A resistor carries 25\ mA when 5.0\ V is across it. Find its resistance. [2 marks]

A 100\ resistor carries 0.20\ A. Find the power it dissipates. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A 12\ V supply is connected to a 470\ resistor in series with a parallel combination of a 1.0\ k and a 1.5\ k resistor. Calculate the total resistance, the supply current, and the power delivered by the supply.

Parallel pair first: 1R_p = 11000 + 11500 = 3 + 23000 = 53000, so R_p = 600\. Total resistance in series: R = 470 + 600 = 1070\. Supply current by Ohm's law: I = VR = 121070 = 1.12 × 10^-2\ A = 11.2\ mA. Power delivered: P = VI = 12 × 1.12 × 10^-2 = 0.135\ W. Markers reward R_p = 600\, the total 1070\, the current 11.2\ mA and the power 0.135\ W (or P = (V^2)/(R)).

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How do Ohm's law and Kirchhoff's two laws let us analyse any resistive electronic circuit?

Circuit fundamentals: charge, current, voltage and resistance, Ohm's law, Kirchhoff's current and voltage laws, combining resistors in series and parallel, and electrical power.

An Eduqas A-Level Electronics answer on circuit fundamentals: charge, current, voltage and resistance, Ohm's law, Kirchhoff's current and voltage laws as conservation of charge and energy, combining resistors in series and parallel, and calculating electrical power in a circuit.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

Current $I$ is the rate of flow of charge, $I = \frac{Q}{t}$ , measured in amperes. Voltage (potential difference) $V$ is the energy transferred per coulomb, measured in volts. Resistance $R = \frac{V}{I}$ is measured in ohms. Ohm's law states that for a metallic conductor at constant temperature the current is proportional to the voltage, so $V = IR$ . Kirchhoff's current law (conservation of charge): currents into a junction equal currents out. Kirchhoff's voltage law (conservation of energy): around any loop the electromotive forces equal the sum of the potential differences. Resistors in series add, $R = R_1 + R_2 + \dots$ ; in parallel they combine by reciprocals, $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ . Power is $P = VI = I^2 R = \frac{V^2}{R}$ .

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What this dot point is asking
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What this dot point is asking

Eduqas wants you to define charge, current, voltage and resistance, state and use Ohm's law, apply Kirchhoff's current and voltage laws, combine resistors in series and parallel, and calculate electrical power. These are the tools every later Electronics topic assumes, so they must be automatic.

The answer

Charge, current, voltage and resistance

Ohm's law

Kirchhoff's laws

Combining resistors

Electrical power

Solving a two-loop network with Kirchhoff's laws

Two cells drive a shared resistor. Cell A ( $6.0\ \text{V}$ ) connects through $10\ \Omega$ and cell B ( $4.0\ \text{V}$ ) through $20\ \Omega$ to a common $30\ \Omega$ resistor to the other rail. Find the current in the $30\ \Omega$ resistor.

step 1: Label currents and apply KCL

Let $I_1$ flow from cell A, $I_2$ from cell B, and $I_3$ through the $30\ \Omega$ resistor. At the junction, $I_3 = I_1 + I_2$ .

step 2: Write KVL for each loop

Loop A: $6.0 = 10 I_1 + 30 I_3$ . Loop B: $4.0 = 20 I_2 + 30 I_3$ .

step 3: Substitute and solve

From loop A, $I_1 = \frac{6.0 - 30 I_3}{10}$ ; from loop B, $I_2 = \frac{4.0 - 30 I_3}{20}$ . Substitute into $I_3 = I_1 + I_2$ : $I_3 = 0.60 - 3 I_3 + 0.20 - 1.5 I_3$ , so $5.5 I_3 = 0.80$ and $I_3 = 0.145\ \text{A} \approx 0.15\ \text{A}$ .

Examples in context

Ohm's law and Kirchhoff's laws are the foundation of every circuit in this course. The potential divider, the op-amp feedback network, the timing resistor of a 555 astable and the pull-up resistor on a microcontroller input are all analysed with exactly these tools. Power calculations decide resistor wattage ratings, heatsink requirements for power transistors, and battery life in a portable system.

Try this

Q1. A resistor carries $25\ \text{mA}$ when $5.0\ \text{V}$ is across it. Find its resistance. [2 marks]

Cue. $R = \frac{V}{I} = \frac{5.0}{0.025} = 200\ \Omega$ .

Q2. Two $330\ \Omega$ resistors are connected in parallel. Find the combined resistance. [2 marks]

Cue. $R = \frac{330}{2} = 165\ \Omega$ (equal resistors in parallel halve).

Q3. A $100\ \Omega$ resistor carries $0.20\ \text{A}$ . Find the power it dissipates. [2 marks]

Cue. $P = I^2 R = (0.20)^2 \times 100 = 4.0\ \text{W}$ .

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 20195 marksA

12\ \text{V}

supply is connected to a

470\ \Omega

resistor in series with a parallel combination of a

1.0\ \text{k}\Omega

and a

1.5\ \text{k}\Omega

resistor. Calculate the total resistance, the supply current, and the power delivered by the supply.

Show worked answer →

Parallel pair first: $\dfrac{1}{R_p} = \dfrac{1}{1000} + \dfrac{1}{1500} = \dfrac{3 + 2}{3000} = \dfrac{5}{3000}$ , so $R_p = 600\ \Omega$ .

Total resistance in series: $R = 470 + 600 = 1070\ \Omega$ .

Supply current by Ohm's law: $I = \dfrac{V}{R} = \dfrac{12}{1070} = 1.12 \times 10^{-2}\ \text{A} = 11.2\ \text{mA}$ .

Power delivered: $P = VI = 12 \times 1.12 \times 10^{-2} = 0.135\ \text{W}$ .

Markers reward $R_p = 600\ \Omega$ , the total $1070\ \Omega$ , the current $11.2\ \text{mA}$ and the power $0.135\ \text{W}$ (or $P = \frac{V^2}{R}$ ).

Eduqas 20214 marksState Kirchhoff's current law and Kirchhoff's voltage law, and for each name the physical quantity that is conserved.

Show worked answer →

Kirchhoff's current law (up to 2 marks): the sum of the currents entering a junction equals the sum of the currents leaving it. The conserved quantity is electric charge, which cannot accumulate at a point.

Kirchhoff's voltage law (up to 2 marks): around any closed loop, the sum of the electromotive forces equals the sum of the potential differences across the components (equivalently, the algebraic sum of the voltage changes around a loop is zero). The conserved quantity is energy: each coulomb returns to its starting point with the same energy.

Markers reward a correct statement of each law and the matching conserved quantity (charge for the current law, energy for the voltage law).

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Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)