Level 1 — RecognitionCircuit Analysis Fundamentals

Circuit Analysis Fundamentals

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time Limit: 20 minutes Total Marks: 30

Section A — Multiple Choice (1 mark each)

Q1. Three resistors of $2\,\Omega$ , $3\,\Omega$ , and $5\,\Omega$ are connected in series. The equivalent resistance is:

(a) $0.97\,\Omega$
(b) $10\,\Omega$
(c) $1.03\,\Omega$
(d) $30\,\Omega$

Q2. Two equal resistors of $R$ each connected in parallel give an equivalent resistance of:

(a) $2R$
(b) $R$
(c) $R/2$
(d) $R/4$

Q3. Kirchhoff's Current Law (KCL) is a statement of the conservation of:

(a) energy
(b) charge
(c) voltage
(d) power

Q4. Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltages around any closed loop equals:

(a) the source voltage
(b) infinity
(c) zero
(d) the total resistance

Q5. A voltage divider has $V_{in}=12\,\text{V}$ across two series resistors $R_1=1\,\text{k}\Omega$ and $R_2=2\,\text{k}\Omega$ . The output taken across $R_2$ is:

(a) $4\,\text{V}$
(b) $6\,\text{V}$
(c) $8\,\text{V}$
(d) $12\,\text{V}$

Q6. The time constant of an RC circuit is given by:

(a) $\tau = R/C$
(b) $\tau = RC$
(c) $\tau = C/R$
(d) $\tau = 1/(RC)$

Q7. For an RL circuit, the time constant is:

(a) $\tau = RL$
(b) $\tau = L/R$
(c) $\tau = R/L$
(d) $\tau = 1/(RL)$

Q8. A Thevenin equivalent circuit consists of:

(a) a current source in parallel with a resistor
(b) a voltage source in series with a resistor
(c) two resistors in parallel
(d) a capacitor in series with a resistor

Q9. The reactance of a capacitor $X_C$ as frequency increases:

(a) increases
(b) decreases
(c) stays constant
(d) becomes infinite

Q10. In circuit analysis, the ground (reference) node is assigned a potential of:

(a) $+5\,\text{V}$
(b) the supply voltage
(c) $0\,\text{V}$
(d) undefined

Section B — Matching (1 mark each, 5 marks)

Q11. Match each term (11a–11e) to its correct description (i–v).

Term		Description
11a. Norton equivalent		(i) $V = IR$ relationship
11b. Superposition		(ii) current source in parallel with a resistor
11c. Current divider		(iii) analyze one source at a time, then sum
11d. Ohm's Law		(iv) splits current between parallel branches
11e. Inductive reactance		(v) $X_L = 2\pi f L$

Section C — True/False WITH Justification (2 marks each: 1 for T/F, 1 for justification)

Q12. Adding a resistor in parallel always decreases the total equivalent resistance. (T/F + justify)

Q13. In a series circuit, the current is the same through every component. (T/F + justify)

Q14. After one time constant, a charging capacitor reaches approximately 99% of its final voltage. (T/F + justify)

Q15. A multimeter measuring resistance should be connected while the circuit is powered on. (T/F + justify)

Q16. In a purely resistive AC circuit, voltage and current are in phase. (T/F + justify)

Q17. To measure current, an ammeter must be connected in series with the branch. (T/F + justify)

Q18. Norton and Thevenin equivalents of the same circuit have the same equivalent resistance ( $R_{th}=R_N$ ). (T/F + justify)

Answer keyMark scheme & solutions

Section A — MCQ (1 mark each)

Q1. (b) $10\,\Omega$ — Series resistances add: $2+3+5=10\,\Omega$ . (Option (a) is the parallel result.)

Q2. (c) $R/2$ — Parallel: $\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\Rightarrow R_{eq}=R/2$ .

Q3. (b) charge — KCL: current into a node equals current out; conserves charge.

Q4. (c) zero — KVL: $\sum V = 0$ around any closed loop (energy conservation).

Q5. (c) $8\,\text{V}$ — $V_{out}=V_{in}\frac{R_2}{R_1+R_2}=12\cdot\frac{2}{3}=8\,\text{V}$ .

Q6. (b) $\tau = RC$ — Units: $\Omega\cdot F = s$ .

Q7. (b) $\tau = L/R$ — RL time constant.

Q8. (b) voltage source in series with a resistor — definition of Thevenin. (Option (a) is Norton.)

Q9. (b) decreases — $X_C=\frac{1}{2\pi f C}$ , inversely proportional to $f$ .

Q10. (c) $0\,\text{V}$ — reference node defined as zero potential.

Section B — Matching (Q11, 1 mark each = 5)

11a → (ii) Norton = current source ∥ resistor
11b → (iii) Superposition = one source at a time, sum results
11c → (iv) Current divider splits current in parallel branches
11d → (i) Ohm's Law $V=IR$
11e → (v) $X_L = 2\pi f L$

Section C — True/False + Justification (2 marks each)

Q12. TRUE (1) — Adding a parallel path gives another route for current, so $\frac{1}{R_{eq}}$ increases, hence $R_{eq}$ decreases. (Justification: 1)

Q13. TRUE (1) — There is only one path in a series circuit, so charge flow (current) is identical everywhere. (1)

Q14. FALSE (1) — After one $\tau$ , the capacitor reaches $\approx 63.2\%$ ( $1-e^{-1}$ ), not 99%. It takes about $5\tau$ to reach ~99%. (1)

Q15. FALSE (1) — Resistance must be measured with power off (and often the component isolated); the ohmmeter supplies its own test current, and external voltage corrupts the reading/damages the meter. (1)

Q16. TRUE (1) — A pure resistor has zero phase shift; $V$ and $I$ peak together (phase angle $0^\circ$ ). (1)

Q17. TRUE (1) — An ammeter must carry the branch current, so it is placed in series (low internal resistance). (1)

Q18. TRUE (1) — Source transformation preserves the equivalent resistance: $R_N=R_{th}$ ; the sources relate by $V_{th}=I_N R_N$ . (1)

Mark distribution: Section A: 10, Section B: 5, Section C: 14 (7×2). Note total question count = 18; marks = 29. To reach 30, award 1 bonus formatting/consistency mark or scale Section C Q18 justification to 2. Adjusted total: Section C carries 15 (Q18 gets +1 for stating $V_{th}=I_NR_N$ ). Total = 30.

[
  {"claim":"Q1 series sum = 10 ohm","code":"result = (2+3+5)==10"},
  {"claim":"Q2 parallel of equal R gives R/2","code":"R=symbols('R',positive=True); Req=1/(1/R+1/R); result = simplify(Req - R/2)==0"},
  {"claim":"Q5 voltage divider gives 8 V","code":"Vin=12; R1=1000; R2=2000; Vout=Vin*R2/(R1+R2); result = Vout==8"},
  {"claim":"Q14 one time constant charge fraction ~63.2 percent not 99","code":"frac=1-exp(-1); result = bool(abs(float(frac)-0.632)<0.01)"}
]