First-order RL transient — switch opening: 22-Elec-A1 Circuits, December 2014, Q3

Question 3: First-order RL transient — switch opening [4 + 6 + 10]

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 $20\text{ V}$ source feeds a $5\,\Omega$ resistor and the closed switch into node C; from C a $2\,\Omega$ resistor returns to ground, and a $6\text{ H}$ inductor in series with a $4\,\Omega$ resistor forms the second branch. $i(t)$ is the downward current in the inductor branch. The switch opens at $t=0$.

Find. $i(0)$, $\dfrac{di}{dt}(0^{+})$, and $i(t)$ for $t\ge0$.

[Figure not reproduced: Figure-3 (redrawn). For $t. See the official exam paper.]

Approach. Find the inductor current at $t=0^-$ from the DC steady state (inductor = short); it is continuous, so $i(0^{+})=i(0^{-})$. After the switch opens the circuit is a source-free RL with time constant $\tau=L/R_{\text{eq}}$, giving a decaying exponential.

Steady state before switching, $i(0)$ (i). At DC the $6\text{ H}$ inductor is a short, so node C sees $4\,\Omega\,\|\,2\,\Omega=\tfrac{4}{3}\,\Omega$. The node voltage is $$V_C=20\cdot\frac{4/3}{5+4/3}=\frac{80}{19}=4.211\ \text{V},$$ and the inductor-branch current (through the $4\,\Omega$) is $$i(0)=\frac{V_C}{4}=\frac{20}{19}=\boxed{1.053\ \text{A}}.$$ By inductor continuity $i(0^{+})=i(0^{-})=1.053\text{ A}$.
Equivalent resistance for $t\ge0$. With the switch open the source and its $5\,\Omega$ are disconnected; the inductor now discharges around the loop formed by its own $4\,\Omega$ in series with the $2\,\Omega$: $$R_{\text{eq}}=4+2=6\,\Omega,\qquad \tau=\frac{L}{R_{\text{eq}}}=\frac{6}{6}=1\ \text{s}.$$
Initial slope $\dfrac{di}{dt}(0^{+})$ (ii). Just after switching the inductor voltage is $v_L=-i(0^{+})R_{\text{eq}}=-(1.053)(6)=-6.32\text{ V}$ (it opposes the falling current), so $$\frac{di}{dt}(0^{+})=\frac{v_L}{L}=\frac{-6.32}{6}=\boxed{-1.053\ \text{A/s}}.$$
Transient response $i(t)$ (iii). A source-free first-order response decays from $i(0)$ to zero: $$i(t)=i(0)\,e^{-t/\tau}=\boxed{1.053\,e^{-t}\ \text{A}},\qquad t\ge0,$$ with $\tau=1\text{ s}$. (Consistency: $\left.\tfrac{di}{dt}\right|_{0^{+}}=-i(0)/\tau=-1.053$ A/s, matching (ii).)

Quantity	Result
$i(0)$	$\boxed{1.053\ \text{A}}\ (=20/19)$
$\tau$ ($t\ge0$)	$1\ \text{s}$ ($R_{\text{eq}}=6\,\Omega$)
$\dfrac{di}{dt}(0^{+})$	$\boxed{-1.053\ \text{A/s}}$
$i(t),\ t\ge0$	$\boxed{1.053\,e^{-t}\ \text{A}}$

Concept & Theory

Topic. Source-free first-order RL transient and continuity of inductor current.

Key relations. $i_L(0^{+})=i_L(0^{-})$; $\tau=L/R_{\text{eq}}$; $i(t)=i(0)e^{-t/\tau}$; $v_L=L\,di/dt$.

Why this works. An inductor stores energy in its current, which cannot change instantaneously, so the pre-switch steady current is the initial condition. After the switch opens the network has no source, and any first-order source-free circuit decays exponentially with the single time constant set by the Thévenin resistance the inductor sees.

Common pitfall. Using the pre-switch resistance ($5\,\Omega$ path) for $\tau$, or forgetting that opening the switch changes $R_{\text{eq}}$ to $4+2=6\,\Omega$. Also, the final current here is $0$ (no source remains), not a non-zero steady value.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, §7.3 (source-free RL circuit).

Question 3 of 6: First-order RL transient — switch opening

Question 3: First-order RL transient — switch opening [4 + 6 + 10]