Cambridge IGCSE0972

Electrical quantities

Physics 0972 Chapter Notes

What this chapter covers

Electrical quantities - Electric chargeElectrical quantities - Electric currentElectrical quantities - Electromotive force and potential differenceElectrical quantities - ResistanceElectrical quantities - Electrical energy and electrical power
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1. Electric Charge and Current

Electricity is all about the movement of electric charge. In metal wires, charge is carried by tiny negatively charged particles called electrons. Electric charge (symbol Q) is a fundamental property of matter and is measured in coulombs (C). When these charges flow through a conductor, we call it an electric current. The size of the current (symbol I) is the rate at which charge flows past a point in the circuit. Think of it like water in a pipe: the charge is the water itself, and the current is how much water flows per second. Current is measured in amperes (A) using an ammeter connected in series. A current of 1 ampere means 1 coulomb of charge is flowing every second. By convention, we talk about 'conventional current' flowing from the positive terminal to the negative terminal, even though in metals it's the electrons that flow the opposite way.

I = Q / t

Key term

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

Examiner insight

Examiners often test the distinction between conventional current (flow of positive charge from + to -) and electron flow (the actual movement of electrons in a metal wire from - to +).

Common pitfall

Confusing charge (the 'stuff' that flows, in coulombs) with current (the 'rate of flow' of that stuff, in amperes).

Worked example 13 marks

A current of 2.5 A flows through a lamp for 3 minutes. Calculate the total charge that flows through the lamp in this time.

  1. 1

    Step 1: State the formula linking charge, current, and time. Q = I × t

  2. 2

    Step 2: Identify the given values. Current I = 2.5 A. Time t = 3 minutes.

  3. 3

    Step 3: Convert the time into SI units (seconds). t = 3 minutes × 60 seconds/minute = 180 s.

  4. 4

    Step 4: Substitute the values into the formula. Q = 2.5 A × 180 s.

  5. 5

    Step 5: Calculate the result. Q = 450 C. The total charge that flows is 450 coulombs.

Recap

  • Electric current is the rate of flow of electric charge.
  • Current (I) is measured in amperes (A) using an ammeter.
  • Charge (Q) is measured in coulombs (C).
  • The formula linking them is Charge = Current × Time (Q = It).
  • Ammeters are always connected in series in a circuit.
  • Conventional current flows from positive to negative.

Quick check

  1. What is the unit of electric charge?1 mark
  2. If 20 C of charge passes a point in 4 s, what is the current?2 marks

2. Potential Difference and E.M.F.

For a current to flow, there needs to be an 'electrical push'. This push is provided by a power source like a battery or power supply. We have two key terms for this 'push': Electromotive Force (e.m.f.) and Potential Difference (p.d.). E.m.f. is the energy supplied per unit charge by the source (e.g., a battery). It's the total energy given to each coulomb of charge. Potential difference (also called voltage) is the energy transferred per unit charge when it passes through a component (e.g., a lamp or resistor). It's the energy 'used' by each coulomb to do work, like creating light and heat. Both e.m.f. and p.d. are measured in volts (V) using a voltmeter connected in parallel across the component or source. One volt is equal to one joule of energy per coulomb of charge (1 V = 1 J/C).

V = W / Q

Key term

Potential Difference (p.d.): The work done or energy transferred per unit charge as it passes through a component, measured in volts (V).

Examiner insight

Marks are awarded for clearly defining e.m.f. as energy supplied by a source and p.d. as energy transferred in a component. Remember to specify 'per unit charge' in your definitions.

Common pitfall

Using the terms 'voltage', 'p.d.', and 'e.m.f.' interchangeably. E.m.f. is the energy source for the whole circuit, while p.d. is the energy used by a part of the circuit.

Worked example 12 marks

A p.d. of 12 V is measured across a resistor. How much energy is transferred to the resistor when 50 C of charge flows through it?

  1. 1

    Step 1: State the formula linking energy (work done), p.d., and charge. W = V × Q.

  2. 2

    Step 2: Identify the given values. Potential difference V = 12 V. Charge Q = 50 C.

  3. 3

    Step 3: Substitute the values into the formula. W = 12 V × 50 C.

  4. 4

    Step 4: Calculate the result. W = 600 J. The energy transferred is 600 joules.

Worked example 22 marks

A battery has an e.m.f. of 9.0 V. It supplies a total of 450 J of energy. Calculate the total charge that it has driven around the circuit.

  1. 1

    Step 1: State the formula for e.m.f. E.m.f. = Work done / Charge (E = W/Q). Rearrange for charge: Q = W / E.

  2. 2

    Step 2: Identify the given values. Work done W = 450 J. E.m.f. E = 9.0 V.

  3. 3

    Step 3: Substitute the values into the rearranged formula. Q = 450 J / 9.0 V.

  4. 4

    Step 4: Calculate the result. Q = 50 C. The total charge driven is 50 coulombs.

Recap

  • E.m.f. is the energy supplied per unit charge by a source.
  • P.d. is the energy transferred per unit charge in a component.
  • Both e.m.f. and p.d. are measured in volts (V).
  • One volt means one joule of energy is transferred per coulomb of charge.
  • Voltmeters are always connected in parallel across a component.

Quick check

  1. State the definition of the volt.1 mark
  2. What is the key difference between e.m.f. and p.d.?2 marks

3. Resistance and Ohm's Law

Resistance (symbol R) is a measure of how much a component opposes the flow of electric current. A component with high resistance will allow less current to flow for a given potential difference compared to a component with low resistance. It's like a narrow pipe resisting the flow of water. Resistance is measured in ohms (Ω). The relationship between resistance, potential difference (V), and current (I) is given by the equation R = V / I. This is the definition of resistance. For some components, like a metal wire at a constant temperature, the resistance is constant. This special case is called Ohm's Law, which states that the current through an ohmic conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions remain unchanged.

R = V / I

Key term

Resistance: A measure of the opposition to current flow, calculated as the ratio of potential difference across a component to the current through it, measured in ohms (Ω).

Examiner insight

Candidates must be able to recall and correctly rearrange the formula R = V / I. Always show your working, including the formula you are using, to secure method marks even if your final calculation is incorrect.

Common pitfall

Believing that all components obey Ohm's Law. A filament lamp, for example, does not have a constant resistance because its temperature changes significantly as it heats up.

Worked example 12 marks

A potential difference of 6.0 V is applied across a resistor, and a current of 0.20 A is measured. Calculate the resistance of the resistor.

  1. 1

    Step 1: State the formula for resistance. R = V / I.

  2. 2

    Step 2: Identify the given values. V = 6.0 V. I = 0.20 A.

  3. 3

    Step 3: Substitute the values into the formula. R = 6.0 V / 0.20 A.

  4. 4

    Step 4: Calculate the result. R = 30 Ω. The resistance is 30 ohms.

Worked example 22 marks

A 150 Ω resistor is connected to a power supply. If a current of 0.080 A flows through it, what is the potential difference across the resistor?

  1. 1

    Step 1: State the resistance formula and rearrange it for potential difference. R = V / I => V = I × R.

  2. 2

    Step 2: Identify the given values. I = 0.080 A. R = 150 Ω.

  3. 3

    Step 3: Substitute the values into the rearranged formula. V = 0.080 A × 150 Ω.

  4. 4

    Step 4: Calculate the result. V = 12 V. The p.d. is 12 volts.

Recap

  • Resistance opposes the flow of current.
  • Resistance is measured in ohms (Ω).
  • The formula R = V / I defines resistance.
  • Ohm's Law states V is proportional to I for a conductor at constant temperature.
  • Good conductors have low resistance; insulators have very high resistance.

Quick check

  1. What is the unit of electrical resistance?1 mark
  2. If the p.d. across an ohmic resistor is tripled, what happens to the current?1 mark

4. Factors Affecting Wire Resistance

The resistance of a wire is not just a random property; it depends on several physical factors. For a wire of a uniform material, the resistance is determined by: 1. Length (L): The longer the wire, the more resistance it has. This is because the electrons have to travel further and collide with more ions in the metal lattice. Resistance is directly proportional to length (R ∝ L). 2. Cross-sectional Area (A): The thicker the wire (larger cross-sectional area), the lower its resistance. A wider path provides more space for electrons to flow, reducing congestion. Resistance is inversely proportional to the cross-sectional area (R ∝ 1/A). 3. Material: Different materials inherently resist current by different amounts. This property is called resistivity (symbol ρ, rho). Materials like copper and silver have very low resistivity, making them excellent conductors. Nichrome has a higher resistivity and is used for heating elements. This is summarised in the equation R = ρL/A.

R ∝ L

R ∝ 1/A

R = ρL / A

Key term

Resistivity (ρ): An intrinsic property of a material that quantifies how strongly it resists electric current, measured in ohm-metres (Ωm).

Examiner insight

Examiners look for a clear understanding of the proportional relationships. For example, explicitly stating 'doubling the length doubles the resistance' or 'doubling the area halves the resistance' will gain credit.

Common pitfall

Confusing diameter with cross-sectional area. If a wire's diameter is doubled, its radius is doubled, and its area (πr²) becomes four times larger, so its resistance drops to one quarter.

Worked example 13 marks

A wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² has a resistance of 4.0 Ω. A second wire is made of the same material, but is 4.0 m long and has a cross-sectional area of 0.5 × 10⁻⁶ m². Calculate the resistance of the second wire.

  1. 1

    Step 1: Analyse the changes. The length is doubled (L → 2L). The area is halved (A → A/2).

  2. 2

    Step 2: Consider the effect of doubling the length. Since R ∝ L, doubling the length will double the resistance. R becomes 4.0 Ω × 2 = 8.0 Ω.

  3. 3

    Step 3: Consider the effect of halving the area on this new resistance. Since R ∝ 1/A, halving the area will double the resistance again. R becomes 8.0 Ω × 2 = 16.0 Ω.

  4. 4

    Step 4: State the final answer. The resistance of the second wire is 16.0 Ω.

Recap

  • Resistance of a wire increases with its length.
  • Resistance of a wire decreases as its cross-sectional area increases.
  • Different materials have different resistivities (ρ).
  • Low resistivity means the material is a good conductor.
  • The formula R = ρL/A combines all these factors (Extended Level).

Quick check

  1. How does the resistance of a wire change if its length is halved?1 mark
  2. To reduce the resistance of a connecting wire, should you use a thicker or thinner wire?1 mark

5. Electrical Power and Energy

Electrical power (P) is the rate at which an appliance transfers energy. A 100 W lamp converts 100 joules of electrical energy into light and heat every second. Power is measured in watts (W). It can be calculated by multiplying the potential difference (V) across an appliance by the current (I) flowing through it. The formula is P = V × I. To find the total electrical energy (E) transferred, we multiply the power by the time(t) the appliance is on for. This gives the formula E = P × t. By substituting P = VI into this, we get a very useful equation: E = V × I × t. Energy is measured in joules (J) when power is in watts and time is in seconds. For billing purposes, electricity companies use a larger unit of energy called the kilowatt-hour (kWh). 1 kWh is the energy used by a 1 kW appliance running for 1 hour.

P = V × I

E = P × t

E = V × I × t

Key term

Power: The rate at which energy is transferred or work is done, measured in watts (W).

Examiner insight

Be prepared to use a combination of formulas. For example, you might be given V and R and asked for power, requiring you to first find I using I=V/R, and then use P=VI.

Common pitfall

Mixing up units in energy calculations. To get energy in joules, power must be in watts and time in seconds. To get energy in kilowatt-hours, power must be in kilowatts and time in hours.

Fun fact

A typical smartphone consumes only about 2-6 kWh of electricity over an entire year, costing just a dollar or two to charge.

Worked example 14 marks

A kettle is rated at 230 V, 2000 W.(a) Calculate the current it draws from the mains.(b) Calculate the energy it transfers in 2 minutes.

  1. 1

    Part (a):

  2. 2

    Step 1: State the power formula and rearrange for current. P = V × I => I = P / V.

  3. 3

    Step 2: Substitute the values. I = 2000 W / 230 V.

  4. 4

    Step 3: Calculate the result. I = 8.70 A (to 3 s.f.).

  5. 5

    Part (b):

  6. 6

    Step 1: State the energy formula. E = P × t.

  7. 7

    Step 2: Identify the values. P = 2000 W. Time t = 2 minutes = 120 s.

  8. 8

    Step 3: Substitute the values. E = 2000 W × 120 s.

  9. 9

    Step 4: Calculate the result. E = 240,000 J or 240 kJ.

Worked example 23 marks

An oven with a power rating of 3.0 kW is used for 90 minutes. If electricity costs 20p per kWh, calculate the total cost of using the oven.

  1. 1

    Step 1: Convert the time to hours. t = 90 minutes / 60 minutes/hour = 1.5 hours.

  2. 2

    Step 2: Calculate the energy transferred in kilowatt-hours. Energy (kWh) = Power (kW) × Time (h).

  3. 3

    Step 3: Substitute the values. Energy = 3.0 kW × 1.5 h = 4.5 kWh.

  4. 4

    Step 4: Calculate the total cost. Cost = Energy used × Cost per unit.

  5. 5

    Step 5: Substitute values. Cost = 4.5 kWh × 20 p/kWh = 90 p.

Recap

  • Power is the rate of energy transfer, measured in watts (W).
  • The formula for electrical power is P = V × I.
  • Energy transferred is calculated by E = P × t or E = V × I × t.
  • The SI unit for energy is the joule (J).
  • The commercial unit of electrical energy is the kilowatt-hour (kWh).

Quick check

  1. What is the formula linking power, p.d., and current?1 mark
  2. How many joules are equivalent to 1 kWh?2 marks

End-of-chapter exercise

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

  1. Define the ampere.1 mark
  2. A steady current of 0.80 A flows in a wire. Calculate the charge that passes through the wire in 2.5 minutes.3 marks
  3. A component has a potential difference of 9.0 V across it when a current of 45 mA flows through it. Calculate its resistance.3 marks
  4. Explain the difference between the electromotive force (e.m.f.) of a battery and the potential difference (p.d.) across a resistor.2 marks
  5. A car headlamp is rated at 12 V, 55 W. Calculate (a) the current that flows through it when it is operating normally, and (b) its resistance.4 marks
  6. Explain, in terms of the movement of electrons and collisions, why a thicker wire has less resistance than a thinner wire of the same length and material.2 marks
  7. A wire has a resistance of 8.0 Ω. The wire is cut in half, and the two halves are twisted together to form a new, shorter, thicker wire. What is the resistance of this new wire?4 marks
  8. A television has a power rating of 150 W. It is used for an average of 5 hours per day. If electricity costs 22 cents per kWh, calculate the cost of running the television for 30 days.4 marks
  9. A battery has an e.m.f. of 6.0 V. When it is connected to a circuit, a current of 1.2 A flows and the p.d. across the battery terminals drops to 5.4 V. Calculate (a) the energy transferred to each coulomb of charge by the battery, and (b) the energy wasted inside the battery for each coulomb of charge that passes.3 marks
  10. A student investigates a component. They measure the p.d. across it and the current through it, obtaining the following results: (V=2.0V, I=0.4A), (V=4.0V, I=0.8A), (V=6.0V, I=1.2A). (a) Calculate the resistance of the component. (b) State whether the component is ohmic or non-ohmic, and give a reason for your answer.3 marks

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