Level 5 — MasteryCircuit Analysis Fundamentals

Circuit Analysis Fundamentals

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Level 5 — Mastery (cross-domain: math + physics + coding, build/prove) Time limit: 90 minutes Total marks: 60

Instructions: Answer all three questions. Show all derivations. Where code is requested, write clean, correct Python (NumPy/SymPy permitted). Units required for full marks.

Question 1 — Thévenin, Norton & Superposition (proof + build) — 22 marks

Consider a linear DC network with two independent sources feeding a load port A–B:

A $12\,\text{V}$ ideal source in series with $R_1 = 4\,\Omega$ connects to node A.
From node A, resistor $R_2 = 6\,\Omega$ goes to node B (the ground/reference).
A $3\,\text{A}$ ideal current source is injected into node A (returning through B).
The load $R_L$ is connected across A–B.

(a) Using superposition, derive a symbolic expression for the open-circuit voltage $V_{AB}$ (load removed) in terms of $V_s, I_s, R_1, R_2$ . Evaluate numerically. (6)

(b) Derive the Thévenin equivalent ( $V_{th}, R_{th}$ ) seen at A–B, and the Norton equivalent ( $I_N, R_N$ ). Prove that $V_{th} = I_N R_N$ . (8)

(c) With $R_L = 5\,\Omega$ attached, compute the load current $I_L$ and power $P_L$ . (4)

(d) Write a Python function thevenin(Vs, R1, Is, R2) returning (Vth, Rth), and show a one-line assertion that reproduces your part (b) numbers. (4)

Question 2 — RC transient: derivation, energy, and numerics — 20 marks

An initially uncharged capacitor $C = 2.2\,\mu\text{F}$ is charged through $R = 47\,\text{k}\Omega$ from a $9\,\text{V}$ step source at $t=0$ .

(a) Starting from KVL and the constitutive relation $i = C\,dv_C/dt$ , derive the ODE and solve for $v_C(t)$ . State the time constant $\tau$ . (6)

(b) Compute the time for $v_C$ to reach $63.2\%$ and $99\%$ of final value. (4)

(c) Prove that, over full charging ( $t: 0 \to \infty$ ), the energy delivered by the source equals twice the energy stored in the capacitor, and identify where the other half went. (6)

(d) Write Python that numerically integrates the ODE (e.g. Euler or odeint) and verifies $v_C(\tau) \approx 0.632\,V_s$ . State the check. (4)

Question 3 — AC reactance & phasor analysis (build/prove) — 18 marks

A series RL circuit ( $R = 100\,\Omega$ , $L = 50\,\text{mH}$ ) is driven by $v(t) = 10\cos(2\pi f t)$ with $f = 1\,\text{kHz}$ .

(a) Compute the inductive reactance $X_L$ , the complex impedance $Z$ , its magnitude $|Z|$ and phase $\phi$ . (6)

(b) Find the current amplitude $I_0$ and write $i(t)$ in the form $I_0\cos(\omega t - \theta)$ . Explain the sign of the phase (lead/lag). (6)

(c) At what frequency $f_c$ does the resistor voltage equal the inductor voltage (i.e. $|Z_R| = |Z_L|$ )? Prove this equals the $-3\,\text{dB}$ corner of the RL low-pass taken across $R$ , and give $f_c$ numerically. (6)

Answer keyMark scheme & solutions

Question 1

(a) Superposition (6) Remove load (open circuit at A–B). Node A referenced to B.

Voltage source acting alone (open the current source): divider not loaded → the $12\,\text{V}$ through $R_1$ then $R_2$ to ground, open port draws no current, so voltage across $R_2$ : $V' = V_s\frac{R_2}{R_1+R_2} = 12\cdot\frac{6}{10} = 7.2\,\text{V}$ (2)
Current source alone (short the voltage source): $R_1 \parallel R_2$ carries $I_s$ : $V'' = I_s (R_1\parallel R_2) = 3\cdot\frac{4\cdot6}{10} = 3\cdot 2.4 = 7.2\,\text{V}$ (2)
Superpose: $V_{AB} = V_s\frac{R_2}{R_1+R_2} + I_s\frac{R_1R_2}{R_1+R_2} = 7.2 + 7.2 = 14.4\,\text{V}$ (2)

(b) Thévenin/Norton (8) $V_{th} = V_{AB,oc} = 14.4\,\text{V}$ . (1) $R_{th}$ : kill sources (short $V_s$ , open $I_s$ ) → $R_1 \parallel R_2 = 2.4\,\Omega$ . (2) Norton: $R_N = R_{th} = 2.4\,\Omega$ . (1) $I_N = V_{th}/R_{th} = 14.4/2.4 = 6\,\text{A}$ (short-circuit current). (2) Proof: $I_N R_N = 6 \times 2.4 = 14.4 = V_{th}$ . Source transformation is an identity because both models must produce the same $V_{oc}$ (open) and $I_{sc}$ (short); by Ohm's law at the port $V_{oc}=I_{sc}R_{th}$ . ∎ (2)

(c) Load (4) $I_L = \frac{V_{th}}{R_{th}+R_L} = \frac{14.4}{2.4+5} = \frac{14.4}{7.4} = 1.9459\,\text{A}$ (2) $P_L = I_L^2 R_L = (1.9459)^2\cdot 5 = 18.93\,\text{W}$ (2)

(d) Code (4)

def thevenin(Vs, R1, Is, R2):
    Rth = R1*R2/(R1+R2)
    Vth = Vs*R2/(R1+R2) + Is*R1*R2/(R1+R2)
    return Vth, Rth
 
assert thevenin(12,4,3,6) == (14.4, 2.4)

(function 2, assertion 2)

Question 2

(a) ODE + solution (6) KVL: $V_s = iR + v_C$ , with $i = C\,dv_C/dt$ : $RC\frac{dv_C}{dt} + v_C = V_s$ (2) Solve (linear 1st-order), $v_C(0)=0$ : $v_C(t) = V_s\left(1 - e^{-t/RC}\right)$ (2) $\tau = RC = 47{,}000 \times 2.2\times10^{-6} = 0.1034\,\text{s} \approx 103.4\,\text{ms}$ (2)

(b) Times (4)

$63.2\%$ : by definition $t = \tau = 0.1034\,\text{s}$ . (2)
$99\%$ : $1 - e^{-t/\tau} = 0.99 \Rightarrow t = \tau\ln 100 = 0.1034\times 4.6052 = 0.476\,\text{s}$ . (2)

(c) Energy proof (6) Charge delivered as $t\to\infty$ : $Q = CV_s$ . Source energy: $W_{src} = \int_0^\infty V_s\, i\,dt = V_s Q = C V_s^2$ (2) Stored energy: $W_C = \tfrac12 C V_s^2$ . (1) Ratio $W_{src}/W_C = 2$ . (1) Dissipated in R: $W_R = \int_0^\infty i^2 R\,dt$ with $i = (V_s/R)e^{-t/\tau}$ : $W_R = \frac{V_s^2}{R}\int_0^\infty e^{-2t/\tau}dt = \frac{V_s^2}{R}\cdot\frac{\tau}{2} = \frac{V_s^2}{R}\cdot\frac{RC}{2} = \tfrac12 CV_s^2$ So $W_{src} = W_C + W_R = CV_s^2$ ; the missing half is dissipated as heat in $R$ , independent of $R$ . ∎ (2)

(d) Code (4)

import numpy as np
Vs, R, C = 9.0, 47e3, 2.2e-6
tau = R*C
dt = tau/2000; t=0.0; v=0.0
while t < tau:
    v += dt*(Vs - v)/tau     # Euler on RC dv/dt = Vs - v
    t += dt
print(v, 0.632*Vs)           # ~5.688 vs 5.688
assert abs(v - 0.632*Vs) < 0.02*Vs

(integration 2, check 2)

Question 3

(a) Impedance (6) $\omega = 2\pi f = 2\pi(1000) = 6283.2\,\text{rad/s}$ . $X_L = \omega L = 6283.2 \times 0.05 = 314.16\,\Omega$ (2) $Z = R + jX_L = 100 + j314.16\,\Omega$ (1) $|Z| = \sqrt{100^2 + 314.16^2} = \sqrt{10000+98696} = 329.7\,\Omega$ (2) $\phi = \arctan(314.16/100) = 72.34^\circ$ (1)

(b) Current (6) $I_0 = \frac{V_0}{|Z|} = \frac{10}{329.7} = 0.03033\,\text{A} = 30.33\,\text{mA}$ (2) $i(t) = 0.03033\cos(\omega t - 72.34^\circ)$ (2) Current lags voltage by $72.34^\circ$ : in an inductive circuit the current cannot change instantly, so it trails the driving voltage; the phase angle equals $\phi$ of $Z$ . (2)

(c) Corner frequency (6) Equal magnitudes: $|Z_R| = |Z_L| \Rightarrow R = \omega_c L \Rightarrow \omega_c = R/L$ . (2) $f_c = \frac{R}{2\pi L} = \frac{100}{2\pi(0.05)} = 318.3\,\text{Hz}$ (2) Low-pass across $R$ : $H = \frac{R}{R+j\omega L}$ , $|H| = \frac{R}{\sqrt{R^2+(\omega L)^2}}$ . At $\omega L = R$ , $|H| = 1/\sqrt2 = -3\,\text{dB}$ . So the equal-magnitude point is the $-3\,\text{dB}$ corner. ∎ (2)

[
  {"claim":"Q1a Voc = 14.4 V by superposition","code":"Vs,R1,Is,R2=12,4,3,6; Voc=Vs*R2/(R1+R2)+Is*R1*R2/(R1+R2); result=abs(Voc-14.4)<1e-9"},
  {"claim":"Q1b IN*RN equals Vth","code":"Rth=Rational(4*6,4+6); Vth=Rational(144,10); IN=Vth/Rth; result=simplify(IN*Rth-Vth)==0 and IN==6"},
  {"claim":"Q1c load power ~18.93 W","code":"Vth,Rth,RL=14.4,2.4,5; IL=Vth/(Rth+RL); PL=IL**2*RL; result=abs(PL-18.93)<0.02"},
  {"claim":"Q2 tau and 99% time","code":"tau=47000*2.2e-6; t99=tau*log(100); result=abs(float(tau)-0.10340)<1e-4 and abs(float(t99)-0.4763)<1e-3"},
  {"claim":"Q3 |Z| and phase for RL","code":"import math; XL=2*math.pi*1000*0.05; Z=math.hypot(100,XL); ph=math.degrees(math.atan2(XL,100)); result=abs(Z-329.7)<0.5 and abs(ph-72.34)<0.1"},
  {"claim":"Q3c corner freq 318.3 Hz","code":"import math; fc=100/(2*math.pi*0.05); result=abs(fc-318.31)<0.1"}
]