First-order RC switching transient: 22-Elec-A1 Circuits, December 2013, Q2

Question 2: First-order RC switching transient [5 + 10 + 5]

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
Left source / series $R$ (pos. a)	20 V, 4 kΩ
Right source / series $R$ (pos. b)	25 V, 5 kΩ
Shunt resistance	$R_p=6$ kΩ
Capacitor	$C=1$ mF

Find. $V_c(0^{+})$, the full response $V_c(t)$, and $V_c(2\,\text{s})$.

Figure 2 — The capacitor sees the 20 V / 4 kΩ branch before switching and the 25 V / 5 kΩ branch after. The 6 kΩ is a permanent shunt across $C$.

Approach. A single capacitor gives a first-order response $V_c(t)=V_c(\infty)+[V_c(0^{+})-V_c(\infty)]e^{-t/\tau}$. Find the two steady states (cap open) and $\tau=R_{th}C$ with the Thévenin resistance the cap sees after switching.

Initial value (switch at a, steady state). With the capacitor fully charged it draws no current, so the 6 kΩ and 4 kΩ form a divider from 20 V: $$V_c(0^{-})=20\cdot\frac{6}{4+6}=\boxed{12\ \text{V}}.$$ Capacitor voltage is continuous, so $V_c(0^{+})=12\ \text{V}$.
Final value (switch at b, steady state). Now the divider is 6 kΩ with 5 kΩ from 25 V: $$V_c(\infty)=25\cdot\frac{6}{5+6}=\frac{150}{11}=13.64\ \text{V}.$$
Time constant. Looking back from the capacitor with the 25 V source dead, it sees $5\,\text{k}\Omega\parallel 6\,\text{k}\Omega$: $$R_{th}=\frac{5\cdot6}{5+6}\,\text{k}\Omega=\frac{30}{11}\,\text{k}\Omega=2.727\ \text{k}\Omega,\quad \tau=R_{th}C=2.727\ \text{s}.$$
Assemble the response. $$V_c(t)=13.64+(12-13.64)e^{-t/2.727}=\boxed{13.64-1.64\,e^{-t/2.727}\ \text{V}},\quad t\ge 0.$$
Evaluate at $t=2$ s. $e^{-2/2.727}=e^{-0.733}=0.480$, so $$V_c(2)=13.64-1.64(0.480)=\boxed{12.85\ \text{V}}.$$

Quantity	Result
$V_c(0^{+})$	$\boxed{12\ \text{V}}$
$V_c(\infty)$	$13.64\ \text{V}$
Time constant $\tau$	$2.727\ \text{s}$
$V_c(t)$	$13.64-1.64\,e^{-t/2.727}\ \text{V}$
$V_c(2\,\text{s})$	$\boxed{12.85\ \text{V}}$

Concept & Theory

Topic. First-order (single-capacitor) switching transient by the initial/final-value method.

Key relations. $x(t)=x(\infty)+[x(0^{+})-x(\infty)]e^{-t/\tau}$; $\tau=R_{th}C$; capacitor voltage is continuous ($V_c(0^{+})=V_c(0^{-})$).

Why this works. Any linear first-order network reduces, from the capacitor’s terminals, to a Thévenin source behind $R_{th}$. The solution of $R_{th}C\,\dot V_c+V_c=V_{oc}$ is a single decaying exponential from the pre-switch value toward the new open-circuit value. Only $C$ stores energy, so no discontinuity in $V_c$ is allowed at the instant of switching.

Common pitfall. Using the pre-switch resistance for $\tau$, or forgetting that the 6 kΩ shunt (not the source resistor alone) sets both the divider and the parallel $R_{th}$.

Source. Sadiku §7.2–7.4 (source-free and step response of RC circuits).

Question 2 of 6: First-order RC switching transient

Question 2: First-order RC switching transient [5 + 10 + 5]