These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: batteries, wires and switches are ideal; resistors are ohmic; capacitors start uncharged unless stated; τ = RC; after one time constant a charging capacitor has about 63% of its final charge and a discharging one about 37% of its initial charge. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A 4.0 μF capacitor and a 12 μF capacitor are connected in series. What is their equivalent capacitance?
- (A) 3.0 μF
- (B) 16 μF
- (C) 8.0 μF
- (D) 0.33 μF
Answer and explanation
Answer: (A). 1/C_eq = 1/4.0 + 1/12 = 3/12 + 1/12 = 4/12, so C_eq = 3.0 μF. It is smaller than 4.0 μF, as a series combination must be.
- (B) adds the capacitances, which is the parallel rule.
- (C) averages the two values. No combination rule does this.
- (D) stops at 1/C_eq = 0.33 μF⁻¹ and forgets to take the reciprocal.
Question 2 (multiple choice · core)
A 2.0 μF capacitor and a 6.0 μF capacitor are connected in series to a 12 V battery and fully charged. Which row is correct?
- (A) Each has 18 μC; the 2.0 μF capacitor has 9.0 V across it.
- (B) Each has 18 μC; the 6.0 μF capacitor has 9.0 V across it.
- (C) Each has 12 V across it; the charges are 24 μC and 72 μC.
- (D) Each has 6.0 V across it; the charges are 12 μC and 36 μC.
Answer and explanation
Answer: (A). C_eq = 1/(1/2.0 + 1/6.0) = 1.5 μF, so Q = (1.5 μF)(12 V) = 18 μC on each capacitor. ΔV = Q/C: 18/2.0 = 9.0 V and 18/6.0 = 3.0 V. The smaller capacitor has the larger ΔV.
- (B) swaps the potential differences, as if ΔV were proportional to C.
- (C) treats the capacitors as parallel, each across the full 12 V.
- (D) splits the 12 V equally, which would give unequal charges. Series capacitors must carry equal charge.
Question 3 (multiple choice · core)
An uncharged capacitor, a 5.0 kΩ resistor, a switch and a 10 V battery are in series. Immediately after the switch is closed, what are the current and the potential difference across the capacitor?
- (A) 2.0 mA; 0 V
- (B) 0 mA; 10 V
- (C) 0 mA; 0 V
- (D) 2.0 mA; 10 V
Answer and explanation
Answer: (A). With no charge yet, ΔV_C = Q/C = 0, so the capacitor acts like a wire. The whole 10 V is across the resistor: I = 10 V ÷ 5000 Ω = 2.0 mA.
- (B) describes the state after a long time, not the start.
- (C) treats the uncharged capacitor as a break. That is only true once it is charged.
- (D) breaks the loop rule: 10 V across the resistor plus 10 V across the capacitor would be 20 V.
Question 4 (multiple choice · core)
A capacitor of 150 μF, charged to 6.0 V, is discharged through a 20 kΩ resistor. Which is closest to the potential difference across the capacitor 3.0 s after the discharge starts?
- (A) 2.2 V
- (B) 3.8 V
- (C) 3.0 V
- (D) 0 V
Answer and explanation
Answer: (A). τ = RC = (20 × 10³ Ω)(150 × 10⁻⁶ F) = 3.0 s, so 3.0 s is one time constant. A discharging capacitor keeps about 37%: 0.37 × 6.0 V ≈ 2.2 V.
- (B) is 63% of 6.0 V, the charging benchmark used for a discharge.
- (C) assumes the voltage halves in one time constant. The fraction left after τ is about 37%, not 50%.
- (D) assumes the capacitor is fully discharged after one time constant.
Question 5 (calculation · core)
C₁ = 3.0 μF and C₂ = 6.0 μF are connected in series. This pair is connected in parallel with C₃ = 5.0 μF. The whole network is connected to a 6.0 V battery and fully charged.
(a) Find the equivalent capacitance. (b) Find the charge and potential difference for each capacitor. (c) Find the total energy stored.
Worked solution and suggested Marlbridge rubric
- (a) Series pair: 1/C₁₂ = 1/3.0 + 1/6.0 = 1/2.0, so C₁₂ = 2.0 μF. In parallel with C₃: C_eq = 2.0 + 5.0 = 7.0 μF.
- (b) The pair and C₃ are in parallel, so each has 6.0 V across it.
- C₃: Q₃ = (5.0 μF)(6.0 V) = 30 μC, ΔV₃ = 6.0 V.
- Pair: Q = (2.0 μF)(6.0 V) = 12 μC, so Q₁ = Q₂ = 12 μC. ΔV₁ = 12/3.0 = 4.0 V; ΔV₂ = 12/6.0 = 2.0 V.
- (c) U = ½C_eq(ΔV)² = ½(7.0 × 10⁻⁶ F)(6.0 V)² = 1.26 × 10⁻⁴ J (126 μJ).
Check. Total charge from the battery: 12 + 30 = 42 μC = (7.0 μF)(6.0 V). ΔV₁ + ΔV₂ = 6.0 V.
| Point | What earns it |
|---|---|
| 1 | C₁₂ = 2.0 μF from the series rule |
| 1 | C_eq = 7.0 μF |
| 1 | Q₃ = 30 μC with ΔV₃ = 6.0 V |
| 1 | Q₁ = Q₂ = 12 μC, with the reason (series, equal charge) |
| 1 | ΔV₁ = 4.0 V and ΔV₂ = 2.0 V |
| 1 | U = 126 μJ (or by summing ½QΔV for each capacitor) |
Question 6 (constructed response · core)
An uncharged 100 μF capacitor, a 30 kΩ resistor, a switch and a 9.0 V battery are in series. The switch is closed at t = 0.
(a) Calculate the time constant, the initial current and the final charge on the capacitor. (b) Sketch graphs of ΔV_C against t and I against t from 0 to 15 s. Label the values at t = 0, at one time constant, and the value approached after a long time. (c) The resistor is replaced by a 60 kΩ resistor and the experiment is repeated. Add dashed curves to both sketches. State which features change and which do not. (d) Outline how a student could measure the time constant using a voltmeter and a timer.
Worked solution and suggested Marlbridge rubric
(a) τ = (30 × 10³ Ω)(100 × 10⁻⁶ F) = 3.0 s. I₀ = 9.0 V ÷ 30 × 10³ Ω = 0.30 mA. Q_final = CΔV = (100 μF)(9.0 V) = 900 μC.
(b) ΔV_C: starts at 0, rises steeply then levels off, about 0.63 × 9.0 ≈ 5.7 V at t = 3.0 s, approaching 9.0 V (nearly there by 15 s, which is 5τ). I: starts at 0.30 mA, falls steeply then levels off, about 0.37 × 0.30 ≈ 0.11 mA at 3.0 s, approaching 0.
(c) With 60 kΩ: τ = 6.0 s, so both dashed curves change more slowly. The initial current halves to 0.15 mA. The final ΔV_C (9.0 V), the final charge (900 μC) and the final stored energy (½(100 × 10⁻⁶ F)(9.0 V)² = 4.05 mJ) do not change; only the time to get there does.
(d) Connect the voltmeter across the capacitor. Close the switch and start the timer together. Record ΔV_C at regular intervals (for example every 1 s) until it stops changing. Plot ΔV_C against t and read the time at which ΔV_C = 0.63 × its final value. Repeat and average; or discharge the capacitor and find when ΔV_C falls to 0.37 of its starting value.
| Point | What earns it |
|---|---|
| 1 | τ = 3.0 s, I₀ = 0.30 mA and Q = 900 μC, with units |
| 1 | ΔV_C sketch: from 0, concave down, levelling towards 9.0 V, about 5.7 V marked at τ |
| 1 | I sketch: from 0.30 mA, concave up, decreasing towards 0, about 0.11 mA marked at τ |
| 1 | Dashed curves rise or fall more slowly, with τ = 6.0 s |
| 1 | Initial current halves; final ΔV_C, charge and energy unchanged |
| 1 | Procedure: voltmeter across C, readings at regular times, graph, 63% (or 37% on discharge) to find τ |
Do not award the sketch points for straight lines or for curves that reach their final values at a definite time.
Question 7 (constructed response · stretch)
An ideal 18 V battery, a switch S and R₁ = 3.0 kΩ are in series. After R₁ the circuit splits at junction J into two branches that rejoin at junction K, which connects back to the battery. Branch 1 is R₂ = 6.0 kΩ. Branch 2 is R₃ = 2.0 kΩ in series with an uncharged capacitor C = 40 μF.
(a) Find the currents in R₁, R₂ and R₃ immediately after S is closed. (b) Find the currents, ΔV_C, the charge on C and the stored energy after S has been closed for a long time. (c) S is then opened. A student says: “With the battery disconnected, the current in R₂ drops to zero immediately.” Evaluate the claim. Include the time constant of the discharge and the current in R₂ just after S opens.
Worked solution and suggested Marlbridge rubric
(a) The uncharged capacitor acts like a wire, so branch 2 is just R₃. R₂ ∥ R₃ = (6.0 × 2.0)/(6.0 + 2.0) = 1.5 kΩ. Total = 4.5 kΩ. I₁ = 18 V ÷ 4.5 kΩ = 4.0 mA. ΔV across the branches = (4.0 mA)(1.5 kΩ) = 6.0 V, so I₂ = 6.0 V ÷ 6.0 kΩ = 1.0 mA and I₃ = 6.0 V ÷ 2.0 kΩ = 3.0 mA. Junction check: 1.0 + 3.0 = 4.0 mA.
(b) The charged capacitor acts like a break: I₃ = 0. Then I₁ = I₂ = 18 V ÷ 9.0 kΩ = 2.0 mA. ΔV across R₂ = 12 V. There is no current in R₃, so no ΔV across it, and ΔV_C = 12 V. Q = (40 μF)(12 V) = 480 μC. U = ½(40 × 10⁻⁶ F)(12 V)² = 2.88 mJ (2.9 mJ).
(c) The claim is wrong. Opening S removes the battery branch, but C is still connected in a loop with R₃ and R₂. The capacitor now drives the current: just after S opens, I = 12 V ÷ (2.0 + 6.0) kΩ = 1.5 mA in R₂ (in the same direction as before, J to K) and in R₃ (reversed). The discharge time constant uses the resistance in the discharge path: τ = (8.0 × 10³ Ω)(40 × 10⁻⁶ F) = 0.32 s. The current in R₂ then falls smoothly towards zero; after about 0.32 s the charge on C is about 37% of 480 μC, roughly 180 μC.
| Point | What earns it |
|---|---|
| 1 | (a) Treats the uncharged capacitor as a wire, giving I₁ = 4.0 mA |
| 1 | (a) I₂ = 1.0 mA and I₃ = 3.0 mA, consistent with the junction rule |
| 1 | (b) I₃ = 0 and I₁ = I₂ = 2.0 mA |
| 1 | (b) ΔV_C = 12 V, justified by zero ΔV across R₃ |
| 1 | (b) Q = 480 μC and U = 2.88 mJ |
| 1 | (c) Rejects the claim: C drives current through R₃ and R₂ after S opens |
| 1 | (c) τ = 0.32 s using R₂ + R₃, and initial current 1.5 mA |
Carry forward an error in ΔV_C from (b) into (c) once. Do not award the τ point for using R₁, which is no longer in a closed loop.
How did you do?
- Q1, Q2 or Q5 wrong: re-read “Capacitors in combination” in the study guide and redo Worked example 1.
- Q3 or Q7 wrong: use the two snapshots in “Capacitors in circuits with branches” and work through Worked example 2 again.
- Q4 or Q6 wrong: revisit “The time constant τ = RC” and Figure 2.
Then tick off the topic checklist.