Electricity 1 · Part C — Combining resistors
⬅ Back to Electricity 1⑤⑥ Combining resistors
Total resistance in series and parallel, then combination circuits — parallel part first.
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L5 — Combining resistors (basic)
Series — resistances add up:
RT = R₁ + R₂ + …
10 Ω and 20 Ω in series: RT = 10 + 20 = 30 Ω.
Adding more in series increases total resistance, so the current gets smaller.
Parallel (equal resistors) — quick method:
RT = R₁ ÷ (number of branches)
Four 12 Ω resistors in parallel: RT = 12 ÷ 4 = 3 Ω.
The total is always smaller than the smallest resistor. Adding a branch gives the current another path, so total current increases.
Q1 (3). A series circuit has a 9 V supply and resistors of 10 Ω and 20 Ω. (a) Total resistance? (b) Current?
Mark scheme
(a) RT = 10 + 20 = 30 Ω
(b) I = V ÷ R = 9 ÷ 30 = 0.30 A
Q2 (4). Two 40 Ω resistors in parallel across 12 V. (a) Total resistance? (b) Current in each branch?
Mark scheme
(a) Equal pair: 40 ÷ 2 = 20 Ω
(b) Each branch: I = 12 ÷ 40 = 0.30 A
Adding resistors in series
Predict, then (in class) measure the total resistance for series and equal-value parallel sets.
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L6 — Combining resistors (tricky)
N5 · Going further Combination circuits and the reciprocal method stretch beyond the core S3 work — useful for N5 revision. The core S3 outcome is series and equal-value parallel sets (Part C, L5).
LearnParallel with different values — reciprocal method:
1 / RT = 1/R₁ + 1/R₂ + …
Which method do I use?
- Resistors one after another → series: add them.
- Parallel, all the same value → divide one value by the number of branches.
- Parallel, different values → reciprocal method, then flip your answer.
Combination worked example (1 kΩ ∥ 1.5 kΩ, then in series with 2.2 kΩ):
Step 1 — parallel pair:
1/Rp = 1/1000 + 1/1500 = 0.001666…
Rp = 600 Ω
Step 2 — add the series resistor:
RT = Rp + R₃ = 600 + 2200 = 2.8 kΩ
Method: always solve the parallel part first, then add anything in series with it.
TryQ1 (4). 9 V across 30 Ω ∥ 15 Ω. (a) Total resistance? (b) Current in the 15 Ω branch?
Mark scheme
(a) 1/RT = 1/30 + 1/15 = 0.1 → RT = 10 Ω
(b) Branch voltage is the full 9 V: I = 9 ÷ 15 = 0.60 A
Q2 (4). 12 V supply, 6 Ω in series with a 12 Ω ∥ 12 Ω pair. Total resistance and supply current?
Mark scheme
Parallel pair = 6 Ω; total = 6 + 6 = 12 Ω; I = 12 ÷ 12 = 1 A.
Check your booklet combination-circuit answers (p.11)
Series rule: RT = R₁ + R₂ + … Parallel rule: 1/RT = 1/R₁ + 1/R₂ + …
① 100 Ω ∥ 200 Ω = 66.7 Ω, + 750 Ω = ≈ 817 Ω
② 3.5 kΩ ∥ 1.5 kΩ = 1.05 kΩ, + 10 kΩ = ≈ 11.05 kΩ
③ (1 kΩ + 2.2 kΩ) = 3.2 kΩ, ∥ 1.5 kΩ = ≈ 1.02 kΩ
④ (7.5 kΩ + 2.2 kΩ) = 9.7 kΩ, ∥ 750 Ω = ≈ 696 Ω
(Always combine the parallel part first.)
🔥 Stretch — power (beyond Block 1)
C1. 0.5 A flows through 20 Ω. Find the power and the energy transferred in 40 s.
Answer
P = I²R = 0.5² × 20 = 5 W; E = P × t = 5 × 40 = 200 J.
C2. A 50 Ω resistor dissipates 12.5 W. Find the voltage across it.
Answer
V = √(P × R) = √(12.5 × 50) = √625 = 25 V.
Adding resistors in parallel
Combination circuit calculations
Can you find the total resistance of a combination circuit, parallel part first?
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Sources & credits: The RS Electricity B1 booklet © R Stewart, 2025. Videos © The Other Mr Stewart (YouTube @mrstewartphysics) and Mr Bell – Practical Electronics, via YouTube.