Question 3 of 7: First-Order RL Transient

Question 3: First-Order RL 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. V_s (in parallel with the distractor R₁) feeds R₂ in series to a node where R₃ hangs to ground, then through switch S to a node where R₄ hangs to ground (v₄ measured here), then through R₅ to a node where L and R₆ hang to ground in parallel. All elements share a common ground rail.

Find. V_R4 and I_L in steady state (switch closed); energy stored at t=0⁻; the post-switching time constant; a plot of I_L(t).

[Figure not reproduced: Figure 3 (redrawn). The switch isolates R1–R2–R3 from R4–R5–L–R6 once open. See the official exam paper.]

Approach. In DC steady state the inductor is a short circuit; solve the resulting resistive network for V_R4 and I_L with the switch closed. After the switch opens, find the Thévenin resistance the inductor sees (R₁,R₂,R₃,V_s are disconnected) to get τ, then write the standard first-order decay.

1. (a) Steady state, switch closed — L is a short. A short at L forces the node after R₅ to 0 V, so R₅ and R₃ and R₄ all effectively run from the switch-closed node (call it node X) to ground, in parallel (R₆ then carries 0 A since it, too, sits directly across the shorted node):
   $$R_3\parallel R_4\parallel R_5 = \left(\frac{1}{6}+\frac{1}{4}+\frac{1}{4}\right)^{-1} = 1.5\ \Omega$$$$V_X = V_s\cdot\frac{R_3\parallel R_4\parallel R_5}{R_2+R_3\parallel R_4\parallel R_5} = 12\cdot\frac{1.5}{3+1.5} = 4.0\text{ V}$$
2. Voltage across R₄. R₄ runs directly from node X to ground, so:
   $$\boxed{V_{R4} = V_X = 4.0\text{ V}}$$
3. Inductor current in steady state. All of R₅'s current flows into L (R₆ carries none, since it is shorted to ground alongside L):
   $$I_L(0^-) = \frac{V_X}{R_5} = \frac{4.0}{4} = \boxed{1.0\text{ A}}$$
4. (b) Energy stored at t=0⁻.
   $$W_L = \tfrac{1}{2}LI_L^2 = \tfrac{1}{2}(0.020)(1.0)^2 = \boxed{10.0\text{ mJ}}$$
5. (c) Time constant, switch open. Opening S disconnects the left-hand R₁–R₂–R₃–V_s branch entirely. The inductor now discharges through R₆ in parallel with the series combination (R₅+R₄):
   $$R_{eq} = R_6\parallel(R_5+R_4) = 8\parallel(4+4) = \frac{8\times 8}{16} = 4.0\ \Omega$$$$\boxed{\tau = \frac{L}{R_{eq}} = \frac{0.020}{4.0} = 5.0\text{ ms}}$$
6. (d) I_L(t). For t<0 the circuit is at steady state, I_L=1.0 A (constant); for t≥0 it decays exponentially toward zero (no source remains in the isolated R₄–R₅–L–R₆ loop):
   $$I_L(t) = \begin{cases} 1.0\text{ A}, & t<0 \\ 1.0\,e^{-t/\tau}\text{ A}, & t\ge 0\end{cases}$$