Electricity 1 · Part C — Combining resistors
⬅ Back to Electricity 1⑤⑥ Combining resistors
Total resistance in series and parallel, then combination circuits — parallel part first.
Download the Full Booklet (PDF)
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.
Combine two resistors
Set the two resistors, the supply and the connection. The circuit, the total and the current all follow.
Predict the total resistance when a 10 Ω and a 20 Ω resistor are joined in parallel.
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.
Rate yourself
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?
Rate yourself
Sources & credits: The RS Electricity B1 booklet © R Stewart, 2025. Videos © The Other Mr Stewart (YouTube @mrstewartphysics) and Mr Bell – Practical Electronics, via YouTube.