Ch 1 · 9 Electricity · A Level Physics 9702
Topic 1 of 2 in Electricity — A Level Physics 9702 — 75 lessons.
Introduction
Every circuit depends on charge moving through materials. This topic explains what current really is at the level of charge carriers, defines p.d., power and resistance precisely, and introduces resistivity, the property that tells you how well a material conducts. (Syllabus topic 9: sections 9.1–9.3.)
Key ideas — 9.1 (a) Current and charge carriers
An electric current is a flow of charge carriers: free electrons in metals, ions in electrolytes and gases.
🧑🏫 Teacher's explanation
Imagine a busy hallway in a school. If you want to count how many students pass a certain point, you can't have half a student; you can only have whole numbers. This is exactly how electric charge works. Charge is 'quantised', meaning it comes in discrete packets. In a metal wire, these packets are electrons.
📖 Definitions
🔍 Worked example — Easy
A constant current of 0.50 A flows through a wire for 2.0 minutes. Calculate the total charge passing a cross-section of the wire.
🔍 Worked example — Standard
A wire has a cross-sectional area of 2.0 × 10^-6 m^2. The number density of electrons n is 1.2 × 10^28 m^-3 and the drift velocity v is 4.0 × 10^-4 m/s. Calculate the current I. (Take e = 1.60 × 10^-19 C)
🔍 Worked example — Exam-style
A total charge of 4.8 × 10^-18 C passes through a point in a circuit. Determine how many electrons have passed this point.
⚠️ Common errors
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🎯 Exam question
Key ideas — 9.1 (b) The transport equation $I = Anvq$
For a conductor of cross-sectional area A, containing n charge carriers per unit volume (the number density), each with charge q, drifting at average speed v:
🧑🏫 Teacher's explanation
Imagine a crowded hallway in a school. If you want to measure the 'flow' of students passing a certain line every second, it depends on three things: how wide the hallway is (A), how many students are packed into every cubic meter (n), how fast they are walking (v), and how much 'charge' (or identity) each student carries (q). Even if the hallway is huge, if everyone is walking at a snail's pace, the flow remains low.
📖 Definitions
⚠️ Common errors
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🎯 Exam question
Key ideas — 9.2 Potential difference and power
The potential difference across a component is the energy transferred per unit charge (from electrical energy to other forms) as the charge passes through it:
🧑🏫 Teacher's explanation
Think of potential difference (p.d.) as 'electrical pressure' or the amount of 'push' given to each unit of charge. Imagine a water slide: the height of the slide represents the potential difference. A higher slide gives each drop of water more energy to convert into motion (kinetic energy) as it falls. In a circuit, the battery provides this 'height', and as charge moves through a resistor, it loses that electrical energy, converting it into heat or light.
📖 Definitions
🔍 Worked example — Easy
A light bulb has a potential difference of 12.0 V across it and a current of 0.500 A flowing through it. Calculate the power dissipated by the bulb.
🔍 Worked example — Standard
A resistor with resistance R = 470 Omega is connected to a 9.00 V battery. Calculate the power dissipated by the resistor.
🔍 Worked example — Exam-style
An electric heater is rated at 2.40 kW and is connected to a 230 V mains supply. Calculate the resistance of the heating element.
⚠️ Common errors
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🎯 Exam question
Key ideas — 9.3 (a) Resistance and Ohm's law
Resistance is the ratio of the p.d. across a component to the current through it:
🧑🏫 Teacher's explanation
Think of electricity like water flowing through a pipe. The current I is the volume of water passing a point per second, and the potential difference V is the pressure pushing it. Resistance R is like the diameter of the pipe or any grit inside it that slows the flow down. If you increase the pressure (voltage), more water flows (current), provided the pipe doesn't change size.
📖 Definitions
🔍 Worked example — Easy
A resistor has a potential difference of 4.5 V across it and a current of 0.50 A flowing through it. Calculate the resistance.
🔍 Worked example — Standard
A filament lamp is connected to a 12.0 V supply. When the current is 2.0 A, the resistance is 6.0\Omega. Calculate the resistance when the current increases to 3.0 A, assuming the lamp follows Ohm's law (constant temperature).
🔍 Worked example — Exam-style
A metallic wire has a resistance of 2.5 Omega at 20^circC. When the current through the wire is 4.0 A, the temperature increases, causing the resistance to rise to 3.5 Omega. Calculate the new potential difference across the wire.
⚠️ Common errors
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🎯 Exam question
Key ideas — 9.3 (b) I–V characteristics
🧑🏫 Teacher's explanation
Think of an I–V characteristic as a 'personality profile' for an electronic component. It tells you exactly how much current will flow for every volt you apply. If you plot current (I) on the y-axis and potential difference (V) on the x-axis, the shape of the graph reveals the component's internal struggle against the flow of charge.
📖 Definitions
🔍 Worked example — Easy
A metallic conductor follows Ohm's law. When a potential difference of 2.0 V is applied, a current of 0.50 A flows. Calculate the resistance of the conductor.
🔍 Worked example — Standard
A filament lamp has a resistance of 10 Omega when the current is 2.0 A. If the current increases to 4.0 A, the resistance increases to 25 Omega due to heating. Calculate the new potential difference across the lamp.
🔍 Worked example — Exam-style
A silicon diode is in forward bias. The current I is given by I = 0.02(e^V/0.6 - 1), where I is in Amperes and V is in Volts. Calculate the current when V = 0.75 V.
⚠️ Common errors
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🎯 Exam question
(a) Sketch the I–V characteristic for the lamp. [2] (b) Explain, in terms of the internal structure of the filament, why the resistance increases as the current increases. [3] [5 marks]
Key ideas — 9.3 (c) Resistivity
The resistance of a uniform wire is proportional to its length and inversely proportional to its cross-sectional area:
🧑🏫 Teacher's explanation
Imagine water flowing through a pipe. If the pipe is very long, there is more friction for the water to overcome, so the flow is harder (higher resistance). If the pipe is very wide, it is much easier for water to pass through (lower resistance). Resistivity is like the 'roughness' of the inside of the pipe; a smooth pipe allows easy flow, while a rough pipe makes it harder, regardless of the pipe's size.
📖 Definitions
🔍 Worked example — Easy
A copper wire has a length of 2.0 m and a cross-sectional area of 0.5\ extmm^2. Calculate its resistance. Resistivity of copper is 1.7 × 10^-8 Omega m.
🔍 Worked example — Standard
A nichrome wire has a resistance of 12 Omega. It is stretched so that its length doubles while its cross-sectional area is halved. Calculate the new resistance. Resistivity is constant.
🔍 Worked example — Exam-style
A cylindrical conductor has a resistance of 0.45 Omega. If the radius of the conductor is doubled, what is the new resistance? Assume resistivity and length remain constant.
⚠️ Common errors
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🎯 Exam question
(a) Calculate the resistance of the wire. [2] (b) The wire is replaced by a wire of the same material and length, but with twice the cross-sectional area. Calculate the new resistance. [2] [4 marks]
Key ideas — 9.3 (d) LDRs and thermistors
(In both, more charge carriers are released as light or temperature increases.)
🧑🏫 Teacher's explanation
Think of resistance as 'friction' for electricity. In a normal wire, the atoms just sit there. But in special materials like LDRs and thermistors, the atoms are a bit more sensitive. Imagine a crowded hallway (the material) where people (charge carriers) are trying to run through.
📖 Definitions
🔍 Worked example — Easy
A light-dependent resistor (LDR) has a resistance of 2.5 × 10^5 Omega in total darkness. When illuminated by a bright lamp, its resistance drops to 4.0 \× 10^3 Omega. Calculate the factor by which the resistance decreased.
🔍 Worked example — Standard
An NTC thermistor is part of a circuit with a 12.0 V battery. At 20^circC, the resistance is 5.0 × 10^3 Omega. At 50^circC, the resistance is 1.2 × 10^3 Omega. Calculate the current through the thermistor at 50^circC.
🔍 Worked example — Exam-style
A potential divider circuit consists of a 10 V source, a fixed resistor of 1000 Omega, and an LDR in series. When light intensity increases, the LDR resistance decreases from 5000 Omega to 500 Omega. Calculate the change in the potential difference across the LDR.
⚠️ Common errors
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🎯 Exam question
Worked example
Example 1: drift speed. A copper wire of cross-sectional area 1.0 × 10^-6 m² carries 2.0 A; n = 8.5 × 10^28 m⁻³.
🔍 Worked example — Standard
A nichrome wire has a length of 3.5 m and a diameter of 0.40 mm. Calculate its resistance if the resistivity rho = 1.5 × 10^-6 Omega m.
🔍 Worked example — Exam-style
An electric heater is connected to a 230 V supply. The resistance of the heating element is 12 Omega. Calculate the power dissipated by the heater and the current flowing through it.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) Charge exists only in whole-number multiples of the elementary charge e. (2) V = IR, so P = VI = (IR)I = I^2R. (3) A curve through the origin, getting flatter at higher p.d. in both directions; the filament heats up as the current increases, so its resistance increases. (4) R ∝ L/d^2: 2/2^2 = tfrac12. (5) The thin section: I = Anvq is the same everywhere, so smaller A means larger v.