Question 3 of 7

Question 3 (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 3. In position 1 the switch ties C1's node directly to the R2–R3 junction, so C1 charges from the Is/R1/R2 source through R3. In position 2 the switch instead ties C1's node, through R4, to the C2‖C3 node (both initially uncharged); the source side is disconnected entirely.

Find. (a) the time constant in position 1; (b) $v_{C1}$ at $t=1\text{ s}$; (c) the time constant in position 2; (d) $v_{C1}$ at $t=6\text{ s}$.

Approach. Convert the Norton pair (Is, R1) to a Thevenin source, add R2 in series, then combine with the shunt R3 to get the Thevenin source charging C1 in position 1 — giving $\tau_1$ and the exponential charging curve. At $t=5\text{ s}$ the switch disconnects the source and connects C1 (charged) through R4 to the uncharged C2‖C3 pair; this is a charge-sharing problem whose difference variable $v_{C1}-v_P$ decays with a single time constant built from R4 and the series combination of C1 with $C_{eq}=C_2+C_3$.

1. (a) Time constant, position 1. Converting the Norton pair to Thevenin: $V_{th1}=I_sR_1=0.2(3000)=600\text{ V}$ in series with $R_1$; adding $R_2$ in series gives $R_{th1}=R_1+R_2=6\text{ k}\Omega$ (Thevenin voltage unchanged by a series addition). Combining this with the shunt $R_3$ at C1's node: $$ V_{th,C1}=V_{th1}\frac{R_3}{R_{th1}+R_3}=600\left(\frac{6000}{12000}\right)=\boxed{300\text{ V}} $$ $$ R_{th,C1}=R_{th1}\Vert R_3=6000\Vert6000=3\text{ k}\Omega \;\Rightarrow\; \tau_1=R_{th,C1}C_1=3000(4\times10^{-6})=\boxed{0.012\text{ s}=12\text{ ms}} $$
2. (b) Voltage across C1 at t = 1 s. With the standard RC charging curve $v_{C1}(t)=V_{th,C1}\left(1-e^{-t/\tau_1}\right)$, and $t=1\text{ s}$ is about 83 time constants after the switch closes, so the capacitor is (to essentially exact precision) fully charged: $$ v_{C1}(1\text{ s})=300\left(1-e^{-1/0.012}\right)=\boxed{300.0\text{ V}} $$
3. (c) Time constant, position 2. By $t=5\text{ s}$ the capacitor is fully settled at $v_{C1}(5\text{ s})\approx300.0\text{ V}$ (again, $5\text{ s}\gg\tau_1$). The switch now disconnects the source branch and ties C1's node, through R4, to the C2‖C3 node; writing $C_1\,dv_{C1}/dt=-(v_{C1}-v_P)/R_4$ and $C_{eq}\,dv_P/dt=(v_{C1}-v_P)/R_4$ with $C_{eq}=C_2+C_3=18\;\mu\text{F}$, the difference $y=v_{C1}-v_P$ decays as $y(t)=y(0)e^{-t/\tau_2}$ with $$ \tau_2=R_4\left(\frac{C_1C_{eq}}{C_1+C_{eq}}\right)=10\left(\frac{4\times18}{22}\right)\mu\text{F}=\boxed{3.27\times10^{-5}\text{ s}=32.73\ \mu\text{s}} $$
4. (d) Voltage across C1 at t = 6 s. Charge is conserved between the two capacitor groups during the redistribution ($C_1v_{C1}+C_{eq}v_P=\text{const}=C_1\,v_{C1}(5\text{ s})$, since $v_P(5\text{ s})=0$). Combining this with the decaying difference $y(t')=v_{C1}(5\text{ s})\,e^{-t'/\tau_2}$ (where $t'=t-5\text{ s}$) gives $$ v_{C1}(t)=v_{C1}(5\text{ s})\,\frac{C_1+C_{eq}\,e^{-t'/\tau_2}}{C_1+C_{eq}} $$ At $t=6\text{ s}$, $t'=1\text{ s}\gg\tau_2$, so the exponential term has fully decayed and the two capacitor groups have reached a common final voltage set purely by the charge-conservation ratio: $$ v_{C1}(6\text{ s})=300.0\left(\frac{4}{4+18}\right)=\boxed{54.55\text{ V}} $$