NivaarExam PrepOfficial exam papers ↗

22-Elec-A1 Circuits · May 2016

Question 3 of 6: First-order RC switching transient

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

Notes on this paper

Reference texts: C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits (7th ed., McGraw-Hill) — series/parallel reduction, Thévenin’s theorem with dependent sources, maximum-power transfer, first-order RC transients, AC phasor mesh analysis, complex power / power factor, and Laplace-domain (s-domain) circuit analysis; W. H. Hayt, J. E. Kemmerly & S. M. Durbin, Engineering Circuit Analysis (9th ed.) — companion treatment of the second-order source-driven RLC network and the initial-condition models used in Q6.

Closed-book national examination, 3 hours, six questions of equal value; any five constitute a complete paper. All six are solved here. A Laplace-transform table and a star–delta conversion table are supplied on the last two pages of the exam. Phasor magnitudes are quoted as magnitude $\angle$ angle with angles in degrees; a common cosine time reference is used, and the source marked rms in Q5 is treated as rms.

Question 3: First-order RC switching transient [5 + 10 + 5]

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. Left network: $20\text{ V}$ source, $40\,\Omega$ series to node P, $60\,\Omega$ from P to ground; right network: $50\,\Omega$ in series with a $25\text{ V}$ source. The capacitor $C=0.1\text{ mF}$ connects the switch pole to ground. For $t<0$ the pole is at a (left network); for $t\ge0$ it is at b (right network).

Find. The capacitor voltage immediately after switching, its full time response, and its value at $t=2\text{ s}$.

[Figure not reproduced: Figure 3 — redrawn at $t=0^+$ (pole thrown to b). Before switching the capacitor charges from the $20\text{ V}/40\,\Omega/60\,\Omega$ divider; after switching it charges toward the $25\text{ V}$ source through $50\,\Omega$. See the official exam paper.]

Approach. Use continuity of capacitor voltage for $V_c(0^+)$; find the new steady state and Thévenin resistance seen by $C$ for the $t\ge0$ exponential.

  1. (i) Initial voltage — steady state at position a. After a long time the capacitor is an open circuit, so node P is a simple divider of the $20\text{ V}$ source across $40\,\Omega$ and $60\,\Omega$: $$V_c(0^-)=20\cdot\frac{60}{40+60}=12\text{ V}.$$ Capacitor voltage cannot change instantaneously, so $$\boxed{V_c(0^+)=12\text{ V}}.$$
  2. Final value at position b. Connected to the right network, at steady state the capacitor again blocks DC, so no current flows in the $50\,\Omega$ and the capacitor charges to the full source voltage: $$V_c(\infty)=25\text{ V}.$$
  3. Time constant. Killing the $25\text{ V}$ source, the resistance seen by $C$ is $R_{th}=50\,\Omega$, so $$\tau=R_{th}C=50\times0.1\times10^{-3}=5\times10^{-3}\text{ s}=5\text{ ms}.$$
  4. (ii) Full response. The standard first-order form $V_c(t)=V_c(\infty)+[V_c(0^+)-V_c(\infty)]e^{-t/\tau}$ gives $$\boxed{V_c(t)=25-13\,e^{-t/0.005}=25-13\,e^{-200t}\text{ V},\quad t\ge0}.$$
  5. (iii) Value at $t=2\text{ s}$. Since $t=2\text{ s}$ is $400$ time constants, $e^{-200(2)}=e^{-400}\approx0$: $$\boxed{V_c(2)\approx 25\text{ V}}.$$ The capacitor is fully charged; the transient (lifetime $\sim5\tau=25\text{ ms}$) has long since decayed.
Check. With $C=0.1\text{ mF}$ the constant is $5\text{ ms}$, so any evaluation time beyond $\sim30\text{ ms}$ returns the steady value; the point of part (iii) is recognising that $t=2\text{ s}\gg\tau$ rather than plugging into the exponential.
QuantityValue
$V_c(0^+)$$12\text{ V}$
Time constant $\tau$$5\text{ ms}$
$V_c(t),\ t\ge0$$25-13e^{-200t}\text{ V}$
$V_c(2)$$\approx25\text{ V}$