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What are current, potential difference and resistance, and how do circuits and mains electricity work?

Standard circuit symbols, current, potential difference and resistance, the I-V characteristics of components, series and parallel circuits, mains electricity and the three-pin plug, and electrical power and energy.

A focused answer to the AQA GCSE Combined Science: Trilogy Electricity topic, covering circuit symbols, current, potential difference and resistance, component characteristics, series and parallel circuits, mains electricity and the plug, and electrical power and energy.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Current, potential difference and resistance
  3. Component characteristics and circuits
  4. Mains electricity, power and energy

What this topic is asking

AQA wants you to use circuit symbols, define current, potential difference and resistance, describe the I-V characteristics of components, compare series and parallel circuits, explain mains electricity and the plug, and calculate electrical power and energy.

Current, potential difference and resistance

The key equation is V=IRV = IR (potential difference = current ×\times resistance). For an ohmic conductor (such as a fixed resistor) at constant temperature, the resistance is constant, so the current is directly proportional to the potential difference and its I-V graph is a straight line through the origin. In the required practical you measure the current and potential difference for a component and plot the I-V graph to find how its resistance behaves.

Component characteristics and circuits

In a series circuit there is only one loop, so the current is the same at every point, the potential differences across the components add up to the supply, and the total resistance is the sum of the individual resistances (adding a resistor increases the total). In a parallel circuit the components are on separate branches, so the potential difference across each branch is the same and equal to the supply, the branch currents add up to the total current from the source, and adding a resistor in parallel decreases the total resistance (because there are more paths for the current).

Mains electricity, power and energy

UK mains supply is alternating current (ac), where the direction of the current reverses repeatedly, at about 230 V and a frequency of 50 Hz; cells and batteries provide direct current (dc), which flows in one direction only. The three-pin plug has three wires: the live wire (brown, carries the alternating potential difference from the supply, around 230 V), the neutral wire (blue, completes the circuit and is near 0 V), and the earth wire (green and yellow, a safety wire that carries current to the ground only if there is a fault, stopping the casing becoming live). Touching the live wire gives a shock even when a switch is off, because it is still at a high potential relative to your body.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA 60 W lamp is connected to the 230 V mains supply. Calculate the current through the lamp and the charge that flows through it in 5 minutes.
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A Physics Paper 1 calculation combining two equations. Method: from P=VIP = VI, current I=PV=60230=0.26I = \dfrac{P}{V} = \dfrac{60}{230} = 0.26 A (to two significant figures). For the charge, first convert 5 minutes to seconds: 5×60=3005 \times 60 = 300 s. Then from Q=ItQ = It, charge =0.26×300=78= 0.26 \times 300 = 78 C. Markers award rearranging the power equation, the time conversion to seconds (a common slip is leaving it in minutes), and the use of Q=ItQ = It. Carrying the unrounded current through avoids a rounding error in the final answer.

AQA 20214 marksTwo resistors are connected in series with a 12 V battery. One has a resistance of 4 ohms and the other 2 ohms. Calculate the total resistance, the current in the circuit, and the potential difference across the 4 ohm resistor.
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A Physics Paper 1 series-circuit calculation. Method: in series the total resistance is the sum, 4+2=64 + 2 = 6 ohms. The current is the same everywhere and is found from V=IRV = IR, so I=VR=126=2I = \dfrac{V}{R} = \dfrac{12}{6} = 2 A. The potential difference across the 4 ohm resistor is V=IR=2×4=8V = IR = 2 \times 4 = 8 V. Markers credit adding the resistances, using the total to find the current, and applying V=IRV = IR to the individual resistor. A check is that the two potential differences (8 V and 4 V) add up to the supply 12 V, as they must in series.

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