Circuit Analysis Fundamentals

Question 1

(a) Branch $R_3+R_4 = 200+400 = 600\ \Omega$. Parallel with $R_2=300$: $R_p = \frac{300\cdot600}{300+600} = \frac{180000}{900} = 200\ \Omega.$ Total $R_{eq} = R_1 + R_p = 100 + 200 = 300\ \Omega$. (4) Why: series adds; parallel uses product/sum; the divider branch is a single series resistance before paralleling.

(b) $I_{tot} = 24/300 = 0.08\ \text{A} = 80\ \text{mA}$. (2)

(c) Voltage across the parallel block $= I_{tot}\cdot R_p = 0.08\cdot200 = 16\ \text{V}$. Current through the $600\ \Omega$ branch (through $R_4$): $16/600 = 0.02667\ \text{A} = 26.7\ \text{mA}$. (3) Why: same voltage across both parallel branches; branch current = branch voltage / branch resistance.

(d) Voltage across $R_2$ equals the parallel-block voltage $= 16\ \text{V}$. (3) Why: $R_2$ is directly across the parallel node pair.

Question 2

(a) Superposition at node A (open circuit, no load current drawn since $R_c$ path carries current to the 6 V source).

With A open, current can still flow through the path $12\text{V}\to R_a\to A\to R_c\to 6\text{V}$ and through $R_b$ to ground. Node equation at A (open output): $\frac{V_A-12}{2} + \frac{V_A-0}{6} + \frac{V_A-6}{3} = 0 \quad (\text{k}\Omega).$ Multiply by 6: $3(V_A-12) + V_A + 2(V_A-6) = 0$ $3V_A - 36 + V_A + 2V_A - 12 = 0 \Rightarrow 6V_A = 48 \Rightarrow V_A = 8\ \text{V}.$ So $V_{th} = 8\ \text{V}$. (6) Why: KCL/nodal analysis (equivalent to superposition of the two sources) gives the open-circuit terminal voltage.

(b) Deactivate sources (short both voltage sources). Then from node A three resistors go to ground: $R_a$, $R_b$, $R_c$ all in parallel: $R_{th} = \left(\tfrac1{2}+\tfrac1{6}+\tfrac1{3}\right)^{-1}\text{k}\Omega = \left(\tfrac{3+1+2}{6}\right)^{-1} = 1\ \text{k}\Omega.$ (3)

(c) With $R_L = 4\ \text{k}\Omega$: $I_L = \frac{V_{th}}{R_{th}+R_L} = \frac{8}{1+4} = 1.6\ \text{mA}.$ $P_L = I_L^2 R_L = (1.6\times10^{-3})^2 \cdot 4000 = 0.01024\ \text{W} \approx 10.24\ \text{mW}.$ (3)

(d) $R_N = R_{th} = 1\ \text{k}\Omega$; $I_N = V_{th}/R_{th} = 8/1 = 8\ \text{mA}$. (2)

Question 3

(a) $\tau = RC = 10^4 \cdot 4.7\times10^{-6} = 0.047\ \text{s} = 47\ \text{ms}$. (2)

(b) $v_C(t) = 5\left(1 - e^{-t/\tau}\right)$. At $t = 30\ \text{ms}$: $t/\tau = 0.030/0.047 = 0.6383$, $e^{-0.6383} = 0.5282$. $v_C = 5(1-0.5282) = 2.359\ \text{V}$. (4)

(c) $4 = 5(1-e^{-t/\tau}) \Rightarrow e^{-t/\tau} = 0.2 \Rightarrow t = \tau\ln 5 = 0.047\cdot1.6094 = 0.0756\ \text{s} \approx 75.6\ \text{ms}$. (3)

(d) At $t=0^+$: capacitor acts as short, $i = 5/10\text{k} = 0.5\ \text{mA}$. At $t\to\infty$: capacitor fully charged, $i = 0$. (3)

Question 4

(a) $\tau = L/R = 0.05/25 = 0.002\ \text{s} = 2\ \text{ms}$. Final current $I_\infty = 10/25 = 0.4\ \text{A}$. (3)

(b) $i(t) = 0.4(1-e^{-t/\tau})$. At $t = 2\ \text{ms} = \tau$: $i = 0.4(1-e^{-1}) = 0.4(0.6321) = 0.2528\ \text{A}$. (4)

(c) $v_L = L\,di/dt = V_{source}\,e^{-t/\tau} = 10\,e^{-1} = 3.679\ \text{V}$. (Check: $v_R = i R = 0.2528\cdot25 = 6.32\ \text{V}$; $v_L + v_R = 10\ \text{V}$ ✓.) (3)

Question 5

(a) $X_C = \dfrac{1}{\omega C} = \dfrac{1}{10^4 \cdot 100\times10^{-9}} = \dfrac{1}{10^{-3}} = 1000\ \Omega$. (3)

(b) $|Z| = \sqrt{R^2 + X_C^2} = \sqrt{1000^2 + 1000^2} = 1000\sqrt2 = 1414.2\ \Omega$. (3)

(c) $I_{rms} = V/|Z| = 5/1414.2 = 3.536\ \text{mA}$. (3)

(d) For series RC, current leads voltage by $\theta = \arctan(X_C/R) = \arctan(1) = 45^\circ$ (current leads). (3)

[
  {"claim":"Q1 Req = 300 ohm and R4 current = 24/300 * 200 / 600 A","code":"Rp=(300*600)/(300+600); Req=100+Rp; Itot=24/Req; Vpar=Itot*Rp; I_R4=Vpar/600; result = (Req==300) and abs(I_R4-0.026666666666)<1e-6"},
  {"claim":"Q2 Vth=8V, Rth=1k, load current 1.6mA","code":"VA=Rational(48,6); Rth=1/(Rational(1,2)+Rational(1,6)+Rational(1,3)); IL=8/(Rth+4); result = (VA==8) and (Rth==1) and (IL==Rational(8,5)/1000*1000/1000 or abs(float(IL)-1.6)<1e-9)"},
  {"claim":"Q3 vC at 30ms and t for 4V","code":"tau=10**4*4.7e-6; vc=5*(1-exp(-0.030/tau)); t4=tau*ln(5); result = abs(float(vc)-2.3592)<1e-3 and abs(float(t4)-0.07564)<1e-4"},
  {"claim":"Q5 |Z|, I, phase","code":"Xc=1/(10**4*100e-9); Z=sqrt(1000**2+Xc**2); I=5/Z; th=atan(Xc/1000); result = abs(float(Xc)-1000)<1e-6 and abs(float(Z)-1414.2135)<1e-2 and abs(float(I)-0.0035355)<1e-6 and abs(float(th)-pi/4)<1e-9"}
]