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Resistors in series and in parallel sheet
5-mark answer frame

A 4 Ω resistor is joined in series with a parallel pair of 6 Ω and 12 Ω resistors across a 12 V battery. Draw the circuit and find the total resistance, the main current and the current in each parallel resistor.

How to lay out the answer. Follow the frame, then write it in your own words.

  1. 1 · Write what is given. R₁ = 4 Ω in series; R₂ = 6 Ω and R₃ = 12 Ω in parallel; V = 12 V.
  2. 2 · Draw the circuit. [Circuit: battery, key, ammeter, 4 Ω resistor, then 6 Ω and 12 Ω as two branches; all values and currents labelled]
  3. 3 · Solve the parallel part first. Rp = (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 Ω.
  4. 4 · Find the total and the main current. R = 4 Ω + 4 Ω = 8 Ω. I = V/R = 12 V ÷ 8 Ω = 1.5 A.
  5. 5 · Split the current. PD across the pair = 12 V − (1.5 A × 4 Ω) = 6 V. Current in 6 Ω = 6 ÷ 6 = 1 A; in 12 Ω = 6 ÷ 12 = 0.5 A. Check: 1 + 0.5 = 1.5 A.
  6. 6 · Real life. House wiring is a mix like this: appliances in parallel, with a fuse in series in the main line.
  7. 7 · Box the answers. Total 8 Ω; main current 1.5 A; 1 A in 6 Ω and 0.5 A in 12 Ω.

Scoring tip: Always solve the parallel part first, then add the series part. Write the unit after every number.