Cambridge IGCSE0625

Electrical quantities

Physics 0625 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 Fields

At the heart of electricity is electric charge. There are two types: positive and negative. Protons in an atom's nucleus are positive, and electrons orbiting the nucleus are negative. Objects become charged by gaining or losing electrons. Gaining electrons makes an object negatively charged; losing them makes it positively charged. Like charges repel each other (positive repels positive), while opposite charges attract (positive attracts negative). Materials that allow electrons to move freely are called electrical conductors (e.g., metals), while materials that don't are electrical insulators (e.g., plastic, rubber). An electric field is a region of space around a charged object where another charge would experience a force. For the Extended syllabus, you need to know the shape of these fields: they radiate outwards from positive charges and inwards towards negative charges.

Key term

Electric Field: A region of space in which an electric charge experiences a force.

Examiner insight

For Extended level, examiners award marks for correctly drawing electric field lines that are radial, do not cross, and have arrows indicating the correct direction (away from positive, towards negative).

Common pitfall

Thinking that protons move to create charge. In solid conductors, only the negative electrons are free to move.

Worked example 14 marks

A plastic rod is rubbed with a dry cloth, causing it to become negatively charged.a) Explain in terms of electrons how the rod became negatively charged.b) The rod is then brought near a small, uncharged piece of paper, which is an insulator. The paper is attracted to the rod. Explain why this happens. (Extended)

  1. 1

    a) Friction between the cloth and the rod causes electrons to be transferred from the cloth to the rod. Since electrons are negatively charged, the rod gains a net negative charge.

  2. 2

    b) When the negative rod is brought near the paper, it repels the electrons in the paper's atoms, pushing them slightly to the far side. This leaves the side of the paper closer to the rod with a net positive charge.

  3. 3

    The attraction between the negative rod and the induced positive side of the paper is stronger than the repulsion between the rod and the induced negative side (which is further away). This results in an overall attractive force.

Worked example 22 marks

Draw the electric field pattern for a single, isolated positive point charge.

  1. 1

    Start by drawing a small circle with a '+' sign in it to represent the positive charge.

  2. 2

    Draw at least six straight lines pointing radially outwards from the charge. These are the electric field lines.

  3. 3

    Add arrows to each line, pointing away from the positive charge, to show the direction of the field.

Recap

  • There are two types of charge: positive and negative.
  • Like charges repel; opposite charges attract.
  • Objects are charged by the transfer of electrons.
  • Conductors allow charge to flow easily; insulators do not.
  • An electric field is a region where a charge feels a force.
  • Electric field lines show the direction of the force on a positive charge.

Quick check

  1. What happens when two negatively charged objects are brought close together?1 mark
  2. Is copper a conductor or an insulator? Explain why.2 marks

2. Current, Potential Difference and Charge

Electric current is the rate of flow of electric charge. Think of it like the amount of water flowing through a pipe per second. We measure current in amperes (A) using an ammeter connected in series. The charge itself is measured in coulombs (C). For current to flow, there needs to be a 'push'. This electrical push is called the potential difference (p.d.) or voltage, measured in volts (V) using a voltmeter connected in parallel. Potential difference is the energy transferred per unit charge. A battery or power supply provides an electromotive force (e.m.f.), also measured in volts. E.m.f. is the total energy supplied per coulomb of charge by the source to the whole circuit, while p.d. is the energy used per coulomb by a specific component.

I = Q / t

V = E / Q

Key term

Potential Difference (p.d.): The work done or energy transferred per unit charge between two points in a circuit.

Examiner insight

A clear distinction between e.m.f. and p.d. is crucial. E.m.f. is the cause (from the battery), and p.d. is the effect (across a component).

Common pitfall

Confusing the direction of conventional current (what we use in circuit diagrams) with the actual flow of electrons. They are in opposite directions.

Worked example 14 marks

A current of 2.5 A flows through a wire for 2 minutes.a) Calculate the total charge that passes through the wire.b) If the potential difference across the wire is 12 V, how much energy is transferred?

  1. 1

    a) First, convert the time into seconds: t = 2 minutes = 2 × 60 = 120 s.

  2. 2

    Use the formula Q = I × t. Substitute the values: Q = 2.5 A × 120 s.

  3. 3

    Calculate the result: Q = 300 C.

  4. 4

    b) Use the formula E = V × Q. Substitute the values: E = 12 V × 300 C.

  5. 5

    Calculate the result: E = 3600 J.

Recap

  • Current (I) is the rate of flow of charge (Q), measured in amperes (A).
  • Potential difference (V) is the energy transferred per unit charge, measured in volts (V).
  • E.m.f. is the energy supplied by the source; p.d. is the energy used by a component.
  • Ammeters are connected in series; voltmeters are connected in parallel.
  • Conventional current flows from positive to negative; electrons flow from negative to positive.

Quick check

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

3. Resistance and Ohm's Law

Resistance is a measure of how difficult it is for current to flow through a component. It's measured in ohms (Ω). A high resistance means a small current will flow for a given potential difference. The relationship between p.d. (V), current (I), and resistance (R) is given by the equation R = V/I. Ohm's Law states that for a conductor at a constant temperature, the current is directly proportional to the potential difference across it. This means its resistance is constant. A graph of current against p.d. (an I-V graph) for an 'ohmic' resistor is a straight line through the origin. However, many components are 'non-ohmic'. For a filament lamp, the I-V graph curves because as it gets hotter, its resistance increases. For a diode, current flows easily in one direction (forward bias) but not at all in the other (reverse bias).

R = V / I

Key term

Resistance: The ratio of the potential difference across a component to the current flowing through it.

Examiner insight

When explaining the I-V graph for a filament lamp, you must link the increased current to increased temperature, and then link increased temperature to increased resistance.

Common pitfall

Incorrectly reading I-V graphs. Remember that resistance R = V/I, so for any point on the graph, you must divide the voltage value by the current value.

Worked example 14 marks

A resistor is connected to a 6.0 V supply and a current of 0.50 A flows through it.a) Calculate the resistance of the resistor.b) The voltage is increased to 12.0 V. Assuming the resistor is ohmic, what is the new current?

  1. 1

    a) Use the formula R = V / I. Substitute the values: R = 6.0 V / 0.50 A.

  2. 2

    Calculate the resistance: R = 12 Ω.

  3. 3

    b) Rearrange the formula to find current: I = V / R. The resistance is constant (12 Ω).

  4. 4

    Substitute the new voltage: I = 12.0 V / 12 Ω.

  5. 5

    Calculate the new current: I = 1.0 A.

Worked example 23 marks

The I-V graph for a component is a curve that gets steeper. Does its resistance increase or decrease as the current increases? Explain your answer.

  1. 1

    Resistance is calculated as R = V/I. This is the gradient of a V-I graph, or 1/gradient of an I-V graph.

  2. 2

    As the I-V graph gets steeper, the gradient (I/V) is increasing.

  3. 3

    Since R = 1 / (gradient of I-V graph), if the gradient is increasing, the resistance R must be decreasing.

Recap

  • Resistance (R) opposes the flow of current, measured in ohms (Ω).
  • The formula linking V, I, and R is R = V/I.
  • Ohm's Law applies to components with constant resistance (at constant temperature).
  • The I-V graph for an ohmic resistor is a straight line through the origin.
  • A filament lamp's resistance increases as it gets hotter.
  • A diode allows current to flow in one direction only.

Quick check

  1. What is Ohm's Law?1 mark
  2. A p.d. of 10 V causes a current of 2 A to flow through a component. What is its resistance?2 marks

4. Series and Parallel Circuits

There are two basic ways to connect components in a circuit: series and parallel. In a series circuit, components are connected end-to-end in a single loop. The current is the same at all points in a series circuit. The total potential difference from the supply is shared between the components. The total resistance is the sum of the individual resistances (R_total = R1 + R2 + ...). In a parallel circuit, components are connected on separate branches. The potential difference across each branch is the same. The total current from the supply is the sum of the currents in each branch (it splits up). The total resistance of a parallel circuit is always less than the smallest individual resistor.

Series: R_total = R1 + R2 + ...

Series: V_supply = V1 + V2 + ...

Parallel: 1/R_total = 1/R1 + 1/R2 + ... (Extended)

Parallel: I_total = I1 + I2 + ...

Key term

Series Circuit: A circuit in which components are connected in a single loop, so the same current flows through each one.

Examiner insight

Be clear about which rules apply to which type of circuit. Examiners often test this by asking you to compare the brightness of bulbs in series vs. parallel arrangements.

Common pitfall

Forgetting to invert the answer when calculating total resistance in parallel. 1/R_total is not the final answer, R_total is.

Worked example 15 marks

Two resistors, 6 Ω and 3 Ω, are connected in series to a 12 V battery. Calculate:a) the total resistance of the circuit,b) the current flowing from the battery,c) the p.d. across the 6 Ω resistor.

  1. 1

    a) For series circuits, R_total = R1 + R2. So, R_total = 6 Ω + 3 Ω = 9 Ω.

  2. 2

    b) Use Ohm's Law for the whole circuit: I = V_total / R_total. So, I = 12 V / 9 Ω = 1.33 A.

  3. 3

    c) The current through the 6 Ω resistor is the same as the total current (1.33 A). Use Ohm's Law for this resistor: V = I × R. So, V = 1.33 A × 6 Ω = 8.0 V.

Worked example 23 marks

The same two resistors, 6 Ω and 3 Ω, are now connected in parallel to the 12 V battery. Calculate the total resistance. (Extended)

  1. 1

    For parallel circuits, use the formula 1/R_total = 1/R1 + 1/R2.

  2. 2

    Substitute the values: 1/R_total = 1/6 + 1/3.

  3. 3

    Find a common denominator: 1/R_total = 1/6 + 2/6 = 3/6.

  4. 4

    Simplify: 1/R_total = 1/2.

  5. 5

    Invert both sides to find R_total: R_total = 2 Ω.

Recap

  • In series, current is constant and voltage is shared.
  • In series, total resistance is R1 + R2 + ...
  • In parallel, voltage is constant and current is shared.
  • In parallel, total resistance is less than the smallest resistor.
  • Lamps in a house are connected in parallel so they can be switched independently and receive the full mains voltage.

Quick check

  1. What happens to the other lamps in a series circuit if one bulb breaks?1 mark
  2. Two 20 Ω resistors are connected in series. What is their total resistance?1 mark

5. Electrical Power and Energy

Power is the rate at which energy is transferred. In electrical circuits, power is the amount of energy an appliance uses per second. It is measured in watts (W). The power of a component can be calculated by multiplying the potential difference across it by the current flowing through it. Since energy is power multiplied by time, the electrical energy transferred can also be calculated directly. The standard unit for energy is the joule (J). However, electricity companies measure energy in a larger unit called the kilowatt-hour (kWh). One kilowatt-hour 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

Kilowatt-hour (kWh): A unit of energy equal to the energy transferred by a 1 kilowatt device operating for 1 hour.

Common pitfall

Mixing up units in energy calculations. For energy in Joules, power must be in Watts and time in seconds. For energy in kWh, power must be in kW and time in hours.

Fun fact

A modern, energy-efficient LED bulb might use only 10W to produce the same amount of light as an old 100W incandescent bulb, demonstrating a huge increase in efficiency and saving 90% of the energy.

Worked example 14 marks

An electric kettle has a power rating of 2300 W and is connected to a 230 V mains supply.a) Calculate the current drawn by the kettle.b) If the kettle takes 3 minutes to boil water, calculate the energy transferred in joules.

  1. 1

    a) Rearrange the power formula: I = P / V. Substitute values: I = 2300 W / 230 V.

  2. 2

    Calculate the current: I = 10 A.

  3. 3

    b) First, convert time to seconds: t = 3 minutes = 3 × 60 = 180 s.

  4. 4

    Use the energy formula: E = P × t. Substitute values: E = 2300 W × 180 s.

  5. 5

    Calculate the energy: E = 414,000 J (or 414 kJ).

Worked example 23 marks

An electric heater is rated at 2 kW. If electricity costs 15p per kWh, how much does it cost to run the heater for 4 hours?

  1. 1

    First, calculate the energy used in kWh. Energy (kWh) = Power (kW) × Time (h).

  2. 2

    Substitute values: Energy = 2 kW × 4 h = 8 kWh.

  3. 3

    Now, calculate the total cost. Cost = Energy used × Cost per unit.

  4. 4

    Substitute values: Cost = 8 kWh × 15 p/kWh = 120 p.

  5. 5

    Convert to pounds: Cost = £1.20.

Recap

  • Power is the rate of energy transfer, measured in watts (W).
  • Electrical power is calculated using P = V × I.
  • Energy transferred is calculated using E = P × t or E = V × I × t.
  • The joule (J) is the standard unit of energy.
  • The kilowatt-hour (kWh) is a larger unit of energy used for electricity bills.

Quick check

  1. What is the equation for electrical power?1 mark
  2. A 100 W light bulb is left on for 10 seconds. How much energy does it use?2 marks

End-of-chapter exercise

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

  1. A charge of 450 C flows through a lamp in 5 minutes. Calculate the current in the lamp.3 marks
  2. Describe the difference between alternating current (a.c.) and direct current (d.c.). You may sketch graphs to illustrate your answer.3 marks
  3. A circuit consists of a 10 Ω resistor and a 20 Ω resistor connected in series with a 6 V battery. Calculate the potential difference across the 10 Ω resistor.4 marks
  4. A student investigates a filament lamp. They measure the current through it for different voltages. Sketch the expected I-V graph and explain its shape.4 marks
  5. A 4 Ω resistor and a 12 Ω resistor are connected in parallel. a) Calculate their combined resistance. b) If they are connected to a 6 V supply, what is the total current drawn from the supply?4 marks
  6. Explain the role of a fuse in a mains plug and why it is important to use a fuse with the correct rating.3 marks
  7. A microwave oven is rated at 800 W. It is used for a total of 30 minutes. If electricity costs 20p per kWh, calculate the cost of using the microwave.4 marks
  8. A copper wire has a resistance of 5.0 Ω. A second wire, made of the same material, is twice as long and has half the cross-sectional area. What is the resistance of the second wire?3 marks
  9. (Extended) A potential divider circuit is made with a 1000 Ω and a 3000 Ω resistor in series with a 12 V supply. The output voltage is taken across the 3000 Ω resistor. Calculate the output voltage.3 marks
  10. (Extended) Describe, with the aid of a diagram, the electric field pattern between two parallel, oppositely charged metal plates.3 marks

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