Electric Circuits: Unit Diagnostic (Physics 2 Unit 11)

What this diagnostic is for

Use this short check to find which topics in Unit 11 (Electric Circuits) you should revisit. It has eight multiple-choice and two short-answer questions. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your total does not predict an exam score. What matters is which questions you miss.

How to take it:

Work without notes, in about 30 minutes. Write a reason for each answer.
A scientific calculator is allowed. Data: e = 1.60 × 10⁻¹⁹ C. Batteries, wires and meters are ideal unless a question says otherwise. Currents are conventional currents.
Open each answer only after you have committed to your own.

Question 1 (multiple choice · 11.1)

In 20 s, 6.0 × 10¹⁹ electrons pass a point in a wire, moving to the left. What is the current in the wire?

(A) 0.48 A, directed to the left
(B) 9.6 A, directed to the right
(C) 0.48 A, directed to the right
(D) 3.0 × 10¹⁸ A, directed to the left

Answer and explanation

Answer: (C). The charge is ΔQ = Ne = (6.0 × 10¹⁹)(1.60 × 10⁻¹⁹ C) = 9.6 C, so I = ΔQ/Δt = 9.6 C ÷ 20 s = 0.48 A. Electrons are negative, so conventional current points opposite to their motion: to the right.

(A) has the right size but follows the electrons.
(B) is the charge, not the charge per second.
(D) counts electrons per second without multiplying by e, and follows the electrons.

If you missed this: read “What current measures” and “Which way is the current?” in the Topic 11.1 study guide.

Question 2 (multiple choice · 11.2)

A battery is connected to bulb X. After X the wire splits into two branches that rejoin before returning to the battery. One branch holds bulb Y. The other holds bulbs Z and W, one after the other. The filament of Z breaks. Which bulbs still carry a current?

(A) X and Y
(B) X only
(C) X, Y and W
(D) None of them, because the circuit is now open

Answer and explanation

Answer: (A). A bulb carries current only if it is in a closed loop with the battery. The loop battery → X → Y → battery is still closed, so X and Y stay lit. Z and W are only in loops that pass through Z’s broken filament, so both go out.

(B) treats the break as opening every branch. Only the loops through Z are broken.
(C) forgets that W’s only path runs through Z.
(D) applies the single-loop rule to a circuit with more than one loop.

If you missed this: read “Closed, open and short circuits” and “Loops that share elements” in the Topic 11.2 study guide.

Question 3 (multiple choice · 11.3)

A uniform metal wire has a resistance of 12 Ω. It is stretched evenly until it is three times as long. Its volume and resistivity do not change. What is its new resistance?

(A) 36 Ω
(B) 12 Ω
(C) 324 Ω
(D) 108 Ω

Answer and explanation

Answer: (D). R = ρL/A. The volume LA is fixed, so tripling L makes A one-third as large. Both changes raise R by a factor of 3: R = 12 Ω × 3 × 3 = 108 Ω.

(A) changes the length but forgets that the wire also gets thinner.
(B) confuses resistance with resistivity, which depends only on the material.
(C) divides the area by 9, as if the diameter fell by a factor of 3. It is the area that falls by 3.

If you missed this: read “Resistance from shape and material” in the Topic 11.3 study guide.

Question 4 (multiple choice · 11.4)

A small motor is connected to an ideal 9.0 V battery. The current in the motor is 0.20 A. How much energy does the battery transfer to the motor in 5.0 minutes?

(A) 9.0 J
(B) 540 J
(C) 1.8 J
(D) 2.7 × 10³ J

Answer and explanation

Answer: (B). P = IΔV works for any circuit element, including a motor: P = (0.20 A)(9.0 V) = 1.8 W. Then ΔE = PΔt = (1.8 W)(300 s) = 540 J. You do not need the motor’s resistance.

(A) uses 5.0 instead of 300 s.
(C) is the power, a rate, not the energy.
(D) multiplies ΔV by time and leaves out the current.

If you missed this: read “Power as a rate of energy transfer” in the Topic 11.4 study guide.

Question 5 (multiple choice · 11.5)

Resistors R₁ and R₂ are in series across an ideal battery. A third resistor, R₃, is then connected in parallel with R₂. What happens to the current from the battery and to the potential difference across R₁?

(A) The current increases; the potential difference across R₁ stays the same.
(B) The current decreases; the potential difference across R₁ decreases.
(C) The current increases; the potential difference across R₁ increases.
(D) Both stay the same, because the battery is ideal.

Answer and explanation

Answer: (C). Adding a parallel path lowers the resistance of that part, so the equivalent resistance of the circuit falls. With a fixed emf the battery current rises. All of it passes through R₁, so ΔV₁ = IR₁ rises, and the potential difference across the R₂–R₃ pair falls.

(A) forgets that ΔV₁ = IR₁ follows the current.
(B) assumes more resistors always mean more resistance. That is true in series, not in parallel.
(D) confuses a fixed emf with a fixed current.

If you missed this: read “Equivalent resistance in parallel” and “Reducing a compound network” in the Topic 11.5 study guide.

Question 6 (multiple choice · 11.6)

Two branches are connected in parallel across an ideal 12 V battery. Branch 1 has a 2.0 Ω resistor next to the positive terminal, then a 4.0 Ω resistor. Branch 2 has a 4.0 Ω resistor next to the positive terminal, then a 2.0 Ω resistor. Point X lies between the resistors in branch 1, and point Y between the resistors in branch 2. What is V_X − V_Y?

(A) +4.0 V
(B) 0 V
(C) −4.0 V
(D) +8.0 V

Answer and explanation

Answer: (A). Each branch has 12 V across 6.0 Ω, so each carries 2.0 A. Take the negative terminal as 0 V. Going down branch 1, the potential falls by (2.0 A)(2.0 Ω) = 4.0 V, so V_X = 8.0 V. Going down branch 2, it falls by (2.0 A)(4.0 Ω) = 8.0 V, so V_Y = 4.0 V. V_X − V_Y = +4.0 V.

(B) assumes both points are “halfway”. The order of the resistors matters.
(C) swaps the potential falls between the branches.
(D) is V_X measured from the negative terminal, not from Y.

If you missed this: read “The loop rule” and “Potential–position graphs” in the Topic 11.6 study guide.

Question 7 (multiple choice · 11.7)

Three wires meet at a junction in a circuit with steady currents. Wire 1 carries 0.75 A into the junction. Each minute, 27 C of charge leaves the junction along wire 2. What is the current in wire 3?

(A) 1.2 A out of the junction
(B) 0.30 A into the junction
(C) 26 A into the junction
(D) 0.30 A out of the junction

Answer and explanation

Answer: (D). First convert wire 2 to a current: 27 C ÷ 60 s = 0.45 A out. Charge cannot build up at the junction, so current in = current out: 0.75 A = 0.45 A + I₃, giving I₃ = 0.30 A out.

(A) adds the two currents instead of balancing them.
(B) has the wrong direction: more arrives than leaves along wire 2.
(C) mixes coulombs per minute with amperes.

If you missed this: read “The junction rule” in the Topic 11.7 study guide.

Question 8 (multiple choice · 11.8)

Three identical capacitors each have capacitance C. Two are connected in parallel, and that pair is connected in series with the third. What is the equivalent capacitance?

(A) 3C
(B) 2C/3
(C) C/3
(D) 3C/2

Answer and explanation

Answer: (B). The parallel pair adds: 2C. In series with C: 1/C_eq = 1/(2C) + 1/C = 3/(2C), so C_eq = 2C/3. It is less than C, as a series combination must be.

(A) puts all three in parallel.
(C) puts all three in series.
(D) is a series pair (C/2) in parallel with the third, the other arrangement.

If you missed this: read “Capacitors in combination” in the Topic 11.8 study guide.

Question 9 (short answer · 11.5)

An ideal voltmeter reads 1.50 V across the terminals of a cell with nothing else connected. When a 4.0 Ω lamp is connected across the cell, the voltmeter reads 1.20 V.

(a) Calculate the current in the lamp. (b) Calculate the internal resistance of the cell. (c) What fraction of the energy transferred by the cell’s emf reaches the lamp?

Worked answer

(a) The lamp has the terminal voltage across it: I = 1.20 V ÷ 4.0 Ω = 0.30 A.
(b) With no current the reading is the emf, so ℰ = 1.50 V. The “lost” 0.30 V is across r: r = (ℰ − ΔV_terminal)/I = 0.30 V ÷ 0.30 A = 1.0 Ω.
(c) Power from the emf: ℰI = (1.50 V)(0.30 A) = 0.45 W. Lamp: I²R = (0.30 A)²(4.0 Ω) = 0.36 W. Fraction = 0.36 ÷ 0.45 = 0.80.

A common slip in (b) is r = 1.50 V ÷ 0.30 A = 5.0 Ω, which is the whole circuit’s resistance.

If you missed this: read “Batteries: emf and internal resistance” in the Topic 11.5 study guide.

Question 10 (short answer · 11.8)

An uncharged capacitor, a 20 kΩ resistor, a switch and an ideal 12 V battery are in series. The switch is closed at t = 0. After a long time the capacitor holds 3.0 mC.

(a) Find the capacitance, the time constant and the starting current. (b) Estimate the potential differences across the capacitor and the resistor at t = 5.0 s. (c) The experiment is repeated with a 24 V battery. A student says: “The capacitor will charge twice as fast.” Evaluate the claim.

Worked answer

(a) After a long time there is no current, so ΔV_C = ℰ and C = Q/ℰ = 3.0 × 10⁻³ C ÷ 12 V = 250 μF. τ = RC = (20 × 10³ Ω)(250 × 10⁻⁶ F) = 5.0 s. At first the capacitor acts like a wire: I₀ = ℰ/R = 0.60 mA.
(b) 5.0 s is one time constant: ΔV_C ≈ 0.63 × 12 V ≈ 7.6 V. The loop rule leaves 12 V − 7.6 V ≈ 4.4 V across the resistor.
(c) The claim is wrong. τ = RC does not contain the emf, so the capacitor still takes 5.0 s to reach 63% of its final charge. The final charge (6.0 mC) and starting current (1.2 mA) both double.

If you missed this: read “Charging a capacitor through a resistor” and “The time constant τ = RC” in the Topic 11.8 study guide.

Your next step

Mark each question. Count a short answer as wrong if any part is missing, and a guess as wrong.

Topic	Question(s)	If you missed it, read
11.1 Electric current	1	Topic 11.1 study guide
11.2 Simple circuits	2	Topic 11.2 study guide
11.3 Resistance, resistivity and Ohm’s law	3	Topic 11.3 study guide
11.4 Electric power	4	Topic 11.4 study guide
11.5 Compound DC circuits	5, 9	Topic 11.5 study guide
11.6 Kirchhoff’s loop rule	6	Topic 11.6 study guide
11.7 Kirchhoff’s junction rule	7	Topic 11.7 study guide
11.8 RC circuits	8, 10	Topic 11.8 study guide

How to use your result

Start with the topic where you missed the most. Read its study guide, then do that topic’s practice questions.
Look at why you missed it. The explanation names the mistake behind each distractor. A units or direction slip (minutes, coulombs per minute, electron flow) needs a habit, not a full re-read.
Missed nothing? Go straight to the Unit 11 mixed review. Its questions combine topics, which is harder.
Come back to this check after a week of revision, and answer again before opening the explanations.