GCSE Physics

Electric circuits

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Current is the rate of flow of electric charge: I = \(\dfrac{Q}{t}\). Potential difference is the energy transferred per unit charge. Resistance describes how strongly a component opposes current, and for an ohmic conductor at constant temperature V = IR.

In a series circuit, current is the same through each component and potential differences add to the supply. In parallel, potential difference is the same across each branch and branch currents add at a junction. Electrical power can be calculated from P = IV, and energy transferred from E = Pt or E = QV. The resistance of a filament lamp increases as it heats; the resistance of an LDR falls as light intensity increases, and the resistance of a thermistor falls as temperature increases.

Charge, current and potential difference

Electric current is the rate of flow of charge, I = \(\dfrac{Q}{t}\). In metals the mobile charge carriers are electrons, although conventional current direction is defined opposite to electron drift. Current is not 'used up' by a component. In a steady series circuit the same current passes through each component because charge is conserved.

Potential difference is energy transferred per unit charge, V = \(\dfrac{E}{Q}\). A battery or power supply gives energy to charge; components transfer energy from the electrical pathway into other stores. In a series circuit, potential differences across components add to the supply. In parallel, each branch is connected across the same two points, so each branch has the same potential difference as the supply.

Resistance and circuit behaviour

Resistance describes how difficult it is for charge to flow: R = \(\dfrac{V}{I}\). For an ohmic conductor at constant temperature, current is proportional to potential difference and resistance is constant. A filament lamp heats as current increases, so its resistance rises. The resistance of a thermistor decreases as its temperature increases, and the resistance of an LDR decreases as light intensity increases. This makes them useful as sensors, for example in thermostats and automatic night lights. A diode lets current flow in one direction only: its resistance is very high in the reverse direction, so its I-V graph shows current only for forward potential differences.

Series resistances add. Adding a resistor in series increases total resistance and usually reduces current for a fixed supply. Parallel branches provide additional paths for charge, so adding branches reduces total resistance and can increase total current from the source. Current splits at a junction and recombines, with total current into a junction equal to total current out.

Electrical power and energy

Power can be found from P = IV. When resistance is known, P = I²R is also useful; it shows why cable heating rises sharply with current. Electrical energy transferred is E = Pt and also E = QV. A high-power appliance transfers more energy each second, but total energy use also depends on how long it operates.

Measuring circuits

An ammeter is connected in series because it measures current through a component. A voltmeter is connected in parallel because it measures potential difference across a component. When investigating I-V characteristics, take a range of paired current and potential-difference measurements and control heating when temperature would alter resistance.

Exam technique

Do not treat current, voltage and energy as interchangeable. Current tells you how quickly charge flows; potential difference tells you energy transferred per coulomb; power tells you energy transferred per second. In calculations, identify which physical meaning the question is asking for before choosing the equation.

Worked example

Worked example 1: A resistor has 12 V across it and a resistance of 4 Ω. I = \(\dfrac{V}{R}\) = \(\dfrac{12}{4}\) = 3 A.

Worked example 2: A 3 Ω resistor and a 9 Ω resistor are in series across 12 V. The total resistance is 12 Ω, so I = \(\dfrac{12}{12}\) = 1.0 A. The potential differences are 3 V and 9 V respectively.

Worked example 3: A 12 V supply across a 4.0 Ω resistor gives I = \(\dfrac{V}{R}\) = 3.0 A.

Misconception check

Current is not 'used up' by components. Charge is conserved; components transfer energy from the charges to other stores.

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Define electric current using an equation.

I = \(\dfrac{Q}{t}\): current is charge flow per unit time.

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