OxfordAQA International AS and A-level Physics: DC circuits, potential dividers, emf and internal resistance (9630) – Revision Notes

These notes are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.4.4, 3.4.5 and 3.4.6 (shared International AS and A-level content), for exams May/June 2020 onwards. The study guide explains each idea in full with longer worked examples. The practice set lets you test yourself. Related pages: course page, checklist and free diagnostics.

Key formulae

Quantity	Relationship
Series resistors	RT = R₁ + R₂ + R₃ + …
Parallel resistors	1/RT = 1/R₁ + 1/R₂ + 1/R₃ + …
Energy	E = IVt
Power	P = IV = I²R = V²/R
Potential divider	Vout = Vin × R₂/(R₁ + R₂), output across R₂
emf	ε = E/Q
Circuit with internal resistance	ε = I(R + r), so V = ε − Ir
Short-circuit current	ε/r

Ammeters are ideal (zero resistance) and voltmeters ideal (infinite resistance) unless a question says otherwise.

3.4.4 Circuits

Series versus parallel

	Series	Parallel
Current	Same everywhere	Splits between branches
Pd	Splits (adds to the supply pd)	Same across each branch
Total resistance	Greater than the biggest single value	Below the smallest single value
Power form to use	I²R (current common)	V²/R (pd common)

Method for a mixed circuit

1. Reduce each parallel group to one resistance.
2. Add the series parts to get RT.
3. Find the supply current, I = V/RT.
4. Work back out: pd across each group, then branch currents.
5. Check: branch currents sum to the total, and pds around the loop sum to the supply pd.

Cells

• In series: emfs add (same direction), internal resistances add.
• Identical cells in parallel: emf unchanged, internal resistance r/n for n cells.
• A parallel bank shares the current, so each cell supplies less and lasts longer.

Conservation laws

• Charge: current into a junction equals current out.
• Energy: round any loop, total emf equals total pd across components. Energy supplied per coulomb equals energy transferred per coulomb.

Worked reminder: cells with internal resistance

Two identical cells (2.0 V, 0.50 Ω each) drive a 3.5 Ω resistor. In series, ε = 4.0 V and r = 1.0 Ω, so I = 4.0/4.5 = 0.889 A. In parallel, ε = 2.0 V and r = 0.25 Ω, so I = 2.0/3.75 = 0.533 A.

Quick reminders

• A 12 V, 3.0 A heater for 300 s: E = 3.0 × 12 × 300 = 1.08 × 10⁴ J.
• For two resistors in series, the larger resistance gets the larger share of the power. In parallel, the smaller resistance gets the larger share.

3.4.5 Potential divider

Method in steps

1. Identify the resistor the output is taken across (call it R₂).
2. Vout = Vin × R₂/(R₁ + R₂).
3. If a load is connected across the output, replace R₂ with R₂ in parallel with the load, then recalculate.

Variable and sensor dividers

Component	What changes	Output across it	Output across the fixed resistor
Variable resistor	Set by the slider or knob	Rises as its resistance rises	Falls as its resistance rises
ntc thermistor	R falls as temperature rises	Falls as temperature rises	Rises as temperature rises
LDR	R falls as light level rises	Falls as light level rises	Rises as light level rises

A variable resistor connected as a divider (the two ends across the supply, the output taken from the slider) gives any pd from 0 V up to the supply pd. A loaded divider gives a smaller output than an unloaded one, and the effect grows as the load resistance gets smaller.

Worked reminder. To get 2.0 V from a 6.0 V supply with a 2.0 kΩ resistor across the output: 2.0 = 6.0 × 2.0/(R₁ + 2.0), so R₁ + 2.0 = 6.0 and R₁ = 4.0 kΩ.

3.4.6 emf and internal resistance

Key ideas

• emf: energy the source supplies per coulomb of charge it drives round the circuit, ε = E/Q.
• Terminal pd V = ε − Ir. The term Ir is the lost volts.
• With no current, V = ε. With more current, V is lower.
• Power: total εI, delivered VI, wasted inside I²r.
• A source with larger r loses more energy internally for the same current.
• Efficiency of the source: VI divided by εI, which equals V/ε, or R/(R + r).
• Largest current from a source is the short-circuit value, ε/r, which is why a real battery heats up if shorted.

Method for internal-resistance problems

1. Write ε = I(R + r).
2. Substitute known values and solve for the unknown (I, R or r).
3. Find V = IR or V = ε − Ir to check consistency.

Worked reminder. ε = 1.80 V, r = 0.60 Ω, R = 2.4 Ω. I = 1.80/3.0 = 0.60 A. V = 0.60 × 2.4 = 1.44 V. Lost volts = 0.36 V.

Required practical 3

• Arrangement: cell, variable resistor, ammeter and switch in series; voltmeter across the cell.
• Take at least six readings of V and I, opening the switch between readings.
• Plot V against I. Gradient = −r. Vertical intercept = ε.
• Use a large triangle for the gradient. Read the intercept from the line.
• Check for a systematic error: a voltmeter that does not read zero, or contact resistance in leads.

Must-know distinctions

• emf versus terminal pd: emf is energy per coulomb supplied by the source; terminal pd is the pd across the terminals when current flows, and it is lower.
• Total resistance versus branch resistance: parallel lowers RT.
• Series cells versus parallel cells: series raises emf, parallel does not.
• Divider unloaded versus loaded: the load lowers the output.
• Rheostat versus divider: a variable resistor in series cannot take the load pd to zero; a divider can.

Quick self-test

1. Find the resistance of 6.0 Ω, 12 Ω and 4.0 Ω resistors in parallel.
2. A 6.0 V supply drives 2.5 A for 40 s through a resistor. How much energy is transferred?
3. What power is dissipated in a 5.0 Ω resistor carrying 0.60 A?
4. A 100 Ω resistor is rated at 0.25 W. What is the largest pd across it, and the largest current?
5. A 9.0 V supply is across 3.0 kΩ in series with 6.0 kΩ. What is the pd across the 6.0 kΩ resistor?
6. In a divider with an ntc thermistor and a fixed resistor, the output is taken across the thermistor. What happens to the output as the temperature rises?
7. A 6.0 V source drives 1.2 A through a 4.0 Ω resistor. What is its internal resistance?
8. What is the terminal pd in question 7?
9. A 1.2 V cell has internal resistance 0.15 Ω. What is the short-circuit current?
10. Two identical 1.2 V cells, each of internal resistance 0.50 Ω, are in parallel. State the emf and internal resistance of the pair.
11. A V against I graph for a cell is a straight line from 1.60 V at zero current to 1.18 V at 0.30 A. State the emf and internal resistance.
12. Define the emf of a cell.

Answers

1. 1/R = 1/6 + 1/12 + 1/4 = 6/12 = 0.5, so 2.0 Ω.
2. E = IVt = 2.5 × 6.0 × 40 = 600 J.
3. P = I²R = 0.60² × 5.0 = 1.8 W.
4. V = √(PR) = √(0.25 × 100) = 5.0 V; I = √(P/R) = 0.050 A.
5. 9.0 × 6.0/9.0 = 6.0 V.
6. It falls, because the thermistor’s resistance falls and so does its share of the total.
7. ε/I = 6.0/1.2 = 5.0 Ω total, so r = 5.0 − 4.0 = 1.0 Ω.
8. V = IR = 1.2 × 4.0 = 4.8 V (check: 6.0 − 1.2 × 1.0 = 4.8 V).
9. I = ε/r = 1.2/0.15 = 8.0 A.
10. emf 1.2 V, internal resistance 0.50/2 = 0.25 Ω.
11. ε = 1.60 V; gradient = (1.18 − 1.60)/0.30 = −1.4 V A⁻¹, so r = 1.4 Ω.
12. The energy the cell supplies per coulomb of charge it drives round the circuit, ε = E/Q.

Where marks are usually lost

• Leaving 1/R as the answer without inverting it.
• Using V²/R with a pd taken from the wrong place in the circuit.
• Calling the terminal pd “the emf” in a circuit that carries a current.
• Forgetting to convert minutes to seconds or kΩ to Ω before using E = IVt.
• Ignoring the load on a divider when the question says one is connected.
• Writing that the “resistance” of a thermistor rises with temperature when the question concerns an ntc thermistor.
• Giving r as a negative number from the gradient, or using ε as the final data point.
• Not saying “per coulomb” or “per unit charge” when asked for the meaning of emf.
• Giving a final answer to far more significant figures than the data supports.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.4.4, 3.4.5 and 3.4.6.