25$ ms (five $\tau$) it is essentially zero (0.007 A).

Plot of I_L(t): flat at 1 A until the switch opens at t = 0, then a single exponential decay with τ = 5 ms toward zero (no source connected after opening).

Question 3 — results
Quantity	Value
v_R4 (closed, steady state)	4 V
I_L (closed, steady state)	1 A
Stored energy W	10 mJ
Time constant (open) τ	5 ms
I_L(t), t ≥ 0	1·e^(−t/5ms) A

Topic. Natural response of a first-order RL circuit through a switching event.

Key relations. Steady-state inductor = short; continuity $I_L(0^+)=I_L(0^-)$; $\tau=L/R_{Th}$; natural response $I_L(t)=I_L(0^+)e^{-t/\tau}$; energy $W=\tfrac12 LI_L^2$.

Why this works. Inductor current cannot change instantaneously, so the pre-switch steady-state current is the initial condition for the decay. After the switch opens, the inductor is the only energy source and it discharges into whatever resistance it now sees — here R6 in parallel with the R5+R4 return path. Because the source is disconnected, the final value is zero and the response is a pure decaying exponential.

Common pitfall. Using the closed-switch resistance for τ, or assuming a non-zero final current. Once S opens, Vs, R1, R2 and R3 play no part; only the resistances still connected to the inductor set τ.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, ch. 7 (first-order transients).

Question 3 of 7: First-Order RL Switching Transient

Question 3: First-Order RL Switching Transient (20 marks)