Switched-Capacitor First-Order Transient: 04-BS-4, December 2013, Q4

Question 4: Switched-Capacitor First-Order Transient (20 marks)

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. The switched network of Figure 4. The switch rests at position 0 (capacitors uncharged) until $t=0$, moves to position 1 for $0\le t\lt5$ s, then to position 2 for $t\ge5$ s.

Given data
$I_s$	$R_1$	$R_2$	$R_3$	$R_4$	$C_1$	$C_2$	$C_3$
200 mA	3 kΩ	3 kΩ	6 kΩ	18 kΩ	10 µF	3 µF	6 µF

Find. (a) the position‑1 time constant; (b) $v_{C1}$ at $t=1$ s; (c) the waveform $i_1(t)$ over $-10$ ms to $200$ ms; (d) the total energy stored in all three capacitors at $t=6$ s.

Figure 4 — Switched-capacitor network for Question 4 (SP3T switch, positions 0/1/2).

Seen from node $X$ (after $R_2$), the current source and $R_1$ Nortonise to a Thévenin source $V_{oc}=I_sR_1$ behind $R_1$; adding $R_2$ in series and the shunt $R_3$ gives the driving voltage and resistance for whichever capacitor branch the switch selects.

Thévenin seen at node $X$. $I_sR_1=600$ V behind $R_1$; with $R_3$ dividing, the branch charges toward $V_{oc}=I_sR_1\dfrac{R_3}{R_1+R_2+R_3}=300$ V through $R_{Th}=R_3\Vert(R_1+R_2)=3 kΩ$.
(a) Position‑1 time constant. The selected branch is $C_1$, so $$\boxed{\tau_1=R_{Th}C_1=30\text{ ms}.}$$
(b) Capacitor voltage at $t=1$ s. Since $1\text{ s}\gg5\tau_1$, $C_1$ is fully charged: $v_{C1}(1\,\text{s})=V_{oc}\big(1-e^{-1/\tau_1}\big)\approx 300$ V.
(c) Current waveform $i_1(t)$. The charging current is $i_1(t)=\dfrac{V_{oc}}{R_{Th}}e^{-t/\tau_1}$: it is $0$ for $t\lt0$, jumps to $i_1(0^+)=100$ mA, then decays with $\tau_1=30$ ms (see plot).
Current i1(t): zero for t < 0, jumps to 100 mA at t = 0, then decays with τ = 30 ms.
(d) Stored energy at $t=6$ s. Moving to position 2 at $t=5$ s isolates $C_1$, which keeps its 300 V. The series pair $C_2,C_3$ ($C_{eq}=2 µF$) then charges to the same $V_{oc}$, splitting as $v_{C2}=200$ V and $v_{C3}=100$ V. Hence $$\boxed{W=\tfrac12C_1V_{oc}^2+\tfrac12C_2v_{C2}^2+\tfrac12C_3v_{C3}^2=0.45+0.06+0.03=0.54\text{ J}.}$$

Final results
Quantity	Value
$\tau_1$ (position 1)	30 ms
$v_{C1}(1\,\text{s})$	300 V
$i_1(0^+)$	100 mA
$v_{C2},v_{C3}$ (at $t=6$ s)	200, 100 V
Total stored energy	0.54 J

Topic: First-order RC transients with a multi-position switch.

Key relations: Thévenin reduction $V_{oc},R_{Th}$; time constant $\tau=R_{Th}C$; step response $v_C=V_{oc}(1-e^{-t/\tau})$, $i_C=(V_{oc}/R_{Th})e^{-t/\tau}$; series capacitance $C_{eq}=C_2C_3/(C_2+C_3)$; energy $\tfrac12CV^2$.

Why this works: Each switch position presents the capacitor(s) with a Thévenin source and resistance, so the response is a single decaying exponential. An isolated capacitor holds its charge indefinitely (ideal), which is why $C_1$ still stores energy long after it is switched out.

Common pitfall: Forgetting that series capacitors carry equal charge (so the larger capacitor holds the smaller voltage); and assuming $C_1$ discharges when it is merely disconnected.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. 6–7 (capacitors, first-order transients).

Question 4 of 7: Switched-Capacitor First-Order Transient

Question 4: Switched-Capacitor First-Order Transient (20 marks)