First-Order RC Transient — Capacitor Voltage and Current: 25-Nav-B6 Ocean Engineering and Offshore Structures, December 2017, Q7

Question 7: First-Order RC Transient — Capacitor Voltage and Current (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.

Given data — Question 7 (component list)
Component	Value	Position
$V_I$	10 V	dc supply
$R_1$	30 kΩ	series, between the supply and the node
$R_2$	30 kΩ	shunt, from the node to the return
$C_1$	3 µF	in series with $S_1$, from the node to the return

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.

Final results — Question 7
Quantity	Result
Thévenin voltage $V_{th}$	5.00 V
Thévenin resistance $R_{th}$	15.0 kΩ
Time constant $\tau=R_{th}C_1$	45 ms
[a] Capacitor voltage	$v_C(t)=5.00\left(1-e^{-t/0.045}\right)$ V
[b] Capacitor current	$i(t)=333.3\,e^{-t/0.045}$ µA
Initial current $i(0^{+})$	333.3 µA
$v_C$ at $t=\tau$	3.161 V (63.2 %)
Settling time $\approx 5\tau$	225 ms (to $v_C=4.966$ V)

Topic. Thévenin reduction and the first-order RC step response.

Key relations. $V_{th}=V_IR_2/(R_1+R_2)$; $R_{th}=R_1\parallel R_2$; $\tau=R_{th}C$; step response $x(t)=x(\infty)+\left[x(0^{+})-x(\infty)\right]e^{-t/\tau}$; $i=C\,dv/dt$; capacitor voltage cannot change instantaneously.

Why this works. Any linear network seen from two terminals is indistinguishable from a single source in series with a single resistance, so the capacitor cannot tell the difference between the original circuit and its Thévenin equivalent. That reduction collapses the problem to the one first-order differential equation whose solution is memorised, and the general step-response form quoted above then requires only three quantities: the initial value, the final value, and the time constant. Every first-order transient in circuits, thermal systems and fluid systems has this same three-parameter shape.

Common pitfall. Using $R_1$ alone as the charging resistance, which gives $\tau=90$ ms — twice the correct value — because it ignores the path the charging current can take through $R_2$. A second, subtler error is using $V_I=10$ V as the final capacitor voltage; $R_2$ remains connected and holds the node at the divider voltage of 5 V forever, not merely during the transient.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., Ch. 4 (Thévenin's theorem) and Ch. 7 (first-order transients).

Question 7 of 8: First-Order RC Transient — Capacitor Voltage and Current

Question 7: First-Order RC Transient — Capacitor Voltage and Current (20 marks)