What this is for. This diagnostic shows which Unit 11 topics to revisit, with at least one question per topic. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your result is not a predicted score.
How to take it. Work without notes for about 30 minutes, writing each answer before opening the explanation. A scientific calculator is assumed. Batteries, wires and meters are ideal unless a question says otherwise, resistors are ohmic, and current means conventional current. All data are invented for practice.
Question 1 (multiple choice · 11.1)
A wire of radius R carries a current whose density is directed along the wire and rises linearly from zero at the centre: J(r) = J₀r/R. What is the total current?
- (A) 2πJ₀R²/3
- (B) πJ₀R²
- (C) πJ₀R²/2
- (D) πJ₀R²/3
Answer and explanation
Answer: (A). Use rings of area 2πr dr: I = ∫₀ᴿ (J₀r/R)(2πr) dr = (2πJ₀/R)(R³/3) = 2πJ₀R²/3.
- (B) multiplies the largest J by the whole area, as if J were uniform.
- (C) uses the average J over the radius, J₀/2. Outer rings have more area, so they count for more.
- (D) is the result for J = J₀(1 − r/R), which is largest on the axis instead.
If you missed this: read “Non-uniform current density: I = ∫J·dA” in the 11.1 study guide.
Question 2 (multiple choice · 11.2)
A battery, a resistor R and an uncharged capacitor are connected in a single loop. A bulb is then connected across the capacitor, joined to the wires on each side of it. Which statement about the closed loops is correct?
- (A) There are three loops, and each element is in exactly two of them.
- (B) There is one loop (battery, R, bulb), because charge cannot cross the capacitor’s gap.
- (C) There are two loops: battery–R–capacitor and battery–R–bulb.
- (D) There are three loops, and the battery is in all three.
Answer and explanation
Answer: (A). The loops are battery–R–capacitor, battery–R–bulb and capacitor–bulb. The battery and R are in the first two; the capacitor is in the first and third; the bulb is in the second and third.
- (B) A capacitor is a loop element. Charge flows round loops containing it while it charges or discharges.
- (C) misses the small capacitor–bulb loop, through which the capacitor can discharge.
- (D) The capacitor–bulb loop does not pass through the battery.
If you missed this: read “Loops: one element, several loops” and “Capacitors, inductors and meters in loops” in the 11.2 study guide.
Question 3 (multiple choice · 11.3)
A solid rod of radius a and a hollow tube with inner radius a and outer radius 2a are made of the same metal and have the same length. Current flows along each. What is R_tube/R_rod?
- (A) 1/3
- (B) 1/4
- (C) 1/2
- (D) 3
Answer and explanation
Answer: (A). R = ρℓ/A, with ρ and ℓ the same. The tube’s metal has area π(2a)² − πa² = 3πa², three times the rod’s πa². So R_tube = R_rod/3.
- (B) uses the tube’s full outer area, π(2a)², and forgets the hole.
- (C) compares radii, not areas.
- (D) inverts the ratio: more area means less resistance.
If you missed this: read “From E = ρJ to R = ρℓ/A” in the 11.3 study guide.
Question 4 (multiple choice · 11.4)
A heating wire connected to a fixed supply dissipates power P. The wire is cut into two equal halves, and the halves are connected in parallel to the same supply. What is the new total power?
- (A) 4P
- (B) 2P
- (C) P
- (D) P/4
Answer and explanation
Answer: (A). Each half has resistance R/2, and two in parallel give R/4. With ΔV fixed, P = ΔV²/R, so the power becomes 4P.
- (B) finds one half’s resistance, R/2, but does not combine the two in parallel.
- (C) The new arrangement has a different resistance.
- (D) uses P = I²R as if I stayed the same. With a fixed supply, ΔV stays the same.
If you missed this: read “Power in a resistor: three forms” in the 11.4 study guide.
Question 5 (multiple choice · 11.5)
A 10 kΩ resistor and a 30 kΩ resistor are in series across an ideal 12 V battery. A voltmeter with resistance 60 kΩ is connected across the 30 kΩ resistor. What does it read?
- (A) 8.0 V
- (B) 9.0 V
- (C) 4.0 V
- (D) 3.0 V
Answer and explanation
Answer: (A). The meter and the 30 kΩ resistor form a 20 kΩ parallel pair, in series with the 10 kΩ. The 12 V divides in the ratio 20 : 10, so the meter reads 12 × 20/30 = 8.0 V.
- (B) is the reading an ideal voltmeter would give. This meter draws current.
- (C) is the potential difference across the other resistor.
- (D) is the ideal potential difference across the 10 kΩ resistor.
If you missed this: read “Measuring current and potential difference” in the 11.5 study guide.
Question 6 (short answer · 11.5 and 11.4)
A battery of emf 12 V and internal resistance 1.0 Ω is connected to two branches in parallel: R₁ = 6.0 Ω, and R₂ = 2.0 Ω in series with R₃ = 4.0 Ω.
(a) Find the battery current and the terminal voltage. (b) Find the current in R₃ and the power dissipated in it. (c) The R₁ branch is disconnected. Find the new power in R₃, and explain why the current in R₃ changes, although its own branch is unchanged.
Answer and explanation
(a) The branches are 6.0 Ω and 2.0 + 4.0 = 6.0 Ω, so in parallel they give 3.0 Ω. I = ℰ/(R_eq + r) = 12 ÷ 4.0 = 3.0 A. Terminal voltage ΔV = ℰ − Ir = 12 − 3.0 = 9.0 V.
(b) The R₂–R₃ branch has 9.0 V across it: I₃ = 9.0 ÷ 6.0 = 1.5 A, and P₃ = I²R₃ = (1.5)²(4.0) = 9.0 W.
(c) Now I = 12 ÷ (6.0 + 1.0) = 1.71 A, all of it in R₃, so P₃ = (1.714)²(4.0) = 11.8 W. The battery current is smaller, so less is lost across r: the terminal voltage rises to 10.3 V and drives more current through R₃. With an ideal battery, I₃ would stay at 2.0 A.
Check yourself: 1 mark each for the current and terminal voltage, I₃ and P₃, the new P₃, and the explanation in terms of Ir (4 in total).
If you missed this: read “Real batteries and real wires” in the 11.5 study guide; for P₃, “Power in a resistor: three forms” in the 11.4 study guide.
Question 7 (multiple choice · 11.6)
A single loop is followed clockwise from point A. It passes through a 9.0 V battery from − to +, then a 2.0 Ω resistor to point B, then a 3.0 V battery from + to −, then a 4.0 Ω resistor back to A. What is V_B − V_A?
- (A) +7.0 V
- (B) +5.0 V
- (C) +9.0 V
- (D) −7.0 V
Answer and explanation
Answer: (A). The batteries oppose each other, so I = (9.0 − 3.0) ÷ (2.0 + 4.0) = 1.0 A, clockwise. From A to B: +9.0 − (1.0)(2.0) = +7.0 V. Check: going from A to B the other way, back through the 4.0 Ω (+4.0 V) and the 3.0 V battery from − to + (+3.0 V), also gives 7.0 V.
- (B) adds the emfs, giving I = 2.0 A.
- (C) forgets the drop across the 2.0 Ω resistor.
- (D) has the sign reversed: potential falls along the current in a resistor.
If you missed this: read “Writing a loop equation” in the 11.6 study guide.
Question 8 (multiple choice · 11.7)
At junction J, 1.20 A arrives from a battery. A student draws I₁, I₂ and I₃ all leaving J along three other wires. Solving Kirchhoff’s equations gives I₁ = 0.80 A and I₂ = −0.30 A. What is I₃?
- (A) 0.70 A, leaving J
- (B) 0.10 A, leaving J
- (C) 0.40 A, leaving J
- (D) 0.70 A, entering J
Answer and explanation
Answer: (A). In = out: 1.20 = 0.80 + (−0.30) + I₃, so I₃ = 0.70 A. It is positive, so it flows the way it was drawn, out of J. The negative I₂ simply means 0.30 A flows into J along that wire.
- (B) treats I₂ as 0.30 A leaving J, ignoring its sign.
- (C) leaves out the I₂ wire.
- (D) A positive result means the drawn direction is right.
If you missed this: read “Using the rule” and “Combining the junction and loop rules” in the 11.7 study guide.
Question 9 (multiple choice · 11.8)
Uncharged 2.0 μF and 8.0 μF capacitors are connected in series across a 10 V battery. Which statement is correct once they are charged?
- (A) Each holds 16 μC; the 2.0 μF capacitor has 8.0 V across it.
- (B) Each holds 16 μC; the 8.0 μF capacitor has 8.0 V across it.
- (C) They hold 20 μC and 80 μC; each has 10 V across it.
- (D) Each holds 100 μC; together they have 10 V across them.
Answer and explanation
Answer: (A). C_eq = (1/2.0 + 1/8.0)⁻¹ = 1.6 μF, so Q = (1.6 μF)(10 V) = 16 μC on each, because charge is conserved on the wire between them. ΔV = Q/C gives 8.0 V on 2.0 μF and 2.0 V on 8.0 μF.
- (B) The smaller capacitance needs the larger ΔV for the same Q.
- (C) is the parallel result.
- (D) adds the capacitances as if they were resistors in series: 10 μF × 10 V.
If you missed this: read “Capacitors in combination” in the 11.8 study guide.
Question 10 (short answer · 11.8)
A 9.0 V battery, a switch and R₁ = 6.0 kΩ are in series with a parallel pair: R₂ = 12 kΩ and an uncharged 25 μF capacitor. The switch closes at t = 0.
(a) Find the current in R₁ and in R₂ just after the switch closes. (b) Find the current in R₂ and the charge on the capacitor a long time later. (c) The switch is then opened, so the capacitor discharges through R₂ alone. Find the time constant, and the time for the current in R₂ to fall to 25% of its starting value. (d) Find the energy dissipated in R₂ during the whole discharge.
Answer and explanation
(a) An uncharged capacitor acts like a wire, so ΔV across R₂ is zero: I₂ = 0. All the current, 9.0 V ÷ 6.0 kΩ = 1.5 mA, flows through R₁ and into the capacitor.
(b) No current flows in the capacitor’s branch, so R₁ and R₂ are in series: I = 9.0 ÷ 18 kΩ = 0.50 mA. The capacitor has the same ΔV as R₂, (0.50 mA)(12 kΩ) = 6.0 V, so Q = CΔV = 150 μC.
(c) τ = R₂C = (12 × 10³)(25 × 10⁻⁶) = 0.30 s. The current is I₀e^(−t/τ), so e^(−t/τ) = 0.25 gives t = τ ln 4 = 0.42 s (0.416 s).
(d) All the stored energy is dissipated: ½C(ΔV)² = ½(25 × 10⁻⁶)(6.0)² = 4.5 × 10⁻⁴ J.
Check yourself: 1 mark each for (a), (b), τ with the time, and the energy (4 in total). Putting all 9.0 V across the capacitor in (b) is the common slip: R₁ takes 3.0 V.
If you missed this: read “Just after and long after”, “Discharging” and “Energy in an RC circuit” in the 11.8 study guide.
Your next step
Count a short answer as missed if you lost more than one mark.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 11.1 Electric current | 1 | 11.1 study guide |
| 11.2 Simple circuits | 2 | 11.2 study guide |
| 11.3 Resistance, resistivity and Ohm’s law | 3 | 11.3 study guide |
| 11.4 Electric power | 4, 6(b)–(c) | 11.4 study guide |
| 11.5 Compound direct current circuits | 5, 6 | 11.5 study guide |
| 11.6 Kirchhoff’s loop rule | 7 | 11.6 study guide |
| 11.7 Kirchhoff’s junction rule | 8 | 11.7 study guide |
| 11.8 Resistor-capacitor (RC) circuits | 9, 10 | 11.8 study guide |
How to use your result
- Missed nothing in a topic? Skip its study guide for now and go straight to the mixed unit review, which joins the topics together.
- Missed one or two topics? Read those study guides, then do their practice sets before the review.
- Missed four or more topics? Work through 11.1 to 11.8 in order; later topics build on the earlier ones.
- Got it right but guessed? Treat it as missed.