Cells and batteries — KCSE Physics

KCSE Physics · 118 practice questions · 8 syllabus objectives · 8 revision lessons

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Last updated · Aligned to the KNEC KCSE syllabus

What You'll Learn

Key learning outcomes for this topic, aligned to the KNEC KCSE syllabus.

Explain why a high-resistance voltmeter is preferred; draw circuits for measuring cell properties

Define electromotive force (EMF) and internal resistance; distinguish between EMF and terminal potential difference

Apply the formula E = V + Ir to calculate EMF, terminal pd, internal resistance or current for a cell circuit

Describe primary and secondary cells; state the advantages of connecting cells in series and in parallel

Distinguish between primary and secondary cells; describe the dry cell and state its components and limitations

Describe the lead-acid accumulator, its maintenance, charging indicators, and advantages of alkaline batteries

Calculate total EMF and current for cells connected in series and parallel; state factors affecting battery capacity

Cells and batteries

Revision Notes

Concise lesson notes for Cells and batteries, written to the KCSE Physics marking standard. Read the first lesson free below.

Using High-Resistance Voltmeters

High-resistance voltmeters are preferred in measuring cell properties because:

  • They minimize current draw from the circuit.
  • They prevent loading effects that can alter the voltage reading.
  • They ensure accurate measurements of the cell’s electromotive force (EMF).

When measuring the voltage across a cell, a high-resistance voltmeter is connected in parallel. This configuration allows the voltmeter to measure the potential difference without significantly affecting the circuit.

Circuit Diagram:

  1. Draw a simple circuit with a cell (battery) connected to a resistor.
  2. Connect the voltmeter in parallel across the cell or resistor.

Example:

  • Cell Voltage Measurement:
    • Connect a high-resistance voltmeter across a 1.5V cell.
    • The voltmeter reading should ideally show 1.5V, indicating the cell's EMF without drawing excessive current.

In summary, using a high-resistance voltmeter ensures accurate voltage measurements by reducing the impact on the circuit being tested.

Key points to remember

  • High-resistance voltmeters minimize current draw from circuits.
  • They prevent loading effects that alter voltage readings.
  • Connected in parallel for accurate potential difference measurement.

Worked example

A circuit has a 1.5V cell and a resistor. Connect a high-resistance voltmeter across the cell. The voltmeter reads 1.5V, confirming accurate EMF measurement.

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More lessons in this topic

Lesson 2: Understanding EMF and Internal Resistance

Objective: Define electromotive force (EMF) and internal resistance; distinguish between EMF and terminal potential difference

Electromotive force (EMF) is defined as the energy provided by a cell or battery per coulomb of charge. It is measured in volts (V). Internal resistance, on the other hand, is the opposition to the flow of current within the cell itself, also measured in ohms (Ω).

Distinction between EMF and Terminal Potential Difference:

  • EMF is the maximum potential difference when no current flows.
  • Terminal potential difference (V) is the potential difference across the terminals when current flows.

The relationship can be expressed as:

V = EMF - Ir

where I is the current and r is the internal resistance. This equation shows that the terminal potential difference decreases when current flows due to internal resistance.

In summary, while EMF indicates the total energy available, terminal potential difference reflects the energy available for external use after accounting for internal losses due to resistance.

  • EMF is energy per coulomb produced by a cell.
  • Internal resistance opposes current flow within the cell.
  • EMF is measured in volts (V).
  • Terminal potential difference is lower than EMF when current flows.
  • V = EMF - Ir explains the relationship.

Define EMF and internal resistance, and explain their difference.

  • EMF is the energy per coulomb from a battery.
  • Internal resistance is the opposition to current in the battery.
  • EMF is the maximum voltage, while terminal potential difference is lower when current flows.
Lesson 3: Understanding Cell Circuits with E = V + Ir

Objective: Apply the formula E = V + Ir to calculate EMF, terminal pd, internal resistance or current for a cell circuit

In electrical circuits, the formula E = V + Ir is essential for analyzing cell behavior. Here, E represents the electromotive force (EMF) of the cell, V is the terminal potential difference (pd), I is the current flowing through the circuit, and r is the internal resistance of the cell.

To effectively use this formula, follow these steps:

  • Identify the known values (EMF, terminal pd, current, internal resistance).
  • Rearrange the formula to solve for the unknown variable.
  • Substitute the values and perform the calculations.

For example, if a cell has an EMF of 12V, a terminal pd of 10V, and a current of 2A, we can find the internal resistance:

  1. Rearranging gives: r = (E - V) / I.
  2. Substituting the values: r = (12V - 10V) / 2A = 1Ω.

Thus, the internal resistance is . Mastering this formula allows you to analyze various components in cell circuits accurately.

  • E represents the electromotive force of the cell.
  • V is the terminal potential difference in the circuit.
  • I denotes the current flowing through the circuit.
  • r is the internal resistance of the cell.
  • Rearranging the formula helps solve for unknowns.

Calculate the current in a circuit with E = 9V, V = 7V, and r = 1Ω. Using E = V + Ir, rearrange to I = (E - V) / r. I = (9V - 7V) / 1Ω = 2A.

Lesson 4: Understanding Primary and Secondary Cells

Objective: Describe primary and secondary cells; state the advantages of connecting cells in series and in parallel

Primary cells are electrochemical cells that cannot be recharged; they are used until depleted. Common examples include alkaline batteries and zinc-carbon batteries. In contrast, secondary cells, such as lead-acid and lithium-ion batteries, can be recharged and reused multiple times.

Advantages of connecting cells in series:

  • Increases the total voltage output, as voltages add up.
  • Useful for applications requiring higher voltage, like flashlights.

Advantages of connecting cells in parallel:

  • Increases the total current capacity while maintaining the same voltage.
  • Provides longer battery life since each cell shares the load.

Understanding these concepts is crucial for efficient energy storage and usage in various devices.

  • Primary cells cannot be recharged and are used once.
  • Secondary cells can be recharged and reused multiple times.
  • Series connection increases total voltage output.
  • Parallel connection increases total current capacity.
  • Choosing the right connection depends on voltage and current needs.

Describe the differences between primary and secondary cells.

  • Primary cells are non-rechargeable; secondary cells are rechargeable.
  • Examples: alkaline batteries (primary) vs. lithium-ion batteries (secondary).

Sample Questions

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1
easySHORT ANSWER4 marks

Identify the main differences between primary and secondary cells, and give one advantage of connecting cells in series. (4 marks)

Answer & marking scheme

Part (a) — 3 marks
A primary cell is non-rechargeable, while a secondary cell can be recharged. (1 mk)
Primary cells have a limited lifespan based on chemical depletion, whereas secondary cells can be reused multiple times. (1 mk)
Primary cells typically have lower energy density compared to secondary cells. (1 mk)
Part (b) — 1 mark
Connecting cells in series increases the total voltage output, allowing for greater energy supply to the circuit. (1 mk)
2
easySHORT ANSWER3 marks

A cell has an electromotive force (EMF) of 12 V and an internal resistance of 2 Ω. If it is connected to a load resistor of 4 Ω, calculate the current flowing through the circuit. (3 marks)

Answer & marking scheme

Part (a) — 3 marks
Total resistance in the circuit = internal resistance + load resistance = 2 Ω + 4 Ω = 6 Ω (1 mk)
Current (I) = EMF / Total resistance = 12 V / 6 Ω = 2 A (1 mk)
Final answer: Current flowing is 2 A (1 mk)
3
easySHORT ANSWER3 marks

Name the difference between electromotive force (EMF) and terminal potential difference in a battery when a current flows. Explain why the terminal potential difference is lower than the EMF. (3 marks)

Answer & marking scheme

Part (a) — 1 mark
EMF is the voltage provided by the battery when no current flows; terminal pd is the voltage across the load when current flows. (1 mk)
Part (b) — 2 marks
Terminal pd is less than EMF due to the voltage drop across the internal resistance of the battery. (1 mk)
As current flows, some energy is converted to heat within the battery, resulting in lower terminal pd. (1 mk)
4

List three characteristics of a dry cell that make it suitable for portable electronic devices. (3 marks)

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Frequently asked questions

What does the KCSE Physics topic "Cells and batteries" cover?

Cells and batteries covers Explain why a high-resistance voltmeter is preferred; draw circuits for measuring cell properties; Define electromotive force (EMF) and internal resistance; distinguish between EMF and terminal potential difference; Apply the formula E = V + Ir to calculate EMF, terminal pd, internal resistance or current for a cell circuit, and more, all aligned to the official KNEC KCSE Physics syllabus.

How many practice questions are available for Cells and batteries?

HighMarks has 118 Cells and batteries practice questions for KCSE Physics, each with a full marking scheme. The first 3 are free; sign up to access the rest, plus all KCSE mock exams and past papers.

Are these aligned with the KNEC KCSE syllabus?

Yes. Every objective on this page is taken directly from the official KNEC KCSE Physics syllabus. Practice questions match the KCSE exam format and are graded against the standard KNEC marking scheme.

How should I revise Cells and batteries for the KCSE exam?

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