Current, Voltage and Resistance

This lesson introduces the three fundamental electrical quantities — current, voltage (potential difference) and resistance — as required by the Edexcel GCSE Combined Science specification (1SC0). You must understand what each quantity means, how it is measured and how the three are related.

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Electric Current

Electric current is the rate of flow of electric charge around a circuit.

$I = \frac{Q}{t}$I=tQ​

where:

• I = current in amperes (A)
• Q = charge in coulombs (C)
• t = time in seconds (s)

Quantity	Symbol	Unit	Measuring Instrument
Current	I	Ampere (A)	Ammeter (connected in series)

Key Points

• Current is carried by electrons in metal wires.
• Electrons are negatively charged and flow from the negative terminal to the positive terminal of the battery.
• Conventional current is defined as flowing from positive to negative — this is the direction used in circuit diagrams and calculations.

Exam Tip: Do not confuse electron flow with conventional current. In exam answers, unless specifically asked about electrons, use conventional current (positive to negative).

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Charge and Current

The equation $Q = It$Q=It lets you calculate the total charge that flows in a given time.

Worked Example 1

A current of 0.5 A flows through a lamp for 2 minutes. Calculate the charge that flows.

Step 1 — Convert time to seconds:

$t = 2 \times 60 = 120 \text{ s}$t=2×60=120 s

Step 2 — Use $Q = It$Q=It:

$Q = 0.5 \times 120 = 60 \text{ C}$Q=0.5×120=60 C

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Potential Difference (Voltage)

Potential difference (often called voltage) is the energy transferred per unit of charge that passes between two points in a circuit.

$V = \frac{E}{Q}$V=QE​

where:

• V = potential difference in volts (V)
• E = energy transferred in joules (J)
• Q = charge in coulombs (C)

Quantity	Symbol	Unit	Measuring Instrument
Potential difference	V	Volt (V)	Voltmeter (connected in parallel)

Key Points

• A 1 volt potential difference means that 1 joule of energy is transferred for every 1 coulomb of charge that passes.
• The battery (or power supply) provides the potential difference that pushes charge around the circuit.
• Components such as lamps and resistors use up potential difference as they transfer energy.

Worked Example 2

A charge of 20 C passes through a resistor. The energy transferred is 100 J. Calculate the potential difference across the resistor.

$V = \frac{E}{Q} = \frac{100}{20} = 5 \text{ V}$V=QE​=20100​=5 V

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Resistance

Resistance is a measure of how much a component opposes the flow of electric current.

$R = \frac{V}{I}$R=IV​

where:

• R = resistance in ohms ($\Omega$Ω)
• V = potential difference in volts (V)
• I = current in amperes (A)

Quantity	Symbol	Unit
Resistance	R	Ohm ($\Omega$Ω)

Key Points

• A higher resistance means less current flows for the same voltage.
• Resistance arises because moving electrons collide with the vibrating ions in the metal lattice.
• Components with a high resistance (like a voltmeter) allow very little current through them.

Exam Tip: The unit of resistance is the ohm, written with the Greek letter omega ($\Omega$Ω). In calculations, always check that your values for V and I are in volts and amperes before substituting.

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The Relationship Between Current, Voltage and Resistance

The three quantities are connected by:

$V = IR$V=IR

This equation can be rearranged:

$I = \frac{V}{R} \qquad R = \frac{V}{I}$I=RV​R=IV​

What the Equation Tells Us

If you increase...	While keeping this constant...	Then...
Voltage (V)	Resistance (R)	Current (I) increases
Resistance (R)	Voltage (V)	Current (I) decreases
Current (I)	Resistance (R)	Voltage (V) increases

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Measuring Current, Voltage and Resistance

To find the resistance of a component experimentally:

1. Set up the component in a circuit with a variable resistor, an ammeter in series and a voltmeter in parallel across the component.
2. Adjust the variable resistor to change the current.
3. Record current (I) from the ammeter and voltage (V) from the voltmeter.
4. Calculate resistance using $R = \frac{V}{I}$R=IV​.

flowchart LR
    A["Battery"] --> B["Variable Resistor"]
    B --> C["Ammeter (A)"]
    C --> D["Component under test"]
    D --> A
    D -. "Voltmeter (V)" .-> C

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Electron Flow vs Conventional Current

Direction	Carrier	Used in...
Conventional current	Positive to negative	Circuit diagrams, calculations
Electron flow	Negative to positive	Describing the actual movement of electrons in metals

The convention of current flowing from positive to negative was established before the discovery of the electron. We continue to use it because all equations and circuit analysis are built around it.

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Summary

• Current (I) is the rate of flow of charge: $I = \frac{Q}{t}$I=tQ​, measured in amperes (A) using an ammeter in series.
• Potential difference (V) is the energy transferred per coulomb: $V = \frac{E}{Q}$V=QE​, measured in volts (V) using a voltmeter in parallel.
• Resistance (R) opposes current flow: $R = \frac{V}{I}$R=IV​, measured in ohms ($\Omega$Ω).
• Conventional current flows from positive to negative; electrons flow from negative to positive.
• These three quantities are linked by $V = IR$V=IR.

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Extended Worked Examples

Worked Example 3 — Chain of Calculations

A 12 V battery drives 0.25 A through a circuit for 3 minutes. Calculate (a) the resistance of the circuit, (b) the charge that flows, and (c) the energy transferred.

(a) Applying $V = IR$V=IR:

$R = \frac{V}{I} = \frac{12}{0.25} = 48 \; \Omega$R=IV​=0.2512​=48Ω

(b) Convert time and use $Q = It$Q=It:

$t = 3 \times 60 = 180 \text{ s}$t=3×60=180 s

$Q = It = 0.25 \times 180 = 45 \text{ C}$Q=It=0.25×180=45 C

(c) Use $E = QV$E=QV:

$E = 45 \times 12 = 540 \text{ J}$E=45×12=540 J