Switched RL Network — Steady State, Stored Energy, and Transient Decay: 04-BS-4, May 2017, Q3

Question 3: Switched RL Network — Steady State, Stored Energy, and Transient Decay (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 in parallel with R₁ (a distractor — the ideal source alone fixes that node) feeds R₂ into a node shunted by R₃, then through switch S into a node shunted by R₄ (v₄(t) measured here), then through R₅ into a node shared by the inductor L and R₆. S is closed for t<0 and opens at t=0.

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

Find. v₄ and I_L in closed-switch steady state; the energy stored in L just before opening; the open-switch time constant; and I_L(t) over −5 ms to 25 ms.

Figure 3: Switched RL network (S closed for t<0, opens at t=0)

Approach. In closed-switch DC steady state the inductor is a short, collapsing the right-hand node to 0 V and leaving a simple series-parallel resistive divider fed by V_s (R₁ never enters, since the ideal source fixes its node regardless of R₁). Once S opens, the source side is isolated entirely and L discharges through whatever resistance remains around it.

Closed-switch steady state (part a). The inductor is a short, so the R₅–R₄ node sees R₃, R₄, and R₅ all in parallel to ground (S closed ties them to one node): $$ R_{eq} = \left(\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\right)^{-1} = \left(\frac{1}{6}+\frac{1}{4}+\frac{1}{4}\right)^{-1} = 1.5\ \Omega. $$ Dividing V_s between R₂ and this parallel combination (V_s is fixed at that node regardless of R₁): $$ v_4 = V_s\cdot\frac{R_{eq}}{R_{eq}+R_2} = 12\cdot\frac{1.5}{1.5+3} = \boxed{4\text{ V}}. $$ The inductor carries whatever flows through R₅ (R₆ sees 0 V across the shorted-L node, so it carries no current): $$ I_L(0^-) = \frac{v_4}{R_5} = \frac{4}{4} = \boxed{1\text{ A}}. $$
Stored energy (part b). $$ W = \tfrac{1}{2}LI_L(0^-)^2 = \tfrac{1}{2}(0.020)(1)^2 = \boxed{10\text{ mJ}}. $$
Open-switch time constant (part c). With S open the source side is disconnected; L sees R₆ in parallel with the series combination R₅+R₄: $$ R_{eq,open} = R_6 \parallel (R_5+R_4) = 8\parallel(4+4) = 4\ \Omega, \qquad \tau = \frac{L}{R_{eq,open}} = \frac{0.020}{4} = \boxed{5\text{ ms}}. $$
I_L(t) plot (part d). For t<0, I_L=1 A (steady state). For t≥0 the inductor decays exponentially from its initial value: $$ I_L(t) = I_L(0)e^{-t/\tau} = e^{-t/0.005}\text{ A}, \qquad t\ge 0, $$ giving I_L(25 ms) = $e^{-5}\approx\boxed{6.7\text{ mA}}$.

I_L(t): 1 A flat for t<0, then exponential decay with τ=5 ms

Final results
Quantity	Value
v₄ (closed)	4 V
I_L(0⁻)	1 A
W stored	10 mJ
τ (open)	5 ms
I_L(25 ms)	6.7 mA

Topic: First-order RL circuit — DC steady state before switching, then source-free exponential decay.

Key relations: inductor short-circuit equivalent in DC steady state, $W=\tfrac12LI^2$, $\tau=L/R_{eq}$, $I_L(t)=I_L(0)e^{-t/\tau}$.

Why this works: steady state means $di_L/dt=0$, so the inductor's voltage is zero and it behaves as a plain wire; once the source is switched away, the inductor's stored current must decay through whatever resistive network remains, at a rate set purely by that network's Thevenin resistance as seen from the inductor's terminals.

Common pitfall: using the closed-switch equivalent resistance (1.5 Ω combination) for the decay time constant instead of re-deriving the open-switch network (R₆ ∥ (R₅+R₄)) — the two topologies are different circuits and must be solved separately.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 7 (first-order RL/RC circuits).

Question 3 of 7: Switched RL Network — Steady State, Stored Energy, and Transient Decay

Question 3: Switched RL Network — Steady State, Stored Energy, and Transient Decay (20 marks)