Question 3 of 6: First-order RC switching transient
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) — series/parallel and Δ–Y reduction, mesh/nodal analysis with dependent sources, first-order RC transients, AC phasor analysis, Thévenin equivalents, maximum-power transfer and Laplace (s-domain) analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of super-node bookkeeping and the initial-condition source models used in Q6.
Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here as a study resource. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are written as magnitude $\angle$ angle with angles in degrees; a sine (peak) time reference is used in Q4 to match the source $100\sin(2\pi 60\,t)$, and the sources marked rms in Q5 are treated as rms.
Figure 3 — RC circuit; the switch connects the $2\,\Omega+5\,\Omega$ branch and capacitor at $t=0$.
Given. $15\text{ V}$ dc source; a $4\,\Omega$ series resistor to node $A$; a $1\,\Omega$ from $A$ to ground; a switch (closing at $t=0$) feeding $2\,\Omega+5\,\Omega$ in series to the capacitor $C=1\text{ mF}$ (the other plate grounded, $V_c$ positive at top).
Find. (a) $V_c(0^{+})$; (b) $V_c(t)$; (c) the sketch.
Approach. Use capacitor-voltage continuity for the initial value, the open-capacitor (dc) rule for the final value, and the Thévenin resistance seen by $C$ for the time constant, then assemble the standard first-order response.
Initial value, part (a). While the switch is open the capacitor branch is disconnected, so no charging path exists and $V_c(0^{-})=0$. Capacitor voltage cannot jump, so
$$\boxed{V_c(0^{+})=V_c(0^{-})=0\text{ V}.}$$
Final value. As $t\to\infty$ the capacitor is fully charged and behaves as an open circuit, so no current flows in the $2\,\Omega+5\,\Omega$ branch and $V_c(\infty)=V_A$. Node $A$ is then a simple divider of the $15\text{ V}$ source across $4\,\Omega$ and $1\,\Omega$:
$$V_c(\infty)=V_A=15\cdot\frac{1}{4+1}=3\text{ V}.$$
Thévenin resistance and time constant. Deactivating the $15\text{ V}$ source (short) and looking back from the capacitor terminals, the $2\,\Omega$ and $5\,\Omega$ are in series with the parallel combination $4\,\Omega\|1\,\Omega$:
$$R_{th}=5+2+\frac{4\cdot1}{4+1}=7.8\,\Omega,\qquad \tau=R_{th}C=7.8\cdot10^{-3}=7.8\text{ ms}.$$
Assemble the response, part (b). The first-order form $V_c(t)=V_c(\infty)+[V_c(0^{+})-V_c(\infty)]e^{-t/\tau}$ gives
$$\boxed{V_c(t)=3\left(1-e^{-t/7.8\text{ ms}}\right)\text{ V},\qquad t\ge0.}$$
Sketch, part (c). A rising exponential from $0\text{ V}$ toward the $3\text{ V}$ asymptote; it reaches $63.2\%$ ($1.90\text{ V}$) at one time constant $t=7.8\text{ ms}$ and is essentially settled ($\approx99\%$) by $5\tau\approx39\text{ ms}$.
Sketch — $V_c(t)=3\left(1-e^{-t/7.8\text{ ms}}\right)$ V: a rising exponential to the 3 V asymptote, 63.2% at one time constant.
