04-BS-4 · Undated paper

Question 3 of 7: First-Order RC Switching Transient

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: First-Order RC Switching 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 switch pole always ties node X (the $R_2$–$R_3$ junction) to either contact "0" or contact "1." Contact "0" is wired straight to the bottom rail, so position 0 short-circuits $R_3$; contact "1" is open, so position 1 puts $R_2$ and $R_3$ in series. $R_4$ and $C$ are always directly across the top and bottom rails, in parallel with whichever branch the switch selects and with the $I_s\Vert R_1$ branch.

Find. $v_C(0^-)$, the post-switching time constant $\tau$, $v_C(200\text{ s})$, and $v_C(t)$ for $0\le t\le100\text{ ms}$.

Approach. Because $R_4$ and $C$ sit directly across the same two rails as the source network, the capacitor always "sees" a Norton-equivalent current source $I_s$ in parallel with an equivalent resistance $R_{eq}=R_1\Vert R_{2,\text{eff}}\Vert R_4$, where $R_{2,\text{eff}}$ is $R_2$ alone (switch at 0, $R_3$ shorted) or $R_2+R_3$ (switch at 1). Solve the standard first-order charging equation $v_C(t)=v_C(\infty)+\big(v_C(0^+)-v_C(\infty)\big)e^{-t/\tau}$ with $\tau=R_{eq}C$ for each regime.

1. Part (a) — initial capacitor voltage (switch at 0, steady state). At steady state the capacitor draws no current, so $R_4$ carries whatever current is left over and $v_C(0^-)$ equals the node voltage across the parallel combination $R_1\Vert R_2\Vert R_4$ (position 0 shorts $R_3$, so it drops out entirely):$$R_{eq,0}=R_1\Vert R_2\Vert R_4=\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}\right)^{-1}\text{k}\Omega=\dfrac{12}{11}\text{ k}\Omega\approx1.0909\text{ k}\Omega$$$$v_C(0^-)=I_s R_{eq,0}=(7\text{ mA})(1.0909\text{ k}\Omega)=\boxed{7.636\text{ V}}$$
2. Part (b) — time constant after switching to position 1. Now $R_{2,\text{eff}}=R_2+R_3=8\text{ k}\Omega$: $$R_{eq,1}=R_1\Vert(R_2+R_3)\Vert R_4=\left(\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{6}\right)^{-1}\text{k}\Omega=\dfrac{24}{19}\text{ k}\Omega\approx1.2632\text{ k}\Omega$$$$\tau=R_{eq,1}C=(1263.2\ \Omega)(5\times10^{-6}\text{ F})=\boxed{6.32\text{ ms}}$$
3. Part (c) — voltage at $t=200\text{ s}$. The capacitor voltage is continuous through the switching instant, so $v_C(0^+)=v_C(0^-)=7.636\text{ V}$; the new final value is $$v_C(\infty)=I_s R_{eq,1}=(7\text{ mA})(1.2632\text{ k}\Omega)=8.842\text{ V}$$ so that $$v_C(t)=8.842+(7.636-8.842)e^{-t/6.32\text{ ms}}\text{ V.}$$ At $t=200\text{ s}$, $t/\tau\approx3.2\times10^{4}$, so $e^{-t/\tau}$ is utterly negligible and the capacitor has settled to its final value to every digit of precision that matters: $$v_C(200\text{ s})=\boxed{8.842\text{ V}}\ (=v_C(\infty),\text{ since }200\text{ s}\gg5\tau\approx31.6\text{ ms})$$
4. Part (d) — $v_C(t)$ for $0\le t\le100\text{ ms}$. Evaluating the same exponential over the requested window (Figure 3b) shows the transient is essentially complete by $t\approx5\tau\approx32\text{ ms}$, well inside the 100 ms plotting window, after which the trace is flat at $8.842\text{ V}$.