IGCSE Physics: electric circuits, current, voltage and resistance

Series and parallel at a glance

Key ideas, board by board

Cambridge 0625 Core covers: positive and negative charges and charging by friction (transfer of electrons only); current as related to the flow of charge; ammeters and voltmeters; the definitions of e.m.f. (work done by a source moving unit charge round a complete circuit) and p.d. (work done by unit charge passing through a component); R = V/I; resistance increasing with wire length and decreasing with cross-sectional area; P = IV and E = IVt; the kilowatt-hour; circuit symbols including thermistors and LDRs; and series and parallel rules. Supplement adds Q = It, conventional current flowing from positive to negative while electrons flow the other way, E = W/Q and V = W/Q, sketching and explaining I-V graphs for a resistor, a filament lamp and a diode, resistance proportional to length and inversely proportional to area, calculating resistance of two resistors in parallel, and the potential divider equation R1/R2 = V1/V2. Mains safety (4.4) is Core.

Edexcel 4PH1 covers mains safety, P = IV and fuse selection (2.2 to 2.6), series and parallel circuits, I-V graphs for wires, resistors, filament lamps and diodes and how to investigate them (2.9), LDRs and thermistors (2.11), V = IR, Q = It, current as a flow of electrons in metals, conservation of current at junctions, calculations for two components in series, and voltage as energy per unit charge with E = QV. Appendix 7 lists Q = It, V = IR, P = IV and E = QV as relationships to recall; E = IVt is to be used but is not on that list. Static electricity (2.22P to 2.28P) is examined only on Paper 2; Cambridge covers charge and electric fields in 4.2.1.

Worked example 1: series circuit

Question: a 6.0 V battery is connected in series with a 4.0 ohm resistor and an 8.0 ohm resistor. Calculate the current and the p.d. across each resistor.

Total resistance = 4.0 + 8.0 = 12 ohms. Current I = V / R = 6.0 ÷ 12 = 0.50 A, the same through both resistors.

p.d. across the 4.0 ohm resistor = I x R = 0.50 x 4.0 = 2.0 V. p.d. across the 8.0 ohm resistor = 0.50 x 8.0 = 4.0 V. Check: 2.0 + 4.0 = 6.0 V, equal to the battery. The larger resistor takes the larger share of the p.d., which is the idea behind a potential divider.

Worked example 2: parallel resistors, charge, energy and fuses

Parallel (Cambridge Supplement): 6.0 ohm and 3.0 ohm resistors in parallel. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 2.0 ohms, less than either resistor alone.

Charge and energy: a current of 0.50 A flows for 2.0 minutes from a 6.0 V supply. Time = 120 s. Q = I x t = 0.50 x 120 = 60 C. Energy transferred E = Q x V = 60 x 6.0 = 360 J. The same answer comes from E = IVt = 0.50 x 6.0 x 120 = 360 J.

Fuse: a 2300 W kettle runs from a 230 V supply. I = P / V = 2300 ÷ 230 = 10 A. Choose the fuse with the lowest rating above the normal current from those offered, for example 13 A rather than 3 A or 30 A. Cost: if the kettle is used for 0.25 hours a day, energy = 2.3 kW x 0.25 h = 0.575 kWh per day; multiply by the price per kWh given in the question.

Common mistakes

• Saying current is "used up" by a lamp. Current is the same before and after a component in series; energy is transferred.
• Connecting a voltmeter in series or an ammeter in parallel in circuit diagrams.
• Adding resistances in parallel as if they were in series.
• Forgetting that 1/R is not R: after adding 1/R1 + 1/R2, take the reciprocal.
• Using minutes instead of seconds in Q = It and E = IVt, or watts instead of kilowatts for kWh.
• Saying a filament lamp has constant resistance. Its resistance increases as it gets hotter, so its I-V graph curves.
• Choosing a fuse rating below the normal working current, or far above it.

Exam technique and mark-scheme language

I-V graph explanations earn marks for cause and effect: "as the current increases, the filament gets hotter, the metal ions vibrate more, so the resistance increases and the gradient of the I-V graph decreases". For a diode: "current flows in one direction only; in reverse the resistance is very high". For thermistors (NTC) and LDRs, state the direction: resistance falls as temperature rises, and falls as light intensity rises.

Safety questions want the mechanism. A fuse melts when the current is too high, breaking the circuit. The earth wire gives a low-resistance path for current if a live wire touches a metal case, so a large current flows and the fuse blows or the trip switch opens, and the case does not stay live. Switches and fuses go in the live wire so the appliance is disconnected from the high voltage when off.

How one-to-one lessons help with this topic

Circuits reward a clear mental model more than memorised rules. A tutor builds the model on the shared whiteboard, drawing circuits, predicting meter readings and then checking with calculation, so the student sees why series and parallel behave differently. Lessons then move to timed past-paper circuit questions, including the I-V practical write-ups and the parallel and potential divider calculations that Cambridge Extended candidates find hardest.

Self-check

1. Define current, p.d., e.m.f. and resistance, with units.
2. State the current and p.d. rules for series and parallel circuits.
3. Use V = IR, P = IV, E = IVt, Q = It and E = QV.
4. Calculate the total resistance of resistors in series and (Cambridge Supplement) two in parallel.
5. Sketch I-V graphs for a resistor, a filament lamp and a diode, and explain their shapes.
6. Describe how the resistance of a thermistor and an LDR changes.
7. Explain how a fuse and an earth wire protect the user, and choose a fuse rating.
8. Calculate the cost of using an appliance in kWh.