22-Mec-A5 Electrical and Electronics Engineering · December 2014
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 2014 — 07-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.
Reference texts. A. S. Sedra and K. C. Smith, Microelectronic Circuits, 8th ed. (BJT current mirrors, 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).
Front-page constants. Note [8] of the examination paper supplies μ0 = 4π × 10−7 H·m−1, π = 3.14159 and 1 hp = 746 W. All three are used exactly as printed; μ0 enters in Question 4 and the horsepower conversion in Question 3.
Check: data-table exponent. The Question 4 specification table reads 3.77 × 10−2 m and 7.54 × 10−2 m, and only those values give a physically sensible core (a few centimetres of flux path rather than several hundred metres) and reluctances that land on the round figures 1.5 × 106 and 3.0 × 106 A/Wb. The negative exponents are used throughout.
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.
