22-Mec-A5 Electrical and Electronics Engineering · December 2019
Question 7 of 8: First-Order RC Transient — Capacitor Voltage and Current
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. PEO / Engineers Canada National Examinations, December 2019 — 16-Mec-A5, Electrical & Electronics Engineering (Mechanical Engineering discipline). Three hours, closed book, one approved Casio or Sharp calculator. Eight questions; any five constitute a complete paper and each question is of equal value, so each counted question carries 20 marks. All eight questions are solved below, because this set is a study resource rather than an exam script. The front page prints the constants $\pi = 3.14159$, $1\text{ hp} = 746\text{ W}$ and $\mu_0 = 4\pi\times10^{-7}\text{ H m}^{-1}$; all three are printed clearly on the front page and are used exactly as given.
Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT biasing and load lines, op-amp frequency response); M. M. Mano and M. D. Ciletti, Digital Design, 6th ed. (universal-gate synthesis, DeMorgan reduction); S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (dc machines, transformer magnetic circuits, induction-motor testing and slip); C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (phasor analysis, power factor, first-order transients); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (magnetic circuits and induced emf).
the gated-logic array, the transformer magnetic circuit, the op-amp band-pass Bode question, the RC transient and the parallel ac network as printed in December 2014; and the induction-motor slip / speed-torque question as printed in December 2016. The full working is transcribed in place below rather than cross-referenced, so this paper stands on its own as a study document.
Question 7: First-Order RC Transient — Capacitor Voltage and Current (20 marks)
The capacitor is initially uncharged, $v_C(0^-)=0$, since $S_1$ has isolated it from the network until $t=0$.
Find. [a] $v_C(t)$ for $t\ge 0$; [b] the current $i(t)$ delivered to $C_1$ for $t\ge 0$.
[Figure not reproduced: Figure 7 redrawn. With $S_1$ closed, the capacitor sees the Thévenin equivalent of the supply and the two resistors. See the official exam paper.]
Approach. Replace everything to the left of the capacitor by its Thévenin equivalent, which turns a two-resistor network into a single-loop first-order circuit; write and solve the resulting differential equation with the initial condition; then differentiate to obtain the current.
Find the Thévenin voltage. With the capacitor branch removed, $R_1$ and $R_2$ form a simple divider across the supply:
$$V_{th}=V_I\,\frac{R_2}{R_1+R_2}=10\times\frac{30\ \text{k}\Omega}{30\ \text{k}\Omega+30\ \text{k}\Omega}=10\times0.5=5.00\ \text{V}$$
Find the Thévenin resistance. Suppressing the ideal voltage source (replacing it by a short) puts $R_1$ in parallel with $R_2$:
$$R_{th}=R_1\parallel R_2=\frac{R_1R_2}{R_1+R_2}=\frac{(30)(30)}{60}\ \text{k}\Omega=15.0\ \text{k}\Omega$$
Form the time constant.
$$\tau=R_{th}C_1=\left(15\times10^{3}\right)\left(3\times10^{-6}\right)=45\times10^{-3}\ \text{s}$$
$$\boxed{\;\tau=45\ \text{ms}\;}$$
Write and solve the circuit equation. KVL around the single Thévenin loop, with $i=C_1\,dv_C/dt$, gives
$$V_{th}=i\,R_{th}+v_C=R_{th}C_1\frac{dv_C}{dt}+v_C \quad\Longrightarrow\quad \tau\frac{dv_C}{dt}+v_C=V_{th}$$
Separating variables and integrating with $v_C(0)=0$ yields the standard first-order step response:
$$\boxed{\;v_C(t)=V_{th}\left(1-e^{-t/\tau}\right)=5.00\left(1-e^{-t/0.045}\right)\ \text{V},\qquad t\ge 0\;}$$
The two limits are the sanity check: $v_C(0)=0$ as required by continuity of capacitor voltage, and $v_C(\infty)=5.00$ V, at which point no current flows in $R_1$ or the capacitor branch, so the node sits at the open-circuit divider voltage.
Differentiate to obtain the current. Since $i=C_1\,dv_C/dt$,
$$i(t)=C_1\frac{d}{dt}\left[V_{th}\left(1-e^{-t/\tau}\right)\right]=\frac{C_1V_{th}}{\tau}e^{-t/\tau}=\frac{V_{th}}{R_{th}}e^{-t/\tau}$$
$$\boxed{\;i(t)=\frac{V_{th}}{R_{th}}e^{-t/\tau}=333.3\,e^{-t/0.045}\ \mu\text{A},\qquad t\ge 0\;}$$
The initial value follows independently: at $t=0^{+}$ the capacitor voltage is still zero, so the whole Thévenin source appears across $R_{th}$ and $i(0^{+})=5.00/15\ \text{k}\Omega=333.3\ \mu$A, matching the expression exactly.
Identify the practical settling time. The exponential is within 1 % of its final value after about five time constants:
$$t_{\text{settle}}\approx5\tau=5\times45=225\ \text{ms},\qquad v_C(5\tau)=4.966\ \text{V}$$
At one time constant the capacitor has reached 63.2 % of its final value, $v_C(\tau)=3.161$ V.
Capacitor voltage rising toward $V_{th}=5$ V and charging current decaying from 333.3 µA, both governed by the same time constant $\tau=45$ ms. The current is plotted on a scaled axis for comparison.
Note the structure of the answer, which generalises well beyond this circuit: the voltage rises toward its final value and the current decays from its initial value, but both are governed by the identical time constant $\tau = R_{th}C_1$. The Thévenin reduction is what makes this immediately visible; attacking the original two-resistor network with node equations would produce the same answer after considerably more algebra, and would obscure the fact that only one resistance — the parallel combination — actually sets the speed of the transient.
