OxfordAQA International GCSE Physics: Series and parallel circuits and heating (9203) – Revision Notes

These revision notes cover the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, section 3.5.1 (electrical circuits), statements p to t: cells in series, series and parallel circuits, mixed circuits and the heating effect of current. Longer explanations and more worked examples are in the study guide.

Other links: practice questions, course hub, printable checklist, the existing energy revision notes and the 10-minute diagnostics.

Key facts at a glance

Specification	Fact
3.5.1 p	The combined voltage of cells in series is the sum of their voltages
3.5.1 q	Components join in series or in parallel; some circuits have both
3.5.1 r	Series: resistances add; same current; supply pd shared
3.5.1 s	Parallel: combined resistance less than any branch; current splits; same pd
3.5.1 t	Charge flowing through a resistor heats it through collisions with stationary atoms; filament bulbs waste a lot of energy as heat, CFLs less

Equation used: V = IR (pd in V, current in A, resistance in Ω).

Series and parallel side by side

	Series	Parallel
Paths for charge	One	Two or more
Combined resistance	Sum of the resistances	Less than the smallest branch
Current	Same everywhere	Splits at the junction; branch currents add to the supply current
Potential difference	Shared; pds add to the supply pd	Same across every branch
One component fails (open)	Everything stops	Other branches carry on
Add another component	Resistance up, current down	Resistance down, supply current up
Larger resistance takes	The larger pd	The smaller current

Cells in series

• Sum the voltages: four 2.0 V cells give 8.0 V.
• Number of cells needed = required voltage / voltage of one cell.
• The cells must point the same way round the loop.

Method boxes

Series circuit, given supply pd and resistances

1. R_total = R1 + R2 + …
2. I = V / R_total.
3. Each pd = I x its own resistance.
4. Check the pds add to the supply pd.

Parallel circuit, given supply pd and branch resistances

1. Every branch has the supply pd.
2. Branch current = V / branch resistance.
3. Supply current = sum of branch currents.
4. Combined resistance = supply pd / supply current.

Mixed circuit

1. Treat the parallel block as one item in series with the rest.
2. pd across the series resistor = supply current x its resistance.
3. pd across the block = supply pd minus that.
4. Divide that pd by each branch resistance for the branch currents.
5. Check the branch currents add to the supply current.

Small worked reminders

Series. A 9.0 V supply drives an 8.0 Ω and a 22 Ω resistor in series. R_total = 30 Ω. I = 9.0 / 30 = 0.30 A. pds: 0.30 x 8.0 = 2.4 V and 0.30 x 22 = 6.6 V. Total 9.0 V.

Parallel. A 24 V supply is across 40 Ω and 60 Ω in parallel. Currents: 24 / 40 = 0.60 A and 24 / 60 = 0.40 A. Supply current 1.0 A. Combined resistance 24 / 1.0 = 24 Ω, which is below 40 Ω.

Equal branches. n equal resistors in parallel: each takes the same current, so the combination has the resistance of one divided by n. Two 80 Ω resistors in parallel give 40 Ω.

Heating effect of current

• Moving charges (electrons) collide with stationary atoms in the wire.
• Energy is transferred from the charges to the atoms, which vibrate more.
• The wire gets hotter and transfers thermal energy to the surroundings.
• A heater element has high resistance; the connecting leads have low resistance and stay cool.
• Same current through two resistors in series: the one with the bigger pd transfers more energy.

Filament bulbs and CFLs

• A filament lamp wastes a lot of energy by heating; only a small fraction becomes light.
• A CFL (power saving lamp) wastes less energy for the same light.
• Efficiency = useful energy transferred / total energy supplied. Give it as a decimal or a percentage.
• Wasted energy = total energy supplied minus useful energy.
• When buying a new appliance you can choose a more energy efficient one.

Example: a lamp is supplied with 50 J and transfers 4.0 J as light. Efficiency = 4.0 / 50 = 0.08 = 8.0%. Wasted energy = 46 J.

Reasoning in words

Examiners often ask for an explanation rather than a number. Use these patterns.

Why do lamps in a house work independently? They are in parallel, so each has the full supply pd across it and its own path for current. Switching one off or removing it leaves the others unchanged.

Why does the combined resistance fall when a branch is added? Charge has another route, so for the same supply pd more current flows from the supply. Since R = V/I, a bigger current for the same V means a smaller resistance.

Why do lamps in a series string get dimmer as you add more? The total resistance rises, so the current falls, and the same supply pd is shared between more lamps, so each gets a smaller pd.

What happens if a lamp in a series string blows? The loop is broken, so there is no current anywhere and all the lamps go out.

Why use a high-resistance wire in a heater? For the same current, a larger share of the supply energy is transferred as heat in the element, not in the leads.

Must-know distinctions

• Current is the same in series; pd is the same in parallel.
• Resistances add in series; they do not add in parallel.
• Parallel branches share the current, not the pd.
• Ammeter in series; voltmeter in parallel.
• Heating comes from collisions, not from the charge being “used up”.
• “Wasted” energy is transferred to the surroundings, not destroyed.

Quick self-test

1. Five 1.5 V cells are joined in series. What is the combined voltage?
2. A 12 V supply is connected to a 6.0 Ω and an 18 Ω resistor in series. Calculate the current and each resistor pd.
3. A 12 V supply is connected across a 30 Ω and a 60 Ω resistor in parallel. Find each branch current, the supply current and the combined resistance.
4. What is the combined resistance of two 80 Ω resistors in parallel?
5. A 9.0 V supply is connected to a 10 Ω resistor in series with a parallel block. An ammeter in the supply lead reads 0.30 A. Find the pd across the block, the resistance of the block and the total resistance.
6. Explain why a resistor gets hot when there is a current in it.
7. A lamp is supplied with 50 J and gives 4.0 J of light. A replacement gives the same light from 16 J. Find both efficiencies as percentages.
8. Three identical lamps are in parallel. One filament breaks. What happens to the other two?
9. A student adds a third resistor in parallel across a supply. What happens to the supply current and the combined resistance?
10. A third resistor is added in series instead. What happens to the current?

Answers

1. 5 x 1.5 = 7.5 V.
2. R_total = 24 Ω; I = 12 / 24 = 0.50 A; pds 3.0 V (0.50 x 6.0) and 9.0 V (0.50 x 18).
3. Each branch has 12 V: 0.40 A (12 / 30) and 0.20 A (12 / 60); supply current 0.60 A; combined resistance 12 / 0.60 = 20 Ω.
4. 40 Ω.
5. pd across the 10 Ω resistor = 0.30 x 10 = 3.0 V, so the block has 9.0 - 3.0 = 6.0 V; block resistance = 6.0 / 0.30 = 20 Ω; total resistance = 10 + 20 = 30 Ω (check: 9.0 / 0.30 = 30 Ω).
6. Moving electrons collide with stationary atoms (ions) in the resistor; energy is transferred to the atoms, which vibrate more, so the resistor’s temperature rises.
7. First lamp: 4.0 / 50 = 0.08, 8.0%. Replacement: 4.0 / 16 = 0.25, 25%.
8. They stay lit and are unchanged, because each still has the full supply pd across it.
9. The supply current increases and the combined resistance decreases.
10. The current decreases, because the total resistance rises.

Where marks are usually lost

• Adding the branch resistances in a parallel circuit.
• Giving a parallel combined resistance larger than the smallest branch, and not noticing.
• Forgetting that the supply current flows through the series part of a mixed circuit.
• Splitting the supply pd between parallel branches.
• Writing “electrons are used up” instead of describing energy transfer.
• Leaving out the word “stationary” or “atoms” in the explanation of heating.
• Dividing useful energy by wasted energy instead of by total energy supplied.
• Missing units, or writing a percentage efficiency as a decimal when a percentage is requested.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.5.1 (electrical circuits), statements p to t.