Level 2 — RecallCircuit Analysis Fundamentals

Circuit Analysis Fundamentals

⏱ 30 minutes50 marksprintable — key stays hidden on paper

Difficulty Level: 2 — Recall (definitions, standard textbook problems, short derivations) Time Limit: 30 minutes Total Marks: 50

Q1. Define Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) in one sentence each. (4 marks)

Q2. Three resistors $R_1 = 100\,\Omega$ , $R_2 = 200\,\Omega$ , $R_3 = 300\,\Omega$ are connected in series. (a) Compute the equivalent resistance. (2 marks) (b) Recompute the equivalent resistance if instead they are all in parallel. (3 marks)

Q3. A voltage divider consists of $R_1 = 4\,\text{k}\Omega$ (top) and $R_2 = 6\,\text{k}\Omega$ (bottom) across a $10\,\text{V}$ supply. Compute the output voltage taken across $R_2$ . State the divider formula used. (4 marks)

Q4. A total current of $6\,\text{A}$ enters a parallel combination of $R_1 = 2\,\Omega$ and $R_2 = 4\,\Omega$ . Using the current divider rule, find the current through $R_1$ . (4 marks)

Q5. An RC circuit has $R = 10\,\text{k}\Omega$ and $C = 100\,\mu\text{F}$ . (a) Compute the time constant $\tau$ . (2 marks) (b) State the capacitor voltage (as a fraction of the final value) reached after one time constant during charging. (2 marks)

Q6. For an RL circuit with $L = 2\,\text{H}$ and $R = 500\,\Omega$ , compute the time constant $\tau$ and state the equation for the current $i(t)$ during energising from a source $V$ . (5 marks)

Q7. Define Thévenin's theorem and Norton's theorem. State the relationship between the Thévenin resistance $R_{Th}$ and the Norton resistance $R_N$ , and between $V_{Th}$ , $I_N$ and this resistance. (6 marks)

Q8. State the superposition theorem. Explain how independent voltage sources and independent current sources are treated ("deactivated") when applying it. (5 marks)

Q9. A capacitor of $C = 2\,\mu\text{F}$ is driven by an AC source at frequency $f = 1\,\text{kHz}$ . Compute its capacitive reactance $X_C$ . State the phase relationship between voltage and current for an ideal capacitor. (5 marks)

Q10. A multimeter is used to measure a resistor. Explain briefly: (a) how the meter should be connected to measure current vs. voltage; (2 marks) (b) what a "reference node" or ground means in a circuit. (2 marks)

Answer keyMark scheme & solutions

Q1. (4 marks)

KCL: The algebraic sum of currents entering a node equals the sum leaving it; i.e. the net current at any node is zero. (2 marks — expresses charge conservation)
KVL: The algebraic sum of voltages around any closed loop is zero. (2 marks — expresses energy conservation)

Q2. (5 marks) (a) Series: $R_{eq} = R_1+R_2+R_3 = 100+200+300 = 600\,\Omega$ . (2) (b) Parallel: $\dfrac{1}{R_{eq}} = \dfrac{1}{100}+\dfrac{1}{200}+\dfrac{1}{300}$ . LCM approach: $= \dfrac{6+3+2}{600} = \dfrac{11}{600}$ . (1) $R_{eq} = \dfrac{600}{11} \approx 54.55\,\Omega$ . (2) Why: series resistances add; parallel conductances add.

Q3. (4 marks) Formula: $V_{out} = V_{in}\cdot \dfrac{R_2}{R_1+R_2}$ . (2) $V_{out} = 10 \cdot \dfrac{6}{4+6} = 10 \cdot 0.6 = 6\,\text{V}$ . (2)

Q4. (4 marks) Current divider: $I_1 = I_{tot}\cdot \dfrac{R_2}{R_1+R_2}$ (opposite resistor over sum). (2) $I_1 = 6 \cdot \dfrac{4}{2+4} = 6 \cdot \dfrac{4}{6} = 4\,\text{A}$ . (2) Check: $I_2 = 6\cdot\frac{2}{6}=2\,\text{A}$ , and $4+2=6$ . ✓

Q5. (4 marks) (a) $\tau = RC = (10\times10^3)(100\times10^{-6}) = 1\,\text{s}$ . (2) (b) After one $\tau$ , $V_C = V_f(1-e^{-1}) = 0.632\,V_f$ → about 63.2% of final. (2)

Q6. (5 marks) $\tau = \dfrac{L}{R} = \dfrac{2}{500} = 0.004\,\text{s} = 4\,\text{ms}$ . (2) Energising current: $i(t) = \dfrac{V}{R}\left(1-e^{-t/\tau}\right) = \dfrac{V}{R}\left(1-e^{-Rt/L}\right)$ . (3) Why: inductor current rises exponentially toward final value $V/R$ .

Q7. (6 marks)

Thévenin: Any linear two-terminal network can be replaced by a single voltage source $V_{Th}$ in series with a resistance $R_{Th}$ . (2)
Norton: Any linear two-terminal network can be replaced by a single current source $I_N$ in parallel with a resistance $R_N$ . (2)
Relationships: $R_{Th}=R_N$ , and $V_{Th}=I_N R_N$ (equivalently $I_N=V_{Th}/R_{Th}$ ). (2)

Q8. (5 marks)

Statement: In a linear circuit with multiple independent sources, the response (voltage/current) at any element equals the algebraic sum of the responses caused by each independent source acting alone. (3)
Deactivation: Independent voltage sources are replaced by a short circuit (0 V); independent current sources are replaced by an open circuit (0 A). Dependent sources are left in place. (2)

Q9. (5 marks) $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi (1000)(2\times10^{-6})}$ . (2) $= \dfrac{1}{4\pi\times10^{-3}} = \dfrac{1000}{4\pi} \approx 79.58\,\Omega$ . (2) Phase: current leads voltage by $90°$ in an ideal capacitor. (1)

Q10. (4 marks) (a) To measure current, connect the ammeter in series (break the branch); to measure voltage, connect the voltmeter in parallel (across the element). (2) (b) The reference node / ground is the node chosen as $0\,\text{V}$ ; all other node voltages are measured relative to it. (2)

[
  {"claim":"Q2b parallel of 100,200,300 = 600/11 ~ 54.545 ohm","code":"Req = 1/(1/100 + 1/200 + 1/300); result = abs(Req - Rational(600,11)) < Rational(1,1000000)"},
  {"claim":"Q3 voltage divider output = 6 V","code":"Vout = 10 * Rational(6,4+6); result = Vout == 6"},
  {"claim":"Q4 current divider I1 = 4 A","code":"I1 = 6 * Rational(4,2+4); result = I1 == 4"},
  {"claim":"Q6 RL time constant = 0.004 s","code":"tau = Rational(2,500); result = tau == Rational(4,1000)"},
  {"claim":"Q9 capacitive reactance ~ 79.577 ohm","code":"Xc = 1/(2*pi*1000*2*10**(-6)); result = abs(float(Xc) - 79.5774715) < 0.001"}
]