IB Physics HL · guided topic map

Electric circuits for IB Physics HL

Electric circuits for IB Physics HL, organized into 1 syllabus topic and 8 mapped concept guides.

Syllabus topics: 1
Mapped concept guides: 8
Educational level: IB Diploma Physics Higher Level

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B.5

Current and circuits

The particulate nature of matter

8 guides

01Electric current and charge carriersSL + HL
02Potential difference and energy per chargeSL + HL
03Resistance and Ohm's lawSL + HL
04Series and parallel circuitsSL + HL
05Electrical powerSL + HL
06Fields driving currentSL + HL
07Heating effect of currentSL + HL
08Electric cellsSL + HL

Diagrams

Electric circuits as IB Physics HL draws it

The figures from the IB Physics HL practice papers that sit on these syllabus points — the apparatus, circuits and graphs an exam question actually puts in front of you.

01Figure 1Current and circuitsIB

Figure comment

Figure 1Two wires of the same material are drawn one above the other, each as a long bar seen from the side. The upper one is labelled wire 1: length L, diameter d, resistance R, with a dimension line under it marking L and a small dimension at its end marking the diameter d. The lower one is drawn twice as long and twice as thick and labelled wire 2: same material, with dimension lines marking 2L along it and 2d across its end.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

B.5 Current and circuits Open IB question · The particulate nature of matter Q4

Guided questions 5 parts

Reading cue. The 2d across the end of the lower wire is a diameter, not an area: doubling it multiplies the cross-section by four, so wire 2 is twice as long but four times as wide in area.

aState State how the resistivity of wire 2 compares with that of wire 1, and give a reason.
recall2 marks
Check answer 2 marks
1. it is the same
2. resistivity is a property of the material and its temperature, not of the length or thickness of the specimen
bDetermine Wire 1 has resistance R. Determine the length of wire of the lower wire's thickness that would be needed to give a resistance of R, and state how it compares with the 2L drawn.
routine3 marks
Check answer 3 marks
1. A ∝ d², so the lower wire's cross-sectional area is four times that of wire 1 and its resistance per unit length is R/(4L)
2. for a resistance of R the length must be 4L
3. that is twice the 2L drawn, so wire 2 as drawn has a resistance of only R/2
cDetermine The two wires are connected in turn across the same battery. Determine the ratio of the current in wire 2 to the current in wire 1, and the ratio of the electron drift speeds in them.
demanding3 marks
Check answer 3 marks
1. the same p.d. is applied to each, so I ∝ 1/R and I₂/I₁ = R/(R/2) = 2
2. I = nAvq with the same n and q in both, so v = I/(nAq)
3. A₂ = 4A₁ and I₂ = 2I₁, so v₂/v₁ = 2/4 = 0.5 — the electrons drift at half the speed in the thicker wire
dExplain The two wires are now joined end to end and the pair is connected across a supply. Explain which wire dissipates the greater power, and what happens to the electron drift speed as the electrons cross the junction.
top of the paper4 marks
Check answer 4 marks
1. in series the current is the same in both, so P = I²R and wire 1 dissipates twice the power of wire 2 (R against R/2)
2. charge is conserved, so the same number of electrons per second passes every cross-section, including the junction
3. I = nAvq with A four times larger in wire 2, so the drift speed there is one quarter of that in wire 1
4. the electrons neither pile up nor speed up at the junction: the current is fixed and only the drift speed adjusts to the area

Transfer challenge

A rectangular block of carbon measures 4.0 cm × 2.0 cm × 1.0 cm and has resistivity 3.5 × 10⁻⁵ Ω m. Determine its resistance between the two 2.0 cm × 1.0 cm faces, and between the two 4.0 cm × 2.0 cm faces.

Check answer 4 marks

between the small faces: L = 0.040 m, A = 0.020 × 0.010 = 2.0 × 10⁻⁴ m², so R = 3.5 × 10⁻⁵ × 0.040 / 2.0 × 10⁻⁴ = 7.0 × 10⁻³ Ω
between the large faces: L = 0.010 m, A = 0.040 × 0.020 = 8.0 × 10⁻⁴ m², so R = 3.5 × 10⁻⁵ × 0.010 / 8.0 × 10⁻⁴ = 4.4 × 10⁻⁴ Ω
the ratio is 16, for the same block of the same material
L and A in R = ρL/A are set by which pair of faces carries the current, not by the shape of the object

02Figure 4Thermodynamics · Current and circuitsIB

Figure comment

Figure 4Circuit diagram of a single series loop. On the left a dashed boundary encloses the cell itself: a cell symbol with its long positive plate uppermost, labelled e.m.f. 6.0 V, in series with a resistor labelled r = 0.75 Ω. The two terminals cross the dashed boundary and the loop continues round to a resistor in the upper wire labelled R = 3.0 Ω. An arrow on the lower wire, labelled I, marks the direction of the conventional current.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

B.4 Thermodynamics B.5 Current and circuits Open IB question · The particulate nature of matter Q10

Guided questions 5 parts

Reading cue. The dashed boundary is not a component: r = 0.75 Ω sits inside the cell, so 6.0 V is the e.m.f. of the whole loop and never the reading across the terminals while current flows.

aState The loop is broken at R and a voltmeter of very high resistance is connected across the cell's terminals. State its reading, and give a reason.
recall2 marks
Check answer 2 marks
1. 6.0 V
2. with no current there is no potential difference across r, so the terminal p.d. equals the e.m.f.
bDetermine The current marked I is 1.6 A. Determine the charge that passes through R in 2.0 minutes, and the chemical energy the cell converts in that time.
routine3 marks
Check answer 3 marks
1. Q = It = 1.6 × 120 = 192 C
2. the e.m.f. is the energy the cell gives to each coulomb, 6.0 J C⁻¹
3. E = εQ = 6.0 × 192 = 1.2 × 10³ J
cDetermine A second 3.0 Ω resistor is connected in parallel with R. Determine the new terminal potential difference of the cell, and state what happens to the e.m.f.
demanding4 marks
Check answer 4 marks
1. external resistance becomes 1.5 Ω
2. I = 6.0 / (1.5 + 0.75) = 2.67 A
3. terminal p.d. = 2.67 × 1.5 = 4.0 V
4. the e.m.f. stays at 6.0 V; only the lost volts Ir has changed, from 1.2 V to 2.0 V
dShow (that) Show that the power delivered to the external resistor is greatest when R equals r, determine that maximum power for the cell drawn, and state the power actually delivered to the 3.0 Ω resistor.
top of the paper4 marks
Show a hint
No calculus is needed: try rewriting (R + r)²/R as (R − r)²/R + 4r.
Check answer 4 marks
1. P = I²R = ε²R/(R + r)², which can be written P = ε²/[(R + r)²/R]
2. (R + r)²/R = (R − r)²/R + 4r, so the denominator is least, and P greatest, when R = r
3. maximum power = ε²/(4r) = 6.0² / (4 × 0.75) = 12 W
4. with R = 3.0 Ω the circuit delivers only I²R = 1.6² × 3.0 = 7.7 W, because R is four times r

Transfer challenge

Four identical cells, each of e.m.f. 1.5 V and internal resistance 0.30 Ω, are connected in series with a lamp of resistance 2.0 Ω. Determine the current and the potential difference across the terminals of one cell.

Check answer 4 marks

e.m.f.s in series add to 4 × 1.5 = 6.0 V, and the internal resistances add to 4 × 0.30 = 1.2 Ω
I = 6.0 / (2.0 + 1.2) = 1.875 A, so 1.9 A to two significant figures
terminal p.d. of one cell = 1.5 − 1.875 × 0.30 = 0.94 V
the four terminal p.d.s sum to 3.75 V across the lamp, so 2.25 V of lost volts is now spread over four internal resistances instead of one