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How are voltage, current, resistance, power and energy calculated for electronic products?

Electronic and systems calculations: Ohm's law, electrical power and energy, the potential divider, resistors in series and parallel, the current-limiting resistor for an LED, and reading and interpreting data, with formulae and units carried through.

A focused answer to Eduqas A-Level Product Design on electronic and systems calculations: Ohm's law, electrical power and energy, the potential divider, resistors in series and parallel, and sizing a current-limiting resistor for an LED, with worked examples and units.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The core relationships: Ohm's law is $V = I \times R$ (voltage = current times resistance), rearranged to $I = \frac{V}{R}$ and $R = \frac{V}{I}$ , in volts, amperes and ohms. Power is $P = V \times I$ (also $P = I^2 R = \frac{V^2}{R}$ ), in watts; energy is $E = P \times t$ , in joules (or watt-hours). The potential divider output is $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ . Resistors in series add ( $R = R_1 + R_2 + \dots$ ); in parallel they combine as $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ . An LED's series resistor is $R = \frac{V_{supply} - V_{LED}}{I}$ . Keep current in amps (convert milliamps), and always state units, because a units slip is the commonest lost mark.

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What this dot point is asking
Ohm's law
Power and energy
The potential divider and series and parallel resistors
Sizing a current-limiting resistor for an LED

What this dot point is asking

Eduqas wants you to perform the electronic and systems calculations of the paper: Ohm's law, electrical power and energy, the potential divider, resistors in series and parallel, and sizing a current-limiting resistor for an LED, with the right formulae and units. A calculator is allowed and working and units carry marks, so this is the quantitative side of the electronic systems topic.

Ohm's law

Key fact

Ohm's law relates the three basic electrical quantities: $V = I \times R$ , where $V$ is voltage (potential difference) in volts (V), $I$ is current in amperes (A), and $R$ is resistance in ohms ( $\Omega$ ). It rearranges to $I = \frac{V}{R}$ (to find current) and $R = \frac{V}{I}$ (to find resistance). Ohm's law lets a designer work out any one of voltage, current or resistance from the other two, which is the foundation of almost every circuit calculation (finding the current a component draws, choosing a resistor, checking a supply). Currents are often given in milliamps (mA) and resistances in kilohms (k $\Omega$ ), so convert to amps and ohms ( $1$ mA $= 0.001$ A, $1$ k $\Omega = 1000$ $\Omega$ ) before substituting, and quote the units.

Power and energy

Definition

Electrical power is the rate at which a component uses energy: $P = V \times I$ (voltage times current), in watts (W). Using Ohm's law it can also be written $P = I^2 R$ or $P = \frac{V^2}{R}$ , all giving the same answer, which is useful when you know different pairs of quantities. Power tells you how hot a component runs and what rating it needs (a resistor must be rated above the power it dissipates). Electrical energy is the total used over time: $E = P \times t$ (power times time), in joules (J) when time is in seconds, or in watt-hours / kilowatt-hours for longer periods (a kilowatt-hour is the energy of a one-kilowatt device for one hour). Energy links electronics to the use phase of a product's life cycle and its running cost, so an efficient product uses less power and energy.

The potential divider and series and parallel resistors

Key fact

A potential divider is two resistors in series across a supply, sharing the voltage in proportion to resistance: the output across the lower resistor is $V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$ . Making one resistor a sensor (an LDR or thermistor) turns a changing resistance into a changing voltage, the basis of sensing circuits. Resistors in series carry the same current and their resistances add: $R_{total} = R_1 + R_2 + \dots$ . Resistors in parallel share the current and combine as $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ , so the total is less than the smallest individual resistor (for two, $R_{total} = \frac{R_1 R_2}{R_1 + R_2}$ ). Knowing how resistors combine lets a designer find the current in a circuit and choose the right values, and the potential divider is the most-calculated circuit in the subject.

Sizing a current-limiting resistor for an LED

An LED must be protected by a series (current-limiting) resistor, because an LED has a roughly fixed forward voltage and would draw a damaging current if connected directly. The resistor drops the leftover voltage (the supply voltage minus the LED's forward voltage) at the LED's required current, so its value is $R = \frac{V_{supply} - V_{LED}}{I_{LED}}$ . For a $2.0$ V LED at $20$ mA on a $9$ V supply, $R = \frac{9 - 2.0}{0.020} = 350$ $\Omega$ (the nearest standard, often preferred, value above this is used so the current does not exceed the rating). The power in the resistor is $P = (V_{supply} - V_{LED}) \times I$ , which sets the resistor's power rating. This is the classic Eduqas electronics calculation, and the two traps are forgetting to subtract the LED voltage and leaving the current in milliamps.

Calculating current, power and an LED resistor

Use Ohm's law for the current

A $9$ V supply drives a $330$ $\Omega$ resistor. The current is $I = \frac{V}{R} = \frac{9}{330} = 0.0273$ A, about $27$ mA.

Find the power dissipated

The power in the resistor is $P = V \times I = 9 \times 0.0273 = 0.245$ W, so a $0.5$ W resistor is needed (above the dissipated power) for safety.

Size a resistor for an LED

For a $2.1$ V LED at $15$ mA on the same $9$ V supply, the resistor drops $9 - 2.1 = 6.9$ V at $0.015$ A, so $R = \frac{6.9}{0.015} = 460$ $\Omega$ (use the nearest higher preferred value).

Check the resistor power and the energy

The resistor dissipates $P = 6.9 \times 0.015 = 0.10$ W (a $0.25$ W resistor is fine), and over an hour of use the LED circuit uses $E = P \times t$ . State every unit (A, W, $\Omega$ , J) and remember to subtract the LED voltage and convert milliamps to amps, because those are the marked steps.

Exam-style practice questions

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

Eduqas 20204 marksA 12 V supply drives a heating element of resistance 8 ohms. Calculate the current through the element and the power it dissipates.

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A Component 1 Ohm's law and power calculation. Marks for the current and the power.

From Ohm's law, current is voltage divided by resistance: $I = \frac{V}{R} = \frac{12}{8} = 1.5$ amperes (A). Power is voltage times current: $P = VI = 12 \times 1.5 = 18$ watts (W). (Equivalently $P = \frac{V^2}{R} = \frac{144}{8} = 18$ W.)

Award marks for $1.5$ A and $18$ W with units. A common dropped mark is omitting the unit or using the wrong power formula; $P = VI$ , $P = I^2 R$ and $P = \frac{V^2}{R}$ all give the same answer here.

Eduqas 20226 marksAn LED needs 2.0 V across it and a current of 20 mA, and is run from a 9 V supply. Calculate the value of the series resistor needed, and the power dissipated in that resistor.

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A Component 1 LED current-limiting resistor calculation. Marks for the voltage across the resistor, its resistance and the power.

The resistor drops the supply voltage minus the LED voltage: $V_R = 9 - 2.0 = 7.0$ V. The current is $20$ mA $= 0.020$ A. By Ohm's law the resistance is $R = \frac{V_R}{I} = \frac{7.0}{0.020} = 350$ ohms (the nearest preferred value, 360 ohms, would be used). The power in the resistor is $P = V_R \times I = 7.0 \times 0.020 = 0.14$ W (so a 0.25 W resistor is fine).

Award marks for $V_R = 7.0$ V, $R = 350$ ohms and $P = 0.14$ W. A common dropped mark is forgetting to subtract the LED voltage, or leaving the current in milliamps instead of amps.

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Sources & how we know this

Eduqas A Level Design and Technology specification (Product Design) — Eduqas (2017)