Cambridge AS & A Level9702

Electricity

Physics 9702 Chapter Notes

What this chapter covers

Electricity - Electric currentElectricity - Potential difference and powerElectricity - Resistance and resistivity
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1. Electric Current and Charge

Electric current is the rate of flow of electric charge. Imagine a river; the water flowing is like the charge, and the speed of the river's flow is the current. In metal wires, the charge carriers are tiny, negatively charged particles called electrons. Current is measured in amperes (A), often called 'amps'. For a current to flow, there must be a complete circuit and a source of potential difference, like a battery. The total charge (Q) that flows is the current (I) multiplied by the time(t) it flows for. Charge is measured in coulombs (C). One coulomb is the charge that passes a point in one second when the current is one amp. A key principle is that charge is 'quantised', meaning it comes in discrete packets. The smallest packet of charge is the 'elementary charge' (e), which is the magnitude of the charge on a single electron or proton, approximately 1.60 x 10⁻¹⁹ C.

Q = It

Q = ne

Key term

Electric Current: The rate of flow of electric charge, measured in amperes (A).

Examiner insight

Examiners award marks for clearly stating the formula Q = It and correctly converting units, such as minutes to seconds or milliamperes to amperes.

Common pitfall

Forgetting to convert time into seconds when using the formula Q = It. Always use SI units (Amperes, Coulombs, Seconds).

Fun fact

A typical lightning bolt can carry a current of up to 200,000 amps, but it only lasts for a fraction of a second. That's enough to transfer about 20-30 coulombs of charge.

Worked example 13 marks

A current of 2.5 A flows through a wire for 4 minutes. Calculate the total charge that passes through the wire.

  1. 1

    Step 1: State the relevant formula: Q = It.

  2. 2

    Step 2: Convert the time from minutes to SI units (seconds). t = 4 minutes = 4 × 60 s = 240 s.

  3. 3

    Step 3: Substitute the values into the formula and calculate Q. Q = 2.5 A × 240 s = 600 C.

Worked example 22 marks

A total charge of 120 C passes a point in a circuit in 90 seconds. What is the current?

  1. 1

    Step 1: Rearrange the formula Q = It to make current (I) the subject: I = Q / t.

  2. 2

    Step 2: Substitute the known values into the rearranged formula. I = 120 C / 90 s = 1.33 A.

Recap

  • Electric current is the rate of flow of charge.
  • The unit of current is the ampere (A) and the unit of charge is the coulomb (C).
  • The formula linking charge, current and time is Q = It.
  • Charge is quantised and the elementary charge 'e' is 1.60 x 10⁻¹⁹ C.
  • Conventional current flows from positive to negative, while electrons flow from negative to positive.

Quick check

  1. What is the definition of the coulomb?1 mark
  2. If a current of 500 mA flows for 20 s, what is the charge transferred?2 marks

2. The Microscopic View of Current

While we imagine current flowing smoothly, on a microscopic level it's about the collective movement of charge carriers, usually electrons in a metal. These electrons don't zoom straight through the wire. They constantly collide with the metal ions in the lattice structure, causing them to move in a chaotic, random way. When a potential difference is applied, they are given a slow, net 'nudge' in one direction. This slow net speed is called the mean drift velocity (v). It's surprisingly slow, often less than 1 mm per second. So why does a light turn on instantly? Because the electric field that pushes the electrons propagates at nearly the speed of light, so all electrons in the wire start drifting almost simultaneously. The current (I) is related to this drift velocity by the equation I = nAve, where 'n' is the number density of charge carriers (how many free electrons per cubic metre), 'A' is the cross-sectional area of the wire, 'v' is the drift velocity, and 'e' is the elementary charge.

I = nAve

Key term

Mean Drift Velocity: The average velocity attained by charged particles in a material due to an electric field.

Examiner insight

Candidates often score well if they can explain the difference between the high speed of the electric field and the low drift velocity of the electrons.

Common pitfall

Confusing the random high speed of individual electrons (around 10⁶ m/s) with their very slow mean drift velocity in the direction of the current.

Fun fact

The drift velocity of electrons in a typical copper wire carrying a 1A current is about 0.1 mm/s. You could walk faster than the electrons that power your phone charger!

Worked example 13 marks

A copper wire has a cross-sectional area of 2.0 x 10⁻⁷ m² and carries a current of 1.5 A. The number density of free electrons in copper is 8.5 x 10²⁸ m⁻³. Calculate the mean drift velocity of the electrons. (Elementary charge e = 1.60 x 10⁻¹⁹ C)

  1. 1

    Step 1: State the formula and rearrange for drift velocity (v): I = nAve => v = I / (nAe).

  2. 2

    Step 2: Substitute the known values into the formula: v = 1.5 / (8.5 x 10²⁸ × 2.0 x 10⁻⁷ × 1.60 x 10⁻¹⁹).

  3. 3

    Step 3: Calculate the result: v = 1.5 / 2720 = 5.5 x 10⁻⁴ m s⁻¹ (or 0.55 mm s⁻¹).

Recap

  • Current is caused by the slow, collective movement of charge carriers.
  • Mean drift velocity is the average speed of charge carriers along the wire.
  • Drift velocity is typically very slow, often less than 1 mm per second.
  • The formula I = nAve links macroscopic current to microscopic properties.
  • 'n' is the number of free charge carriers per unit volume (number density).

Quick check

  1. If the cross-sectional area of a wire is halved, what happens to the drift velocity for the same current?2 marks
  2. What do the letters in I = nAve stand for?1 mark

3. Potential Difference and EMF

What makes charge flow? An 'electrical push' is needed. This push is provided by a source like a battery or power supply. We have two key terms for this: electromotive force (e.m.f.) and potential difference (p.d.). Both are measured in volts (V). Electromotive Force (e.m.f.): This is the energy supplied *by the source* to each coulomb of charge. Think of it as the total energy given to each charge to make a full lap of the circuit. It is the work done per unit charge in converting chemical or other energy into electrical energy. Potential Difference (p.d.): This is the energy converted *by a component* from electrical energy into other forms (like light, heat, sound) for each coulomb of charge that passes through it. It is the work done per unit charge as it passes through the component. The definition of the volt links them: 1 Volt is 1 Joule per Coulomb (1 V = 1 J/C). So, a 6V battery gives 6 joules of energy to every coulomb of charge. A lamp with a 2V p.d. across it turns 2 joules of electrical energy into light and heat for every coulomb that passes through.

V = W/Q

ε = W/Q

Key term

Electromotive Force (e.m.f.): The work done by a source in driving a unit charge around a complete circuit.

Examiner insight

Examiners look for the key distinction: e.m.f. is energy supplied *to* the charge by the source, while p.d. is energy dissipated *by* the charge in a component.

Common pitfall

Using the terms e.m.f. and p.d. interchangeably. E.m.f. is a cause (the battery's push), while p.d. is an effect (the energy drop across a resistor).

Worked example 12 marks

A battery has an e.m.f. of 9.0 V. It supplies a charge of 30 C to a circuit. Calculate the total energy supplied by the battery.

  1. 1

    Step 1: State the formula relating energy, e.m.f., and charge: W = εQ.

  2. 2

    Step 2: Substitute the values: W = 9.0 V × 30 C = 270 J.

Worked example 22 marks

A resistor has a potential difference of 5.0 V across it. In 10 seconds, a charge of 2.0 C passes through it. How much energy is converted into heat in the resistor?

  1. 1

    Step 1: State the formula relating energy, p.d., and charge: W = VQ.

  2. 2

    Step 2: Substitute the values: W = 5.0 V × 2.0 C = 10 J. The time is extra information not needed for this part.

Recap

  • E.m.f. is the energy supplied per unit charge by the source.
  • Potential difference (p.d.) is the energy converted per unit charge by a component.
  • Both e.m.f. and p.d. are measured in volts (V).
  • One volt is defined as one joule per coulomb (1 V = 1 J/C).
  • In a simple circuit, the sum of p.d.s across components equals the source e.m.f.

Quick check

  1. What is the difference between e.m.f. and p.d. in terms of energy transfer?2 marks
  2. If 120 J of energy is converted in a component when 20 C of charge flows through it, what is the p.d. across it?1 mark

4. Resistance and Ohm's Law

Resistance is a measure of how much a component opposes the flow of electric current. Think of it as electrical friction. A component with high resistance will allow less current to flow for a given potential difference compared to a component with low resistance. Resistance (R) is measured in ohms (Ω). The relationship between potential difference (V), current (I), and resistance (R) for many components is described by Ohm's Law. It states that the current through a conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions remain constant. This relationship is summarised in the famous equation V = IR. A component that obeys Ohm's Law is called an 'ohmic' conductor (e.g., a resistor at constant temperature), and its V-I graph is a straight line through the origin. Components like filament lamps or diodes are 'non-ohmic' as their resistance changes, and their V-I graphs are curved.

V = IR

R = V/I

Key term

Resistance: A measure of the opposition to current flow, defined as the ratio of potential difference across a component to the current flowing through it.

Examiner insight

Marks are often awarded for correctly rearranging V=IR to solve for I or R. Be careful to use the p.d. *across the component*, not the e.m.f. of the whole circuit, unless it's the only component.

Common pitfall

Assuming that all components obey Ohm's Law. Filament lamps are a classic example of a non-ohmic component where resistance increases as it gets hotter.

Fun fact

The 'resistors' in your electronic devices are often tiny ceramic cylinders with coloured bands that tell you their resistance value. It's a universal code that technicians can read at a glance.

Worked example 12 marks

A resistor of 15 Ω is connected to a 6.0 V battery. Calculate the current flowing through the resistor.

  1. 1

    Step 1: State Ohm's Law and rearrange for current: V = IR => I = V/R.

  2. 2

    Step 2: Substitute the values: I = 6.0 V / 15 Ω = 0.40 A.

Worked example 23 marks

A current of 200 mA flows through a component when a potential difference of 4.0 V is applied across it. What is the resistance of the component?

  1. 1

    Step 1: State Ohm's Law and rearrange for resistance: R = V/I.

  2. 2

    Step 2: Convert the current into Amperes: I = 200 mA = 0.200 A.

  3. 3

    Step 3: Substitute the values: R = 4.0 V / 0.200 A = 20 Ω.

Recap

  • Resistance is the opposition to the flow of current, measured in ohms (Ω).
  • Ohm's Law states V is proportional to I for an ohmic conductor at constant temperature.
  • The formula for Ohm's Law is V = IR.
  • A component's resistance is the ratio of the voltage across it to the current through it (R = V/I).
  • Not all components obey Ohm's Law (e.g., filament lamps).

Quick check

  1. What is the resistance of a component if a p.d. of 12V causes a current of 3A to flow?1 mark
  2. What happens to the current in a circuit if the resistance is doubled but the voltage stays the same?1 mark

5. Electrical Power and Energy

Electrical power is the rate at which electrical energy is transferred or converted. Since power is energy per unit time (P = E/t) and potential difference is energy per unit charge (V = E/Q), we can combine these ideas. We know current is charge per unit time (I = Q/t). By substituting, we get the fundamental equation for electrical power: P = VI. This tells you the power dissipated in a component is the product of the potential difference across it and the current flowing through it. Power is measured in watts (W), where 1 watt is 1 joule per second. Using Ohm's Law (V=IR), we can derive two other useful power equations: P = I²R and P = V²/R. These are great for when you don't know V or I respectively. Electrical energy (E) is simply power multiplied by time (E = Pt). Combining this with our power equations gives us E = VIt, E = I²Rt, and E = (V²/R)t.

P = VI

P = I²R

P = V²/R

E = Pt

E = VIt

Key term

Electrical Power: The rate at which electrical energy is converted into other forms of energy, measured in watts (W).

Examiner insight

Students who can select the most appropriate power equation (P=VI, P=I²R, or P=V²/R) based on the information given in the question often solve problems more quickly and with fewer steps.

Common pitfall

Mixing up energy and power. Power is the *rate* of energy use (in Watts), while energy is the total amount used over a period of time (in Joules).

Fun fact

Your electricity bill measures energy consumption in kilowatt-hours (kWh), not joules. One kWh is the energy used by a 1000W device running for one hour, which is equal to a massive 3.6 million joules!

Worked example 14 marks

A 230 V kettle has a power rating of 3.0 kW. Calculate the current it draws and the resistance of its heating element.

  1. 1

    Step 1: Calculate the current using P = VI. First convert kW to W: P = 3000 W. Rearrange for I: I = P/V.

  2. 2

    Step 2: Substitute values: I = 3000 W / 230 V = 13.0 A.

  3. 3

    Step 3: Calculate resistance. We can use P = V²/R. Rearrange for R: R = V²/P.

  4. 4

    Step 4: Substitute values: R = (230 V)² / 3000 W = 17.6 Ω.

Worked example 23 marks

A resistor with resistance 5.0 Ω has a steady current of 2.0 A flowing through it for 1 minute. Calculate the energy dissipated as heat in the resistor.

  1. 1

    Step 1: Choose an appropriate energy formula. We have I, R, and t, so E = I²Rt is best.

  2. 2

    Step 2: Convert time to SI units (seconds): t = 1 minute = 60 s.

  3. 3

    Step 3: Substitute values: E = (2.0 A)² × 5.0 Ω × 60 s = 1200 J (or 1.2 kJ).

Recap

  • Power is the rate of energy transfer, measured in watts (W).
  • The main formula for power is P = VI.
  • Alternative power formulas are P = I²R and P = V²/R.
  • Energy is power multiplied by time (E = Pt).
  • Energy transferred can also be calculated using E = VIt.
  • Always check your units: Power in W, Energy in J, Time in s.

Quick check

  1. A 12V car headlamp draws a current of 4A. What is its power rating?1 mark
  2. If a 100W light bulb is left on for 50 seconds, how much electrical energy has it converted?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A current of 150 mA flows for 2 minutes. Calculate the total charge transferred.3 marks
  2. A component has a resistance of 25 Ω. What potential difference is needed to drive a current of 0.5 A through it?2 marks
  3. An electric motor connected to a 12 V supply draws a current of 3.5 A. Calculate its power.2 marks
  4. A 9.0 V battery is connected to a 18 Ω resistor. Calculate the energy dissipated in the resistor in 5 minutes.4 marks
  5. Define potential difference and state its unit.2 marks
  6. A copper wire has a cross-sectional area of 1.5 x 10⁻⁶ m² and the number density of free electrons is 8.5 x 10²⁸ m⁻³. Calculate the current if the drift velocity of electrons is 0.2 mm s⁻¹. (e = 1.60 x 10⁻¹⁹ C)3 marks
  7. A power source has an e.m.f. of 6.0 V. When it delivers a current of 0.5 A to a lamp, the potential difference across the lamp is 5.5 V. a) How much energy is supplied by the source per coulomb of charge? b) How much energy is converted by the lamp per coulomb of charge? c) What accounts for the difference between these two values?3 marks
  8. A hairdryer is rated at 1800 W for use on a 230 V supply. Fuses are available with values 1A, 3A, 5A, 10A, 13A. Calculate which fuse should be used and explain your choice.4 marks
  9. Two copper wires, A and B, are connected in series, so the same current flows through both. Wire A has twice the diameter of wire B. What is the ratio of the drift velocity of electrons in wire A to that in wire B (v_A / v_B)?4 marks
  10. An electron is accelerated from rest through a potential difference of 5.0 kV in a vacuum. Calculate its final kinetic energy in joules. (e = 1.60 x 10⁻¹⁹ C)3 marks

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