Term 1 review: circuits, magnetism, induction and AC
Week 15 · Term 1 review (Chapters 1–4) and Darslab’s own review tasks · 3 lessons
1
Review: current electricity and magnetic effects
Physics Sec 3 Term 1, Chapters 1–4: Review
GoalCombine the tools of Chapters 1 and 2 in multi-step tasks: circuits with internal resistance, Kirchhoff’s laws, magnetic fields, forces, torque and the three meters.
New words
terminal voltage · V = VB − Ir: the voltage across the source while it delivers a currentsign rule · fix the assumed currents and the loop direction first, then write every term with its signfull-scale current Ig · the current that deflects a galvanometer pointer to the end of the scale
Explanation
Chapters 1 and 2 give a toolbox, shown in the table. For a circuit, find the equivalent resistance in steps: combine parallel parts, add series parts, find the total current I = VB ÷ (R + r), then go back to the branches. For a two-loop circuit use the Week 3 rules: assume a current in each branch, choose a direction round each loop, count IR as positive when the assumed current agrees with the loop direction, and count an emf as positive when the loop crosses the battery from its negative to its positive pole. A negative answer means that the real current runs opposite to the assumed one. For magnetism, keep μ0 = 1.256 × 10⁻⁶ T·m/A and say which angle you use: θ is the angle that the wire makes with the field in F = BIℓ sin θ, but the angle that the normal to the coil makes with the field in τ = BIAN sin θ. For the meters, the shunt goes in parallel and is tiny, the multiplier goes in series and is large, and an ohmmeter pointer rests at half scale when Rx equals the instrument’s own internal resistance Ri. Always convert units first, write the formula, substitute with units, and check the size of the answer.
Topic
Tool
series, parallel
R = R₁ + R₂ + …; 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + …
source with internal resistance r
I = VB ÷ (R + r); V = VB − Ir
Kirchhoff
ΣI(in) = ΣI(out); ΣE = ΣIR with the Week 3 sign rules
fields
B = 2 × 10⁻⁷ I ÷ d; B = μ0NI ÷ 2r; B = μ0NI ÷ ℓ; μ0 = 1.256 × 10⁻⁶
force and torque
F = BIℓ sin θ; τ = BIAN sin θ
instruments
Rs = IgRg ÷ (I − Ig); Rm = V ÷ Ig − Rg; Ri = E ÷ Ig, I = E ÷ (Ri + Rx)
Worked examples
A battery of VB = 18 V and r = 1 Ω feeds a 10 Ω and a 15 Ω resistor in parallel, in series with a 3 Ω resistor. The pair: (10 × 15) ÷ 25 = 6 Ω. R = 6 + 3 = 9 Ω. I = 18 ÷ (9 + 1) = 1.8 A. Terminal voltage V = 18 − 1.8 × 1 = 16.2 V (also 1.8 × 9 = 16.2 V). Voltage across the pair: 1.8 × 6 = 10.8 V, so the branch currents are 10.8 ÷ 10 = 1.08 A and 10.8 ÷ 15 = 0.72 A, and 1.08 + 0.72 = 1.8 A ✓.
Branch
Contains
Assumed current
Result
1 (left)
E₁ = 8 V, R₁ = 1 Ω
I₁ upward
+4 A
3 (middle)
R₃ = 2 Ω
I₃ downward
+2 A
2 (right)
E₂ = 2 V, R₂ = 1 Ω
I₂ upward
−2 A
Two-loop circuit, laid out as in Week 3: two junctions P (top) and Q (bottom); branch 1 has battery E₁ = 8 V (+ pole up) and R₁ = 1 Ω; branch 2 has E₂ = 2 V (+ pole up) and R₂ = 1 Ω; the middle branch has R₃ = 2 Ω. Assumed currents: I₁ and I₂ upward, I₃ downward, so I₃ = I₁ + I₂. Loop A: 8 = I₁ + 2I₃ = 3I₁ + 2I₂. Loop B: 2 = I₂ + 2I₃ = 2I₁ + 3I₂. Multiply A by 3 and B by 2: 24 = 9I₁ + 6I₂ and 4 = 4I₁ + 6I₂. Subtract: 20 = 5I₁, so I₁ = 4 A. Then 2I₂ = 8 − 12, so I₂ = −2 A and I₃ = 2 A. Check in B: 2 × 4 + 3 × (−2) = 2 ✓. I₂ is negative: the real current in branch 2 flows downward, into the + pole, so battery 2 is being charged. The voltage across R₃ is 2 × 2 = 4 V.
A long solenoid has N = 800 turns on ℓ = 0.4 m and carries I = 2.5 A in air. B = μ0NI ÷ ℓ = 1.256 × 10⁻⁶ × 800 × 2.5 ÷ 0.4 = 6.28 × 10⁻³ T, along the axis. A straight wire 0.15 m long carries 4 A inside it at 90° to the field: F = BIℓ = 6.28 × 10⁻³ × 4 × 0.15 = 3.77 × 10⁻³ N. At 30° to the field the force is half of that, 1.88 × 10⁻³ N. If the field points east and the current flows north, Fleming’s left-hand rule (forefinger east, middle finger north) gives a force downward.
Class activity
Paper only. Make your own one-page summary of Chapters 1 and 2: the table of tools above, plus for each tool one line of “watch out” (for example: convert cm and mA first; which angle; parallel or series; the sign of a current). Then pick one task of this lesson and solve it again without looking at the worked example.
Practice
1
A galvanometer has Rg = 40 Ω and Ig = 5 mA. Find the shunt for a range of 1 A, and the multiplier for a range of 20 V.
Rs = IgRg ÷ (I − Ig) = 0.005 × 40 ÷ 0.995 = 0.201 Ω, in parallel. Rm = V ÷ Ig − Rg = 20 ÷ 0.005 − 40 = 4000 − 40 = 3960 Ω, in series.
2
Find B at 0.06 m from a long straight wire carrying 30 A, and at the centre of a circular coil of N = 20 turns, radius 0.05 m, carrying 1.5 A (μ0 = 1.256 × 10⁻⁶ T·m/A).
Wire: B = 2 × 10⁻⁷ × 30 ÷ 0.06 = 1.0 × 10⁻⁴ T. Coil: B = μ0NI ÷ 2r = 1.256 × 10⁻⁶ × 20 × 1.5 ÷ 0.10 = 3.77 × 10⁻⁴ T.
3
A coil of N = 60 turns measuring 0.08 m × 0.05 m carries 2.5 A in a field of 0.4 T. Find the torque when its plane is parallel to the field, and when the normal makes 30° with the field.
A = 0.08 × 0.05 = 4 × 10⁻³ m². Plane parallel (θ = 90°): τ = 0.4 × 2.5 × 4 × 10⁻³ × 60 = 0.24 N·m. Normal at 30°: τ = 0.24 × sin 30° = 0.12 N·m.
4
An ohmmeter has a cell of E = 3 V and a full-scale current Ig = 400 µA. Find Ri, and the pointer position (as a fraction of full scale) for Rx = 15 000 Ω.
Ri = E ÷ Ig = 3 ÷ (400 × 10⁻⁶) = 7500 Ω. I = 3 ÷ (7500 + 15 000) = 133 µA, which is 133 ÷ 400 = one third of full scale (Rx is two times Ri).