Question 6 of 6: Laplace-domain solution of a switched RL circuit
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Elec-A1 Circuits, December 2014. Closed-book, 3-hour paper; a Laplace-transform table and Y–Δ formulas are supplied in the appendix. Six questions of equal value; any five constitute a complete paper. Full worked solutions to all six questions are given below so the set can be used for study regardless of which five a candidate chose.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction and the Wheatstone bridge, mesh/supermesh and nodal analysis, first-order transients, AC steady-state, Thévenin’s theorem and maximum-power transfer, and Laplace-domain circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the s-domain and initial-condition source models.
each network below is redrawn element-by-element from the original drawing. The Wheatstone bridge of Q1 is shown in the equivalent rectangular form (electrically identical to the diamond on the paper). Every reference node, mesh direction and source polarity used is stated with its solution so a reader can reproduce every sign. Phasor answers are quoted in magnitude∡angle form with angles in degrees; the source of Q5 is rms.
Question 6: Laplace-domain solution of a switched RL circuit [10 + 10]
Given. For $t<0$ the switch is at A, connecting a $5\text{ A}$ current source in a loop with the series $R=5\,\Omega$, $L=0.2\text{ H}$. At $t=0$ it moves to B, connecting a $10\text{ V}$ source (+ toward $R$) in a loop with the same $R$–$L$. $i(t)$ is the downward inductor current.
Find. The $s$-domain circuit for $t\ge0$ and $i(t)$.
[Figure not reproduced: Figure-6 (redrawn), shown at position B ($t\ge0$): the $10\text{ V}$ source drives the series $R=5\,\Omega$ and $L=0.2\text{ H}$. For $t. See the official exam paper.]
Approach. Get the initial current from the position-A steady state, then transform the position-B loop to the $s$-domain (source $\to10/s$, inductor $\to sL$ with an initial-condition source $L\,i(0)$), solve for $I(s)$, and invert by partial fractions.
Initial current. For $t<0$ the $5\text{ A}$ source is in series with the R–L loop, so in DC steady state the inductor simply carries the source current: $$i(0^{-})=5\ \text{A}=i(0^{+}).$$
Laplace-transformed circuit (i). For $t\ge0$ the $10\text{ V}$ step becomes $10/s$; the resistor stays $R=5$; the inductor becomes an impedance $sL=0.2s$ in series with an initial-condition voltage source $L\,i(0)=0.2\times5=1\text{ V}$ (polarity aiding the initial current). See the $s$-domain schematic below.
Laplace ($s$-domain) equivalent for $t\ge0$: source $10/s$, $R=5$, $sL=0.2s$, and the initial-condition source $L\,i(0)=1\text{ V}$.
Loop equation. KVL around the single loop gives $$\frac{10}{s}+L\,i(0)=\big(R+sL\big)I(s)\;\Rightarrow\;\frac{10}{s}+1=(5+0.2s)\,I(s).$$
Solve for $I(s)$. $$I(s)=\frac{10/s+1}{0.2s+5}=\frac{10+s}{0.2\,s(s+25)}=\frac{5(s+10)}{s(s+25)}.$$
Invert to time domain (ii). Using $1/s\!\to\!1$ and $1/(s+25)\!\to\!e^{-25t}$, $$i(t)=\boxed{2+3e^{-25t}\ \text{A}},\qquad t\ge0.$$ Checks: $i(0)=2+3=5\text{ A}$ (matches the initial condition) and $i(\infty)=2\text{ A}=10/5$ (the final DC current), with $\tau=L/R=0.04\text{ s}$ ($25=1/\tau$).
