Potential divider practice
Calculate outputs, choose a resistor and predict LDR, thermistor and loading effects.
DC circuit questions
Solve series, parallel and mixed networks; calculate charge, energy and power.
Practise
Series and parallel togetherA 50 Ω resistor is in series with a parallel pair of 70 Ω and 60 Ω. An ideal 12 V supply is connected across the complete combination. Find the supply current. Give your answer to 3 significant figures.
Ideal DC supply, wires and meters. Resistors have constant resistance; heating does not change their values.
Larger diagram
Treat the given values as exact unless an uncertainty is stated.
Hint
Find the parallel pair’s resistance, add the series resistor, then divide supply voltage by total resistance.
Worked solution
- R_parallel = (70 × 60)/(70+60) Ω.
- I = 12/[50 + (70 × 60)/(70+60)] A.
- Answer: 0.146 A (3 significant figures).
Question type: Series and parallel together
Report a problem
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.