First-order RC switching transient: 23-Mechatronics-A2 Circuits and Electronics, 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.

Given.

Quantity	Value
Source $E$	10 V dc
$R_1$	5 kΩ
$R_2$ (position A)	15 kΩ
$R_3$ (position B)	10 kΩ
$C_1$	0.5 mF

Find. $v_c(0^+)$, $\dfrac{dv_c}{dt}(0^+)$, $v_c(\infty)$, and $v_c(t)$ for $t\ge0$.

Figure-3: 10 V_dc in series with R₁=5 kΩ, feeding a switch (position A → R₂=15 kΩ, position B → R₃=10 kΩ) that shares its pole node with capacitor C₁=0.5 mF (voltage V_c) to the common return rail.

Approach. Use dc steady-state (capacitor open) before and after switching to fix the boundary values, continuity of capacitor voltage to link them at $t=0$, then solve the first-order equation for the post-switch network.

Steady state at position A ($t<0$). The capacitor is open at dc, so the loop current is $I=\dfrac{E}{R_1+R_2}=\dfrac{10}{20\,\text{k}}=0.5\ \text{mA}$, and $$v_c(0^-)=I\,R_2=0.5\,\text{mA}\times15\,\text{k}=7.5\ \text{V}.$$ Capacitor voltage cannot jump, so $$v_c(0^+)=\boxed{7.5\ \text{V}}.$$
Rate of change at $0^+$ (position B). Current from the source through $R_1$ into the switch node: $I_{R_1}=\dfrac{E-v_c(0^+)}{R_1}=\dfrac{10-7.5}{5\,\text{k}}=0.5\ \text{mA}$. Current out through $R_3$ to the rail: $I_{R_3}=\dfrac{v_c(0^+)}{R_3}=\dfrac{7.5}{10\,\text{k}}=0.75\ \text{mA}$. The capacitor current is $i_C=I_{R_1}-I_{R_3}=-0.25\ \text{mA}$, so $$\frac{dv_c}{dt}(0^+)=\frac{i_C}{C_1}=\frac{-0.25\times10^{-3}}{0.5\times10^{-3}}=\boxed{-0.5\ \text{V/s}}.$$
Final value ($t\to\infty$, position B). New steady state: $$v_c(\infty)=E\cdot\frac{R_3}{R_1+R_3}=10\times\frac{10\,\text{k}}{15\,\text{k}}=\boxed{6.667\ \text{V}}.$$
Time constant. With $E$ replaced by a short for the Thévenin view, the resistance seen by $C_1$ is $R_1\|R_3=\dfrac{5\,\text{k}\times10\,\text{k}}{15\,\text{k}}=3.333\ \text{k}\Omega$, so $$\tau=(R_1\|R_3)\,C_1=3.333\,\text{k}\times0.5\,\text{mF}=1.667\ \text{s}.$$
Assemble $v_c(t)$. $$v_c(t)=v_c(\infty)+\big[v_c(0^+)-v_c(\infty)\big]e^{-t/\tau}=\boxed{6.667+0.833\,e^{-0.6\,t}\ \text{V}},\quad t\ge0.$$ Check: $-\big[v_c(0^+)-v_c(\infty)\big]/\tau=-0.833/1.667=-0.5\ \text{V/s}$, matching Step 2.

Quantity	Result
$v_c(0^+)$	$\boxed{7.5\ \text{V}}$
$dv_c/dt(0^+)$	$\boxed{-0.5\ \text{V/s}}$
$v_c(\infty)$	$\boxed{6.667\ \text{V}}$
$v_c(t)$, $t\ge0$	$\boxed{6.667+0.833e^{-0.6t}\ \text{V}}$

Concept & Theory

Topic. First-order RC transient with a switched Thévenin source.

Key relations. Capacitor voltage continuity $v_c(0^+)=v_c(0^-)$; $i_C=C\,dv_c/dt$; general solution $v_c(t)=v_c(\infty)+[v_c(0^+)-v_c(\infty)]e^{-t/\tau}$, $\tau=R_{th}C$.

Why this works. A capacitor cannot pass an instantaneous impulse of current in a finite network, so its voltage is continuous across the switching instant; from that fixed starting point, the network relaxes exponentially toward the new dc steady state at a rate set by the Thévenin resistance the capacitor sees after switching.

Common pitfall. Using $R_2$ (the pre-switch resistor) instead of $R_3$ when computing $\tau$ or $v_c(\infty)$ after the switch has moved; forgetting the source is shorted (not removed) when finding $R_{th}$.

Source. Sadiku, Fundamentals of Electric Circuits §7.2–7.4 (source-free and step-forced RC circuits).

Question 3 of 11: First-order RC switching transient

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