Circuit Analysis Fundamentals

Question 1 (10 marks)

(a) Innermost series: $4 + 2 = 6\,\Omega$. (1) Parallel with $12\,\Omega$: $\dfrac{12 \cdot 6}{12+6} = \dfrac{72}{18} = 4\,\Omega$. (2) Series with the $6\,\Omega$: $6 + 4 = 10\,\Omega$. (2) Parallel with $18\,\Omega$: $\dfrac{10 \cdot 18}{10+18} = \dfrac{180}{28} = 6.4286\,\Omega$. (2)

$R_{eq} = \frac{45}{7} \approx 6.43\,\Omega$

(b) Series = same current path, elements share one node only and no branch between them; add resistances. Parallel = share both nodes (same voltage across); use reciprocal/product-over-sum. Decide by tracing whether a node between two elements has a third connection (branch out ⇒ not simple series). (3)

Question 2 (10 marks)

(a) Parallel pair: $R_p = \dfrac{3\cdot 6}{3+6}=2\,\text{k}\Omega$. (1) Divider: $V_x = 12 \cdot \dfrac{R_p}{R_1+R_p} = 12\cdot\dfrac{2}{1+2}=8\,\text{V}$. (3)

(b) Total current $I = \dfrac{12}{3\text{k}} = 4\,\text{mA}$. Current divider into $R_3$: $I_3 = I\cdot\dfrac{R_2}{R_2+R_3} = 4\text{mA}\cdot\dfrac{3}{9}= \tfrac{4}{3}\,\text{mA}\approx1.33\,\text{mA}$. (4) (Check: $I_3 = V_x/R_3 = 8/6\text{k} = 1.33\,\text{mA}$.)

(c) KCL: $I_{R2}+I_{R3}$ should equal $I$. $I_{R2}=8/3\text{k}=2.667\,\text{mA}$, $I_{R3}=1.333\,\text{mA}$, sum $=4\,\text{mA}=I$. ✓ (2)

Question 3 (12 marks)

(a) Let $I_1$ (from $E_1$), $I_2$ (from $E_2$) flow into the top node; $I_3$ flows down through $R_3$. KCL: $I_1 + I_2 = I_3$. (1) KVL Loop A: $E_1 = I_1 R_1 + I_3 R_3 \Rightarrow 10 = 2I_1 + 6I_3$. (2) KVL Loop B: $E_2 = I_2 R_2 + I_3 R_3 \Rightarrow 6 = 3I_2 + 6I_3$. (2)

(b) Substitute $I_1 = I_3 - I_2$: $10 = 2(I_3-I_2)+6I_3 = 8I_3 - 2I_2$. From loop B: $I_2 = (6-6I_3)/3 = 2 - 2I_3$. $10 = 8I_3 - 2(2-2I_3) = 8I_3 -4 +4I_3 = 12I_3 -4 \Rightarrow I_3 = \dfrac{14}{12}=1.1667\,\text{A}$. (3) $I_2 = 2 - 2(1.1667) = -0.3333\,\text{A}$ (actually flows opposite assumed). $I_1 = I_3 - I_2 = 1.1667 + 0.3333 = 1.5\,\text{A}$. (2)

(c) Ground = bottom reference node (junction of both source returns). Branch currents depend only on potential differences, so shifting the reference by a constant leaves all currents unchanged. (2)

Question 4 (12 marks)

(a) Remove $R_3$. Open circuit ⇒ no current through the middle. The two source branches ($E_1,R_1$) and ($E_2,R_2$) now form one loop across the open terminals. Loop current $I = \dfrac{E_1 - E_2}{R_1+R_2} = \dfrac{10-6}{5}=0.8\,\text{A}$. (2) $V_{th}$ = voltage at top terminal ref bottom $= E_1 - I R_1 = 10 - 0.8(2) = 8.4\,\text{V}$ (check via other branch: $E_2 + IR_2 = 6 + 0.8(3)=8.4$ ✓). (3)

(b) Deactivate sources (short EMFs): $R_1 \parallel R_2 = \dfrac{2\cdot3}{5}=1.2\,\Omega = R_{th}$. (3)

(c) $I_N = V_{th}/R_{th} = 8.4/1.2 = 7\,\text{A}$, $R_N = 1.2\,\Omega$. (2) Reconnect $R_3=6\,\Omega$: current-divide $I_{R_3} = I_N\cdot\dfrac{R_N}{R_N+R_3}=7\cdot\dfrac{1.2}{7.2}=1.1667\,\text{A}$, matching Q3(b). (2)

Question 5 (10 marks)

(a) KVL: $E = iR + v_C$, $i = C\,dv_C/dt$ ⇒ $RC\dfrac{dv_C}{dt}+v_C = E$. Solution: $v_C(t) = E(1-e^{-t/\tau})$, $\tau = RC$. (3) $\tau = 10^4\cdot 100\times10^{-6} = 1\,\text{s}$. (1)

(b) $4 = 5(1-e^{-t}) \Rightarrow e^{-t} = 0.2 \Rightarrow t = \ln 5 = 1.609\,\text{s}$. (3)

(c) (3)

import numpy as np
E, tau = 5.0, 1.0
t = np.linspace(0, 5*tau, 100)
vC = E*(1 - np.exp(-t/tau))

Question 6 (6 marks)

(a) $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi(1000)(10^{-6})} = 159.15\,\Omega$. (2) $|Z| = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 159.15^2} = 187.9\,\Omega$. (1)

(b) Superposition with two frequencies: analyze each source separately (other source replaced by its internal impedance — voltage source shorted). Because reactance is frequency-dependent, compute a separate $X_C$ for each frequency, solve each single-source phasor circuit, then sum the time-domain responses (cannot add phasors of different frequencies). (3)

[
  {"claim":"Q1 equivalent resistance = 45/7",
   "code":"inner=4+2; p1=(12*inner)/(12+inner); s=6+p1; Req=(s*18)/(s+18); result = abs(Req - Rational(45,7)) < Rational(1,100000)"},
  {"claim":"Q2 Vx=8V and I_R3=4/3 mA",
   "code":"Rp=(3*6)/(3+6); Vx=12*Rp/(1+Rp); I3=Vx/6; result = (Vx==8) and (abs(I3-Rational(4,3))<Rational(1,10000))"},
  {"claim":"Q3 branch currents I3=7/6, I2=-1/3, I1=3/2",
   "code":"I1,I2,I3=symbols('I1 I2 I3'); sol=solve([I1+I2-I3, 10-2*I1-6*I3, 6-3*I2-6*I3],[I1,I2,I3]); result = (sol[I3]==Rational(7,6)) and (sol[I2]==Rational(-1,3)) and (sol[I1]==Rational(3,2))"},
  {"claim":"Q4 Vth=8.4, Rth=1.2, Norton current in R3 = 7/6",
   "code":"I=(10-6)/5; Vth=10-I*2; Rth=(2*3)/(5); IN=Vth/Rth; IR3=IN*Rth/(Rth+6); result = (Vth==Rational(42,5)) and (Rth==Rational(6,5)) and (abs(IR3-Rational(7,6))<Rational(1,100000))"},
  {"claim":"Q5 time to reach 4V is ln(5)",
   "code":"t=symbols('t'); eq=5*(1-exp(-t/1))-4; sol=solve(eq,t); result = abs(sol[0]-log(5))<Rational(1,100000)"}
]