OCR GCSE Physics A: Series and parallel circuit calculations and electrical power (J249) – Revision Notes

These revision notes condense the full study guide for the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), learning outcomes P3.2i to P3.2m. All of these statements are assessed in both tiers. Use them with the practice questions, the OCR GCSE Physics hub, the printable checklist and the free diagnostics.

Equations to know

Equation	Words	Units
V = I × R	p.d. = current × resistance	V, A, Ω
Q = I × t	charge = current × time	C, A, s
E = Q × V	energy transferred = charge × p.d.	J, C, V
P = V × I	power = p.d. × current	W, V, A
P = I² × R	power = current squared × resistance	W, A, Ω
E = P × t	energy = power × time	J, W, s or kW h, kW, h

Rearrangements to be fluent in: I = V / R, R = V / I, I = P / V, V = P / I, t = E / P, V = E / Q, Q = E / V.

P3.2i: qualitative explanation

• Series: one path. Every resistor opposes the same current, so the opposition adds. Net resistance goes up. For the same supply p.d. the current goes down.
• Parallel: each new branch is a new path. The supply p.d. is across every branch, so the supply current is the sum of the branch currents. For the same p.d. the supply current goes up, so net resistance goes down.
• Only a qualitative explanation is required for series and parallel net resistance.
• Bulbs: in series they share the p.d. and are dimmer. In parallel each gets the full p.d. and each is as bright as it would be alone.

Model sentence: “Adding a resistor in parallel gives the charge another path, so the total current from the supply increases for the same p.d., which means the net resistance has decreased.”

P3.2j: the rules

	Series	Parallel
Current	Equal at all points	Branch currents sum to the supply current
p.d.	Shares add up to the supply p.d.	Same across each branch
Resistance	R = R₁ + R₂ + …	Net R less than the smallest branch

Method in steps: series circuit

1. Sum the resistor values to find the equivalent resistance.
2. Find the current: I = V / R using the supply p.d.
3. Find each p.d.: V = I × R for each resistor.
4. Check the p.d.s add to the supply p.d.

Method in steps: parallel circuit

1. Write the supply p.d. against every branch.
2. Find each branch current: I = V / R.
3. Add the branch currents for the supply current.
4. If needed, net R = supply V / supply I.

Worked reminder: series

A 10 V supply, 2.0 Ω and 3.0 Ω in series. R = 5.0 Ω, I = 10 / 5.0 = 2.0 A. p.d.s: 2.0 × 2.0 = 4.0 V and 2.0 × 3.0 = 6.0 V. They add to 10 V.

Worked reminder: parallel

A 6.0 V supply, 4.0 Ω and 12 Ω in parallel. Currents: 6.0 / 4.0 = 1.5 A and 6.0 / 12 = 0.50 A. Supply current = 2.0 A. Net R = 6.0 / 2.0 = 3.0 Ω, which is below 4.0 Ω.

Finding a missing resistance

A resistor has 5.0 V across it and 0.25 A through it. R = V / I = 5.0 / 0.25 = 20 Ω. In a series circuit, a missing resistor’s value is the equivalent resistance minus the known ones.

P3.2k: measurement and testing circuits

Item	Rule
Ammeter	In series, in the path of the current. Very low resistance.
Voltmeter	In parallel, across the component. Very high resistance.
Resistance measurement	Read V and I, then R = V / I.
Variable resistor	Changes the current so that several readings can be taken.
Thermistor or LDR with a fixed resistor	Changes in temperature or light change the voltmeter reading across the fixed resistor.
Fault in a series lamp circuit	A voltmeter across a broken lamp reads about the supply p.d.

Worked reminder: sensor reading

An LDR is in series with a 400 Ω fixed resistor on an 8.0 V supply. In bright light the LDR is 100 Ω, so R = 500 Ω, I = 8.0 / 500 = 0.016 A, and the p.d. across the fixed resistor = 0.016 × 400 = 6.4 V. In the dark the LDR is 1200 Ω, so R = 1600 Ω, I = 0.0050 A, and the fixed resistor p.d. = 2.0 V. The voltmeter reading falls as it gets darker.

Design checklist for an explain question

• Where does the ammeter go and why (series)?
• Where does the voltmeter go and why (parallel)?
• What is varied, what is measured, and which equation is applied?
• How is the reading made more reliable (repeat, take a range, average)?

P3.2l and P3.2m: power, energy and charge

• Power is the rate of energy transfer. 1 W = 1 J transferred per second.
• Bigger p.d. and bigger current both mean a greater power.
• Energy transferred = power × time. Convert minutes to seconds (× 60) unless you use kW with hours.
• One volt means one joule is transferred per coulomb of charge.
• In a series circuit, the total power is the sum of the powers in the resistors, and it equals the supply p.d. × the current.

Worked reminder: power

A torch bulb has 4.5 V across it and 0.40 A through it. P = 4.5 × 0.40 = 1.8 W. In 2.0 minutes (120 s), E = 1.8 × 120 = 216 J. Charge passed = 0.40 × 120 = 48 C, and E = 48 × 4.5 = 216 J.

Worked reminder: P = I² × R

A 4.0 Ω resistor carries 3.0 A. P = 3.0² × 4.0 = 9.0 × 4.0 = 36 W.

Worked reminder: kilowatt-hours

A 3.0 kW oven runs for 90 minutes (1.5 h). E = 3.0 × 1.5 = 4.5 kW h. In joules this is 4.5 × 3 600 000 = 16 200 000 J.

Must-know distinctions

• Current and p.d.: current flows through a component, p.d. is measured across it.
• Power and energy: power is a rate (W), energy is an amount (J or kW h).
• Series and parallel net resistance: up in series, down in parallel.
• Equivalent resistance: the single resistance that would draw the same current from the same supply.
• kW h and J: both measure energy. 1 kW h = 3 600 000 J.

Quick self-test

1. Two resistors, 7.0 Ω and 5.0 Ω, are in series. What is the equivalent resistance?
2. A 12 V supply is across a 6.0 Ω resistor. What current flows?
3. Why does net resistance fall when you add a second resistor in parallel?
4. A 16 V supply is across two resistors in series. One has 10 V across it. What is the p.d. across the other?
5. Two branches in parallel carry 0.30 A and 0.70 A. What is the supply current?
6. Where does a voltmeter go in a circuit?
7. A heater uses 230 V and 4.0 A. What is its power?
8. A 5.0 Ω resistor carries 2.0 A. What is its power?
9. A 60 W lamp is on for 10 s. How much energy is transferred?
10. 20 C of charge passes through a p.d. of 3.0 V. How much energy is transferred?
11. What is the energy in kW h for a 2.0 kW fire on for 3.0 h?
12. An 18 V supply drives 0.60 A through a series chain of one known 10 Ω resistor and one unknown resistor. What is the unknown resistance?

Answers

1. 12 Ω.
2. I = 12 / 6.0 = 2.0 A.
3. It gives the charge another path, so the total current from the same supply p.d. increases.
4. 16 − 10 = 6.0 V.
5. 0.30 + 0.70 = 1.0 A.
6. In parallel, across the component being measured.
7. 230 × 4.0 = 920 W.
8. 2.0² × 5.0 = 20 W.
9. 60 × 10 = 600 J.
10. 20 × 3.0 = 60 J.
11. 2.0 × 3.0 = 6.0 kW h.
12. Equivalent R = 18 / 0.60 = 30 Ω, so unknown = 30 − 10 = 20 Ω.

Where marks are usually lost

• Writing “resistance goes down because there are more wires” in a parallel explanation without mentioning the extra path or the increased current.
• Giving a series explanation that says current is “used up”. The current is the same around the loop.
• Quoting a number without a unit, or with the wrong unit.
• Forgetting to convert minutes to seconds before using E = P × t in joules.
• Using the supply p.d. in V = I × R for a single resistor in a series chain.
• Drawing a voltmeter in series in a circuit diagram.
• Not showing the equation and substitution, so the reader cannot follow your method.
• Rounding too early in a multi-step problem.
• Mixing up P = V × I and P = I² × R by squaring the wrong quantity.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment in 2018, published by OCR. Learning outcomes P3.2i to P3.2m.