Question 6 of 6: Laplace-domain analysis of a second-order network
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — mesh/supermesh and nodal analysis, first-order RC transients, AC steady-state phasors, complex power, 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 initial-condition source models in the s-domain.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table is supplied with the paper. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms.
Question 6: Laplace-domain analysis of a second-order network [8 + 4 + 6]
Find. The transformed network, $V_L(s)$, and the time-domain inductor voltage $V_L(t)$.
Figure 6 — after the switch closes, the $20\text{ V}$ source feeds the parallel $C$–$L$ pair through $4\,\Omega$. Because $C$ and $L$ share the node, $V_L(t)=V_c(t)=v_N(t)$.
Approach. Replace each element by its $s$-domain model carrying its initial condition, write one nodal equation for the shared node $V(s)$ (which is $V_L(s)$), then invert with a damped-sinusoid pair.
Transformed nodal equation. With node voltage $V(s)=V_L(s)$, the capacitor draws $sCV-Cv_c(0)$ and the inductor draws $V/(sL)+i_L(0)/s$; KCL at the node fed through $4\,\Omega$ from $20/s$:$$\frac{20/s-V}{4}=\Big(0.1sV-0.4\Big)+\Big(\frac{V}{2s}+\frac{1}{s}\Big).$$
Collect terms. Multiplying by $4s$ and grouping,$$16+1.6s=V\,(0.4s^2+s+2)\;\Rightarrow\;V_L(s)=\frac{4s+40}{s^{2}+2.5s+5}.$$
Complete the square. $s^2+2.5s+5=(s+1.25)^2+3.4375$, so $\alpha=1.25$, $\omega_d=\sqrt{3.4375}=1.854\text{ rad/s}$. Writing $4s+40=4(s+1.25)+35$:$$V_L(s)=\frac{4(s+1.25)}{(s+1.25)^2+\omega_d^2}+\frac{35}{(s+1.25)^2+\omega_d^2}.$$
Invert. Using the $e^{-\alpha t}\cos$ and $e^{-\alpha t}\sin$ pairs ($35/\omega_d=18.88$):$$\boxed{\,V_L(t)=e^{-1.25t}\big[\,4\cos(1.854t)+18.88\sin(1.854t)\,\big]\text{ V}\,}$$or equivalently $V_L(t)=19.30\,e^{-1.25t}\cos(1.854t-78.0^\circ)\text{ V}$.
Check endpoints. $V_L(0^+)=4\text{ V}$ (matches $V_c(0)$, since the node holds the capacitor voltage) and $V_L(\infty)=0$ (the inductor becomes a short), both consistent with the initial- and final-value theorems.
