First-order RC switching transient: 22-Elec-A1 Circuits, December 2018, Q3

Question 3: First-order RC switching transient [4 + 6 + 2 + 8]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Figure‑3 — switch on A (long time) then to B at t=0. The pole node feeds R1, the selected leg (R2 or R3) and the top plate of C1; all lower ends share the ground rail.

Given. $E=10\text{ V}$ dc through $R_1=5\text{ k}\Omega$ to the switch pole; the pole also feeds the top plate of $C_1=0.5\text{ mF}$. Position-A connects the pole to $R_2=15\text{ k}\Omega$ (to ground); position-B connects it to $R_3=10\text{ k}\Omega$ (to ground). The lower plate of $C$ is grounded, $v_c$ positive at the top.

Quantity	Value	Quantity	Value
Source $E$	$10\text{ V}$	$C_1$	$0.5\text{ mF}$
$R_1$ (series)	$5\text{ k}\Omega$	$R_2$ (pos. A)	$15\text{ k}\Omega$
$R_3$ (pos. B)	$10\text{ k}\Omega$

Find. initial value $v_c(0^+)$, initial slope $\dfrac{dv_c}{dt}(0^+)$, final value $v_c(\infty)$, and the full response $v_c(t)$.

Approach. Get the pre-switch capacitor voltage from the position-A steady state (cap = open circuit), use continuity of $v_c$, find the post-switch Thévenin time constant, and assemble the standard first-order response.

Initial voltage (position-A steady state). In dc steady state the capacitor draws no current, so $R_1$ and $R_2$ form a divider: $$v_c(0^-)=E\frac{R_2}{R_1+R_2}=10\frac{15}{5+15}=7.5\text{ V}.$$ By continuity $$\boxed{v_c(0^+)=v_c(0^-)=7.5\text{ V}.}$$
Final voltage (position-B steady state). Now $R_1$ and $R_3$ divide: $$\boxed{v_c(\infty)=E\frac{R_3}{R_1+R_3}=10\frac{10}{5+10}=6.667\text{ V}.}$$
Time constant. For $t>0$, deactivating $E$ (short) leaves $R_1\|R_3$ across the capacitor: $$R_{th}=R_1\|R_3=\frac{5\cdot10}{15}=3.333\text{ k}\Omega,\qquad \tau=R_{th}C=3333(0.5\times10^{-3})=1.667\text{ s}.$$
Initial slope. From $\dfrac{dv_c}{dt}(0^+)=-\dfrac{v_c(0^+)-v_c(\infty)}{\tau}$: $$\boxed{\frac{dv_c}{dt}(0^+)=-\frac{7.5-6.667}{1.667}=-0.5\text{ V/s}.}$$ Check by node current at $t=0^+$: $i_C=\dfrac{E-v_c}{R_1}-\dfrac{v_c}{R_3}=\dfrac{10-7.5}{5\text{k}}-\dfrac{7.5}{10\text{k}}=-0.25\text{ mA}$, and $\dfrac{dv_c}{dt}=i_C/C=-0.5\text{ V/s}.$ ✓
Full response. The first-order form $v_c(t)=v_c(\infty)+[v_c(0^+)-v_c(\infty)]e^{-t/\tau}$ gives $$\boxed{v_c(t)=6.667+0.833\,e^{-0.6t}\text{ V},\quad t\ge0.}$$

Quantity	Result
$v_c(0^+)$	$7.5\text{ V}$
$\dfrac{dv_c}{dt}(0^+)$	$-0.5\text{ V/s}$
$v_c(\infty)$	$6.667\text{ V}$
Time constant $\tau$	$1.667\text{ s}$
$v_c(t)$	$6.667+0.833\,e^{-0.6t}\text{ V}$

Concept & Theory

Topic: first-order (single-capacitor) switched dc transient.

Key relations: $v_c(t)=v_c(\infty)+[v_c(0^+)-v_c(\infty)]e^{-t/\tau}$, $\tau=R_{th}C$, capacitor voltage continuous across the switching instant, dc steady state $\Rightarrow$ capacitor open.

Why this works: a single storage element with a resistive surrounding always relaxes exponentially toward its new equilibrium; only the initial value, final value and Thévenin resistance (seen by $C$ with sources killed) are needed. The pole node ties $R_1$, the selected leg and $C$ together, so the same node carries all three currents.

Common pitfall: using the pre-switch resistor ($R_2$) in the time constant — $\tau$ must come from the post-switch network ($R_1\|R_3$) — or forgetting that $v_c$ (not $i$) is the continuous variable.

Source: Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., §7.2–7.4 (source-free and step RC response).

Question 3 of 6: First-order RC switching transient

Question 3: First-order RC switching transient [4 + 6 + 2 + 8]