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DC circuit questions

Solve series, parallel and mixed networks; calculate charge, energy and power.

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Power in a resistor

A 70 Ω resistor has 12 V across it. Find its electrical power dissipation. Give your answer to 3 significant figures.

Ideal DC supply, wires and meters. Resistors have constant resistance; heating does not change their values.

Treat the given values as exact unless an uncertainty is stated.

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Use P = VI and I = V/R, giving P = V²/R.

Worked solution
  1. P = 12²/70 W.
  2. Answer: 2.06 W (3 significant figures).
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Question type: Power in a resistor
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Follow current and potential difference

Series components carry the same current; their potential differences add. Parallel branches have the same potential difference; their currents add. Kirchhoff’s junction rule expresses conservation of charge, and the loop rule expresses conservation of energy.

Reduce a network in stages

Series resistances add. For two parallel resistors, R = R1R2/(R1 + R2); this result is smaller than either branch resistance. Find a parallel pair’s resistance before adding a series resistor. Supply current and branch current need different resistance values.

Charge, energy and power

Q = It, E = VQ, and P = VI = I²R = V²/R for the same resistor. Convert minutes to seconds before finding charge. These tasks use an ideal supply and constant resistances. Internal resistance, sensors and dividers have separate practice pages; multiple-source loops and nonlinear components are not covered here.

A-level relevance

AQA 7408: 3.5.1.1, 3.5.1.4 (selected); OCR A H556: 4.1.1; 4.2.1–2; 4.3.1 (selected). Tasks cover the selected skills listed above.

AQA electricity specification · OCR A specification

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