Question 3 of 7: Switched RC Network — Time Constants and Stored Energy

Question 3: Switched RC Network — Time Constants and Stored Energy (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. Position 0 (t<0) leaves the switch open (no stored energy, as stated). Position 1 (0<t<5 s) connects the Is-R1-R2-R3 source network to node M, where C1 hangs to ground and R4 leads on to the series pair C2-C3. Position 2 (t≥5 s) disconnects the source network again, leaving C1-R4-C2-C3 as an isolated loop.

Find. The circuit time constant in position 1, V_C1 at t=1 s, the time constant in position 2, and the total capacitor energy at t=6 s.

Approach. Reduce position 1 to a single dominant time constant using the Thevenin resistance at node M and the merged capacitance, verify the reduction against the exact 2-pole system, evaluate the DC steady state (t=1 s is far beyond τ1), then check whether switching to position 2 launches any new transient at all before computing the stored energy.

1. Dominant time constant in position 1 (part a). With Is open-circuited, the Thevenin resistance seen by node M is R3 in parallel with (R1+R2): $$ R_{th,M} = R_3 \parallel (R_1+R_2) = 6\text{k} \parallel 6\text{k} = 3\text{ k}\Omega. $$ Because R4 « R_{th,M}, the series pair C2-C3 ($C_{eq,23}=1/(1/C_2+1/C_3)=2\ \mu\text{F}$) behaves as if it hangs directly in parallel with C1 for the slow response: $$ C_{tot} = C_1+C_{eq,23} = 10+2 = 12\ \mu\text{F}, \qquad \tau_1 = R_{th,M}\,C_{tot} = (3000)(12\times10^{-6}) = \boxed{36\text{ ms}}. $$ Solving the exact 2-state linear system (state variables $V_{C1}$ and the C2-C3 pair voltage) confirms two well-separated eigenvalues, $-27.78\ \text{s}^{-1}$ (i.e. $\tau=36.0$ ms) and $-6.0\times10^5\ \text{s}^{-1}$ ($\tau\approx1.67\ \mu\text{s}$) — the fast mode decays to nothing within microseconds, leaving 36 ms as the circuit's one meaningful time constant.
2. Voltage across C1 at t=1 s (part b). Since $t=1\text{ s} \approx 28\,\tau_1$, C1 has settled to its DC steady-state value: at steady state no current flows into any capacitor, so the R4 branch (and both capacitor branches) draw zero current and can be treated as open. Node A then sees Is split between R1 and the series path (R2+R3): $$ V_A = I_s\big[R_1 \parallel (R_2+R_3)\big] = (0.2)\!\left[\frac{(3000)(9000)}{12000}\right] = 450\text{ V}, $$ and the divider through R2-R3 gives $$ V_M = V_A\cdot\frac{R_3}{R_2+R_3} = 450\cdot\frac{6000}{9000} = \boxed{300\text{ V}}. $$ Numerically integrating the exact 2-state system to t=1 s gives $V_{C1}=300.0000000\text{ V}$, matching the steady-state value to 8 significant figures.
3. State at t=5 s (needed for parts c, d). Position 1 has run for 5 s ≫ τ1, so the network is fully settled: $V_{C1}(5\text{s})=300\text{ V}$ as above. At steady state R4 carries no current either (zero drop), so the series pair sees the same 300 V: by charge conservation ($Q_{C2}=Q_{C3}$ in series), $V_{C2}=300\cdot\dfrac{1/C_2}{1/C_2+1/C_3}=200\text{ V}$ and $V_{C3}=100\text{ V}$ (check: $200+100=300$ V ✓).
4. Time constant in position 2 (part c). Opening the switch to position 2 disconnects Is-R1-R2-R3 entirely, leaving C1, R4, C2, C3 as a single isolated series loop (one mesh, one state variable). Its equivalent series capacitance and time constant are $$ C_{eq,123} = \left(\frac1{C_1}+\frac1{C_2}+\frac1{C_3}\right)^{-1} = 1.667\ \mu\text{F}, \qquad \tau_2 = R_4\,C_{eq,123} = (1)(1.667\times10^{-6}) = \boxed{1.667\ \mu\text{s}}. $$
5. Energy at t=6 s (part d). The initial loop current the instant the switch reaches position 2 is $$ i(5\text{s}^+) = \frac{V_{C1}-V_{C2}-V_{C3}}{R_4} = \frac{300-200-100}{1} = 0\text{ A}: $$ the position-1 steady state already satisfies the position-2 loop's own equilibrium condition ($V_{C1}=V_{C2}+V_{C3}$), so switching to position 2 launches no transient at all — the three capacitor voltages are unchanged from t=5 s all the way to t=6 s (one second is $\approx6\times10^5\,\tau_2$ regardless, so even a nonzero initial current would have relaxed instantly). The total stored energy is therefore $$ E = \tfrac12C_1V_{C1}^2+\tfrac12C_2V_{C2}^2+\tfrac12C_3V_{C3}^2 = 0.450+0.060+0.030 = \boxed{0.54\text{ J}}. $$