Question 4 of 6: First-order RC switching transient
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Elec-A1 Circuits, May 2013. Closed-book, 3-hour paper; a Laplace-transform table is supplied. Six questions (Q1 has two parts); any five of equal value constitute a complete paper. Full worked solutions to all six questions are given below.
Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — nodal/mesh analysis, Thevenin and maximum-power transfer, first-order transients, AC power and power-factor correction, 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 models.
every network shown here was reconstructed element-by-element from the original drawing. Reference-node and source-polarity choices are stated with each solution so a reader can reproduce every sign.
Given. A 10 V source; a switch in parallel with a 10 Ω resistor; a 5 Ω series resistor; then a 25 Ω resistor in parallel with a 0.05 F capacitor, across which $v$ is measured. For $t<0$ the switch is closed (shorting the 10 Ω); at $t=0$ it opens.
Find. $v(0^+)$, $\dfrac{dv}{dt}(0^+)$, $v(\infty)$, the time constant $\tau$, and $v(t)$ for $t\ge 0$.
Figure 5 — RC switching network for Q4. Closing the switch shorts the 10 Ω; opening it at $t=0$ inserts the 10 Ω in series.
Approach. Use capacitor-voltage continuity for the initial value, DC steady states (capacitor open) for $v(0^-)$ and $v(\infty)$, and the source-free look-back resistance for $\tau$; assemble the single-time-constant response.
Initial value $v(0^+)$. For $t<0$ the closed switch shorts the 10 Ω and the capacitor is fully charged (open). The current $10/(5+25)$ flows through the 25 Ω, so $$v(0^-)=10\cdot\frac{25}{5+25}=\boxed{8.33\ \text{V}}=v(0^+)\ \text{(continuity)}.$$
Final value $v(\infty)$. With the switch open the 10 Ω is now in the path; at steady state the capacitor is again open: $$v(\infty)=10\cdot\frac{25}{10+5+25}=\boxed{6.25\ \text{V}}.$$
Time constant. Deactivate the 10 V source; the capacitor sees $25\parallel(10+5)$: $$R_{eq}=\frac{25\cdot 15}{40}=9.375\ \Omega,\qquad \tau=R_{eq}C=9.375(0.05)=\boxed{0.469\ \text{s}}.$$
Initial slope. The capacitor current at $t=0^+$ is $i_C=\dfrac{10-v(0^+)}{15}-\dfrac{v(0^+)}{25}=0.111-0.333=-0.222\ \text{A}$, so $$\frac{dv}{dt}(0^+)=\frac{i_C}{C}=\frac{-0.222}{0.05}=\boxed{-4.44\ \text{V/s}}.$$ (Equivalently $-(v(0^+)-v(\infty))/\tau$.)
Complete response. A single-time-constant circuit gives $$v(t)=v(\infty)+\bigl[v(0^+)-v(\infty)\bigr]e^{-t/\tau}=\boxed{6.25+2.08\,e^{-2.13\,t}\ \text{V}},\quad t\ge 0.$$
