First-Order RL Transient Following a Switch Opening: 04-BS-4, May 2013, Q3

Question 3: First-Order RL Transient Following a Switch Opening (equal value)

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.

[Figure not reproduced: Figure 3 — redrawn: Vs ∥ R1 across the source terminals; R2 leads to a node shared by R3 (shunt) and, through switch S, R4 (shunt); R5 then leads to the L ∥ R6 pair. Opening S disconnects R3's node from R4's, isolating the R4–R5–L–R6 sub-networ. See the official exam paper.]

Given.

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

Find. $V_{R4}$ and $I_L$ in steady state (S closed); energy stored in L; the post-switching time constant; and a plot of $I_L(t)$ over $-5\ \text{ms}\le t\le 25\ \text{ms}$.

Approach. In DC steady state an inductor behaves as a short circuit, so reduce the closed-switch circuit to a resistive network first. Then, for $t\gt0$, find the Thévenin resistance the inductor sees (with the switch open) to get the time constant, and write the standard first-order decay $I_L(t)=I_L(0)e^{-t/\tau}$.

Steady state, switch closed — reduce the network. R1 sits directly across the ideal source Vs and does not affect the rest of the circuit. With S closed, R3 and R4 share the same two nodes (both shunt from the R2–R5 junction to ground), so $R_{34}=R_3\parallel R_4 = 6\parallel4 = 2.4\ \Omega$. The inductor is a short in DC steady state, so the R5 branch is just $R_5=4\ \Omega$ to ground. These two paths are themselves in parallel: $$R_{shunt} = R_{34}\parallel R_5 = 2.4\parallel 4 = 1.5\ \Omega$$ Total resistance seen by the source (excluding R1): $R_2+R_{shunt}=3+1.5=4.5\ \Omega$.
Junction voltage and $V_{R4}$. By the divider between $R_2$ and $R_{shunt}$: $$V_{R4} = V_s\cdot\dfrac{R_{shunt}}{R_2+R_{shunt}} = 12\cdot\dfrac{1.5}{4.5} = \boxed{4.0\ \text{V}}$$ (R4 shares this same node, so $V_{R4}$ equals this junction voltage directly.)
Inductor current in steady state. The R5–L branch carries $$I_L(0^-) = \dfrac{V_{R4}}{R_5} = \dfrac{4.0}{4} = \boxed{1.0\ \text{A}}$$ (cross-check via current divider: $I_{tot}=12/4.5=2.667$ A splits $2.667\times\frac{2.4}{6.4}=1.00$ A into the R5 branch ✓.)
Energy stored before switching. $$W_L = \tfrac12 L I_L^2 = \tfrac12(0.020)(1.0)^2 = \boxed{10\ \text{mJ}}$$
Time constant, switch open. Opening S disconnects the R2–R3 side entirely from R4's node, so the source and R1–R2–R3 play no further role. Looking into the inductor's terminals, R6 shunts directly to ground while R5+R4 forms the other path back to ground (R4 now shunting the now-isolated node to ground): $$R_{th} = R_6 \parallel (R_5+R_4) = 8\parallel(4+4) = 8\parallel 8 = 4\ \Omega$$ $$\tau = \dfrac{L}{R_{th}} = \dfrac{0.020}{4} = \boxed{5\ \text{ms}}$$
Plot. Current continuity keeps $I_L=1.0$ A right up to $t=0$; for $t\ge0$ it decays as $I_L(t)=1.0\,e^{-t/5\text{ms}}$, reaching $\approx1\%$ of its initial value by $t=25\text{ ms}=5\tau$.

IL(t): flat at 1.0 A for t < 0 (steady state), then exponential decay with τ = 5 ms once the switch opens.

Question 3 — final results
Quantity	Value
V(R4), switch closed	4.0 V
I_L, switch closed (steady state)	1.0 A
Energy stored in L	10 mJ
τ, switch open	5 ms
I_L(t), t ≥ 0	$1.0\,e^{-t/5\text{ms}}$ A

Topic: First-order RL transient response following a switching event.

Key relations: DC steady state: inductor = short circuit. Source-free response: $i(t)=i(0)e^{-t/\tau}$ with $\tau=L/R_{th}$, where $R_{th}$ is the resistance the inductor sees with all independent sources deactivated.

Why this works: An inductor's current cannot change instantaneously (it stores energy in its magnetic field), so the value just before switching becomes the initial condition for the transient just after. Once the switch removes the source's path to the L–R4–R5–R6 sub-network, that sub-network relaxes toward zero current with a single natural time constant set by the equivalent resistance it now sees.

Common pitfall: Forgetting that opening the switch changes which resistors are in the Thévenin-resistance calculation — R4 flips from being in parallel with R3 (switch closed) to being in series with R5 (switch open), so the two 4 Ω time-constant calculations are easy to conflate if not redrawn.

Source: Sadiku, Fundamentals of Electric Circuits, Ch. 7 (first-order circuits).

Question 3 of 7: First-Order RL Transient Following a Switch Opening

Question 3: First-Order RL Transient Following a Switch Opening (equal value)