Electric current

1. Overview

Electric current is the fundamental process by which energy is transferred through electrical circuits to power appliances. Understanding how charge moves through materials and how to measure it is essential for mastering electronics and circuit design.

Key Definitions

• Electric Current: The rate of flow of electric charge.
• Electric Charge: A property of matter (measured in Coulombs) that causes it to experience a force in an electromagnetic field.
• Conductor: A material (usually a metal) that allows electricity to flow through it easily.
• Direct Current (d.c.): Current that flows in one constant direction only.
• Alternating Current (a.c.): Current that continuously changes direction and magnitude.
• Ammeter: A device used to measure the size of the electric current in a circuit.

Core Content

Electric Current and Charge

Electric current is the result of the movement of charge. In a circuit, this charge is carried by tiny particles. Without the movement of these charges, there is no current.

Electrical Conduction in Metals

Metals are excellent conductors because of their internal structure:

• Metals consist of a lattice of positive ions surrounded by a "sea" of free electrons (delocalized electrons).
• When a power source is connected, these free electrons are pushed away from the negative terminal and toward the positive terminal.
• The movement of these free electrons constitutes the electric current.

Using Ammeters

To measure current, an ammeter must be connected in series with the component you are investigating.

• Analogue Ammeters: Use a needle and a scale. You must look directly at the scale to avoid parallax error and check the range (e.g., 0-1A or 0-5A) before reading.
• Digital Ammeters: Provide a direct numerical reading. They are generally more accurate and easier to read.
• Ranges: Always start with the highest range setting to avoid damaging the meter if the current is unexpectedly high.

Direct Current (d.c.) vs. Alternating Current (a.c.)

• d.c. (Direct Current):
  • Flows in one direction only (from positive to negative).
  • Produced by cells and batteries.
• a.c. (Alternating Current):
  • Direction of flow reverses back and forth periodically.
  • Produced by mains electricity (e.g., wall sockets).

Extended Content (Extended Only)

Defining Current Mathematically

Electric current ($I$) is defined as the charge ($Q$) passing a point per unit time ($t$). $$I = \frac{Q}{t}$$

Worked Example: A charge of 12 C passes through a bulb in 4 seconds. Calculate the current.

• $Q = 12 \text{ C}$
• $t = 4 \text{ s}$
• $I = 12 / 4 = 3 \text{ A}$

Conventional Current vs. Electron Flow

There is a historical distinction between the direction we draw current and how it actually moves:

• Conventional Current: Flows from the positive terminal to the negative terminal. (This is what we use when drawing circuit diagrams).
• Electron Flow: Free electrons (which are negatively charged) actually flow from the negative terminal to the positive terminal.

Key Equations

• The Equation: $I = \frac{Q}{t}$ or $Q = I \times t$
• Symbols:
  • $I$ = Current
  • $Q$ = Charge
  • $t$ = Time
• Units:
  • Current: Amperes (A)
  • Charge: Coulombs (C)
  • Time: Seconds (s)

Common Mistakes to Avoid

• ❌ Wrong: Thinking current is "used up" by a bulb or resistor.
• ✓ Right: Current is conserved; the amount of current entering a component is the same as the amount leaving it.
• ❌ Wrong: Calculating current using time in minutes.
• ✓ Right: Always convert time to seconds before using the $I = Q/t$ formula. (e.g., 2 minutes = 120 seconds).
• ❌ Wrong: Connecting an ammeter in parallel (like a voltmeter).
• ✓ Right: Ammeters must always be in series; their low resistance allows current to flow through them without changing the circuit's behavior.
• ❌ Wrong: Confusing current with Potential Difference (Voltage).
• ✓ Right: Current is the rate of flow of charge; Voltage is the energy transferred per unit charge.

Exam Tips

1. Check the units: If a question gives you charge in millicoulombs (mC) or time in minutes, convert them to standard units (C and s) immediately.
2. Series Rule: Remember that in a single-loop (series) circuit, the current is the same at all points. If an ammeter at the start reads 2A, an ammeter at the end will also read 2A.
3. Definitions: When asked to define current, always use the phrase "rate of flow of charge" or "charge per unit time" to get full marks.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) Define electric current. [2]

(b) State the conventional direction of electric current. [1]

(c) Describe, in terms of free electrons, how electric current flows in a metal wire. [2]

Worked Solution:

(a)

1. Electric current is the rate of flow of charge. This is the basic definition.

2. or: Electric current is the amount of charge passing a point per unit time. Alternative correct definition.

How to earn full marks:

• Mention "rate of flow of charge" or equivalent.
• Alternatively, "amount of charge passing a point per unit time" is also acceptable.

(b)

1. From positive to negative. This is the conventional direction.

How to earn full marks:

• State "positive to negative" explicitly.

(c)

1. Free electrons move through the metal. Electrons are the charge carriers in a metal.
2. The electrons move from the negative terminal to the positive terminal of the power supply. Electrons are repelled by the negative terminal and attracted to the positive terminal.

How to earn full marks:

• Mention "free electrons" or "delocalised electrons".
• State that the electrons move from negative to positive (or the reverse direction compared to conventional current).

Common Pitfall: Many students confuse current with voltage or energy. Remember that current is the rate at which charge flows, not the "push" behind it (voltage) or the total energy involved. Also, don't say that current "gets used up" in a circuit; it's the energy carried by the charge that is transferred.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) An electric kettle draws a current of 10 A from a 230 V mains supply. Calculate the amount of electric charge that flows through the kettle in 30 seconds. [3]

(b) An electric heater has a label indicating that it uses alternating current (a.c.). State two differences between alternating current (a.c.) and direct current (d.c.). [3]

Worked Solution:

(a)

1. Recall the formula relating current, charge, and time: $I = \frac{Q}{t}$. This is the fundamental equation for electric current.

2. Rearrange the formula to solve for charge: $Q = I \times t$. Isolate the variable of interest.

3. Substitute the given values: $Q = 10 \text{ A} \times 30 \text{ s} = 300 \text{ C}$. Plug in the values and calculate.

How to earn full marks:

• State the correct formula $I = Q/t$ or $Q = It$.
• Correct substitution of $I=10$ and $t=30$ into the formula.
• Correct final answer with unit: $\boxed{300 \text{ C}}$

(b)

1. In a.c., the direction of the current changes periodically; in d.c., the direction of the current is constant. Direction of current is a key difference.

2. In a.c., the magnitude of the current changes periodically; in d.c., the magnitude of the current is constant. Magnitude of current is also a key difference.

How to earn full marks:

• State that a.c. changes direction, while d.c. does not.
• State that a.c. changes magnitude (or voltage), while d.c. does not.

Common Pitfall: In part (a), be sure to use the correct units for time (seconds). Many students forget to convert minutes or hours into seconds, leading to a wrong answer. In part (b), avoid saying that AC is "stronger" or "weaker" than DC; focus on the changing direction and magnitude.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A student connects a lamp to a power supply using connecting wires. The student increases the potential difference (voltage) across the lamp and measures the current flowing through it. The results are recorded in a table.

Potential Difference (V)	Current (A)
0.0	0.00
1.0	0.10
2.0	0.19
3.0	0.27
4.0	0.34
5.0	0.40

(a) On the grid provided, plot a graph of current (y-axis) against potential difference (x-axis). Draw a line of best fit. [4]

(b) Describe the relationship between the potential difference across the lamp and the current flowing through it, as shown by your graph. [2]

(c) The student notices that the lamp gets brighter as the potential difference increases. Explain why the lamp gets brighter in terms of electrical energy and heat. [2]

Worked Solution:

(a)

1. 📊A graph with Potential Difference (V) on the x-axis (0 to 5, labelled) and Current (A) on the y-axis (0 to 0.4, labelled). The points (0,0), (1, 0.1), (2, 0.19), (3, 0.27), (4, 0.34), and (5, 0.4) are plotted correctly. A smooth curve of best fit is drawn through the points, starting at (0,0) and curving upwards slightly.

How to earn full marks:

• Correctly label both axes with quantities and units.
• Accurately plot all points (allow +/- half a small square tolerance).
• Draw a smooth curve of best fit that passes reasonably close to all points.
• The line of best fit should start from the origin.

(b)

1. As the potential difference increases, the current increases. This is the basic relationship.

2. The current increases non-linearly with the potential difference. The graph is not a straight line, indicating a non-linear relationship.

How to earn full marks:

• State that current increases as potential difference increases.
• State that the relationship is non-linear (or that the graph is a curve).

(c)

1. Electrical energy is converted into heat energy in the lamp filament. This is the energy conversion that occurs.

2. As the current increases, more electrical energy is converted into heat energy per unit time, increasing the temperature of the filament, causing it to glow brighter. Higher current = more energy converted = hotter filament = brighter light.

How to earn full marks:

• Mention the conversion of electrical energy to heat energy.
• Explain that higher current leads to more energy conversion, a hotter filament, and therefore brighter light.

Common Pitfall: When plotting graphs, always double-check that your axes are labeled correctly with both the quantity and the unit. Also, remember that a "line of best fit" doesn't necessarily have to go through every point, but should represent the overall trend of the data. In part (c), be specific about the energy conversions taking place.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A scientist is investigating the current-carrying capacity of different metal wires before they melt and act as fuses.

(a) Plan an experiment to investigate how the melting current of a wire depends on its diameter. Your plan should include: - a list of the apparatus required - a clear description of the procedure, including how you would control any relevant variables - a description of how you would analyse the data to reach a conclusion. [7]

(b) State one safety precaution that should be taken when carrying out this experiment. [1]

(c) The scientist finds that a wire with a diameter of 0.5 mm melts at a current of 10 A. Predict the melting current for a wire made of the same metal with a diameter of 1.0 mm, assuming the melting current is proportional to the cross-sectional area of the wire. Show your working. [1]

Worked Solution:

(a)

1. Apparatus: Power supply (variable), ammeter, connecting wires, different lengths of wires of the same metal but with different diameters (e.g., 0.2 mm, 0.4 mm, 0.6 mm, 0.8 mm, 1.0 mm), ruler, micrometer screw gauge (to accurately measure the wire diameter). List the necessary equipment.

2. Procedure:

   • Set up the circuit with the wire connected in series with the ammeter and the power supply.
   • Start with the lowest diameter wire.
   • Gradually increase the potential difference (voltage) of the power supply, observing the current on the ammeter.
   • Record the current at which the wire melts and breaks the circuit.
   • Repeat the measurement three times for each diameter of wire, and calculate the average melting current.
   • Repeat the entire procedure for the other wires with different diameters.
   • Keep the length of each wire the same for each measurement to ensure fair testing. Use a ruler to measure the length. Describe the experimental steps, including how to control length.

3. Data Analysis:

   • Calculate the cross-sectional area $A$ of each wire using the formula $A = \pi r^2$, where $r$ is the radius of the wire (half of the diameter).
   • Plot a graph of the average melting current (y-axis) against the cross-sectional area (x-axis).
   • If the melting current is proportional to the cross-sectional area, the graph will be a straight line through the origin. Describe how to calculate area and plot the graph, and how to interpret the graph.

How to earn full marks:

• List appropriate apparatus including a variable power supply, ammeter, wires of different diameters, and a way to measure diameter (e.g., micrometer).
• Describe a clear procedure for increasing the current and recording the melting current.
• Describe repeating the experiment for different wire diameters and taking multiple readings to calculate an average.
• State that the length of the wire should be kept constant.
• Describe calculating the cross-sectional area of the wires.
• Describe plotting a graph of melting current against cross-sectional area.
• State that a straight line through the origin would indicate proportionality.

(b)

1. Wear safety goggles to protect the eyes from flying pieces of the wire when it melts. Safety first!

How to earn full marks:

• Mention wearing safety goggles or safety glasses.
• OR mention that the circuit should be switched off immediately after the wire melts.
• OR mention that the wires may become hot and should not be touched immediately after the experiment.

(c)

1. The cross-sectional area of the 1.0 mm wire is four times the cross-sectional area of the 0.5 mm wire, since area is proportional to the diameter squared: $(\frac{1.0}{0.5})^2 = 4$. Area is proportional to diameter squared.

2. If the melting current is proportional to the area, the melting current for the 1.0 mm wire will be four times that of the 0.5 mm wire: $10 \text{ A} \times 4 = 40 \text{ A}$. Melting current for 1.0 mm wire.

How to earn full marks:

• Show that the ratio of the areas is 4 (or that the area is proportional to the diameter squared).
• Calculate the melting current for the 1.0 mm wire: $\boxed{40 \text{ A}}$.

Common Pitfall: In experimental design questions, remember to identify and control all relevant variables. In this case, the length of the wire is crucial. Also, don't forget basic safety precautions. In part (c), many students forget that area is proportional to the square of the diameter.