Question 3 of 6: Source-free LC circuit — initial conditions and damping
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) — nodal/mesh analysis, first- and second-order transients, AC steady-state phasors, complex power and power-factor correction, Thévenin’s theorem and maximum-power transfer, and Laplace-domain analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the source-free RLC circuit and s-domain transfer functions.
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 and a star–delta conversion table are supplied on the last two pages. Phasor results are quoted as magnitude∡angle with angles in degrees; sources marked rms are treated as rms and a common cosine time reference (ω = 377 rad/s) is used throughout Q2.
Given. $18\text{ V}$ DC source, $R_1=5\,\Omega$ (in series with the switch), $L=0.6\text{ H}$, $C=0.01\text{ F}$, $R_2=4\,\Omega$ (below the inductor to ground) and $R_3=2\,\Omega$ (below the capacitor to ground).
Find. Initial values, their first derivatives, the governing ODE for $i$, and the damping class.
[Figure not reproduced: Figure 3 — redrawn. In position A the $18\text{ V}$ source charges the network through $R_1$; moving to B disconnects the source, leaving $L,\ C,\ R_2,\ R_3$ in a single source-free loop. See the official exam paper.]
Approach. Use DC steady state (inductor short, capacitor open) to get the initial store; apply continuity of $i_L$ and $v_C$ at the switching instant; then write KVL round the source-free series loop and differentiate to obtain the second-order ODE.
Initial store, part (i). In position A for a long time the inductor is a short and the capacitor an open. The source current flows $18\text{ V}\to R_1\to L(\text{short})\to R_2$:$$i(0)=\frac{18}{R_1+R_2}=\frac{18}{5+4}=\boxed{\,2\text{ A}\,}$$ With the capacitor branch open, no current flows in $R_3$, so its lower plate sits at ground; the upper plate is the node between $L$ and $R_2$, at $V=i(0)\,R_2=2\times4$:$$V_C(0)=\boxed{\,8\text{ V}\,}$$
Continuity at $t=0$. Inductor current and capacitor voltage cannot jump, so $i(0^+)=2\text{ A}$ and $V_C(0^+)=8\text{ V}$ carry into the position-B loop.
Derivatives, part (ii). In position B the elements form one series loop; KVL (taking $i$ downward through $L$) gives $L\tfrac{di}{dt}=V_C-i(R_2+R_3)$:$$\frac{di}{dt}(0^+)=\frac{V_C(0^+)-i(0^+)(R_2+R_3)}{L}=\frac{8-2(6)}{0.6}=\boxed{\,-6.67\ \text{A/s}\,}$$ The same loop current discharges the capacitor, $i=-C\,\tfrac{dV_C}{dt}$, hence$$\frac{dV_C}{dt}(0^+)=-\frac{i(0^+)}{C}=-\frac{2}{0.01}=\boxed{\,-200\ \text{V/s}\,}$$
Second-order ODE, part (iii). KVL round the loop with $v_C=\frac{1}{C}\int i\,dt$:$$L\frac{di}{dt}+(R_2+R_3)\,i+\frac{1}{C}\int i\,dt=0.$$ Differentiating once removes the integral:$$\boxed{\,0.6\frac{d^2i}{dt^2}+6\frac{di}{dt}+100\,i=0\quad\Longleftrightarrow\quad \frac{d^2i}{dt^2}+10\frac{di}{dt}+166.7\,i=0\,}$$
Damping class, part (iv). The characteristic equation $s^2+10s+166.7=0$ has$$\alpha=\frac{R_2+R_3}{2L}=\frac{6}{1.2}=5\ \text{s}^{-1},\qquad \omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.6\times0.01}}=12.91\ \text{rad/s}.$$ Since $\alpha<\omega_0$ the discriminant is negative and the roots are complex, $s=-5\pm j11.90$:$$\boxed{\,i(t)\ \text{is UNDERDAMPED}\,}$$ with damped frequency $\omega_d=\sqrt{\omega_0^2-\alpha^2}=11.90\ \text{rad/s}$.
