Question 2 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) — 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.
Find. The capacitor voltage immediately after switching, at steady state, its full time response, and its value at $t=\tau$.
Figure 2 — the capacitor node is common to $C$, the $10\,\Omega$ shunt and (through switch A) $V_{s2}$. In position A the node is clamped to $V_{s2}$; in position B that branch is open and $V_{s1}$ drives the node through $15\,\Omega$.
Approach. Use continuity of capacitor voltage for $V_c(0^+)$, a DC steady-state (capacitor open) for $V_c(\infty)$, and the standard single-time-constant form $V_c(t)=V_c(\infty)+[V_c(0^+)-V_c(\infty)]e^{-t/\tau}$.
Initial value ($t<0$, switch A). With $V_{s1}=0$ and the switch at A, the capacitor node is wired directly to the ideal $10\,\text{V}$ source $V_{s2}$, so$$V_c(0^-)=10\text{ V}\;\Rightarrow\;\boxed{\,V_c(0^+)=10\text{ V}\,}$$ (capacitor voltage cannot jump).
Final value ($t>0$, switch B). $V_{s2}$ is now disconnected and $V_{s1}=25\text{ V}$ drives the node through $15\,\Omega$ with the $10\,\Omega$ to ground; at steady state the capacitor is open, so a simple divider gives$$V_c(\infty)=25\cdot\frac{10}{15+10}=\boxed{\,10\text{ V}\,}$$
Time constant. For $t>0$ the Thévenin resistance seen by $C$ is $15\parallel 10=6\,\Omega$, so$$\tau=R_{th}C=(6)(0.5)=3\text{ s}.$$
Full response. Because the operating point is unchanged by the switching — the node sat at $10\text{ V}$ before and settles at $10\text{ V}$ after — the transient amplitude is zero:$$V_c(t)=10+(10-10)e^{-t/3}\;\Rightarrow\;\boxed{\,V_c(t)=10\text{ V},\ t\ge 0\,}$$
Value at one time constant. Since the response is flat, $V_c(\tau)=V_c(3\text{ s})=10\text{ V}$.
Check / insight. This network is deliberately balanced: the incoming divider $25\times 10/25 = 10\text{ V}$ equals the pre-switch clamp voltage $V_{s2}=10\text{ V}$. The correct engineering answer is therefore to recognise that no transient is excited — the capacitor is already at its final voltage — rather than to report a decaying exponential.
