Series and Parallel Circuits: the two patterns behind every network
Get these two right and half the paper's numericals become arithmetic.
How to tell series from parallel, without guessing
Almost every mistake in this topic is misreading which is which. There is a test that always works.
🔍 The test
SeriesThe components are joined end to end, with nothing else connected at the joint. There is only one road — every electron must pass through all of them.
ParallelThe components share BOTH ends — both left ends joined together, both right ends joined together. The current reaches a fork and splits.
The physical consequence follows immediately, and is worth trusting over any formula. In series the current has nowhere else to go, so the current is the same through each and the voltage divides. In parallel both ends are at the same two points, so the voltage is the same across each and the current divides.
SERIES R_total = R₁ + R₂ + R₃ …
always BIGGER than the largest one
PARALLEL 1/R_total = 1/R₁ + 1/R₂ + …
always SMALLER than the smallest one
Two in parallel, the shortcut worth memorising:
R_total = (R₁ × R₂) / (R₁ + R₂)
💡 Use those two sanity checks on every answer. If you combine 6 Ω and 3 Ω in parallel and get anything above 3, you have made a mistake — parallel resistance must come out below the smallest branch, because you have added another road for the current to take.
Why do Christmas lights all go dark when one bulb fails, while your house keeps working when one room's light burns out? Same electricity, two different connection patterns — and every complicated network you will ever solve reduces to repeated application of these two.
Series: one path
Elements are in series when they carry the same current, joined end to end with no branch point between them.
Series:
R_eq = R₁ + R₂ + R₃ + … (resistances add)
I is the SAME through all
V total = V₁ + V₂ + V₃ + … (voltages divide)
Parallel: many paths
Elements are in parallel when they have the same voltage across them, both ends connected to the same pair of nodes.
Parallel:
1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + …
V is the SAME across all
I total = I₁ + I₂ + I₃ + … (currents divide)
Two resistors only — the shortcut worth memorising:
R_eq = (R₁ × R₂) / (R₁ + R₂) "product over sum"
In parallel, R_eq is always smaller than the smallest resistor in the group. If your answer comes out bigger than any individual branch, you have inverted something — check before moving on. This one sanity check catches most arithmetic slips.
The divider rules — use them, don't derive them each time
V₂ = V × R₂ / (R₁ + R₂) With V = 12 V, R1 = 30 Ω, R2 = 60 Ω, V2 = 8 V, V1 = 4 V, I = 0.13 A.
Make R₁ and R₂ equal: the supply splits in half. Now make R₂ twice R₁ — it takes two thirds, not two times. The divider shares voltage in proportion, and the two shares always add back to V.
Voltage divider (series):
V₁ = V_total × R₁ / (R₁ + R₂)
→ the LARGER resistor takes the LARGER share of voltage
Current divider (two parallel branches):
I₁ = I_total × R₂ / (R₁ + R₂) ← note: the OTHER resistor on top
→ the SMALLER resistor takes the LARGER share of current
The current divider has the opposite resistor in the numerator. Writing R₁/(R₁+R₂) for I₁ is the single most common slip in these problems. Sanity check: current prefers the easier path, so the smaller resistance must get more current.
Worked numerical 1 — series with divider
A 24 V supply feeds 2 kΩ and 4 kΩ in series. Find the current and the voltage across each.
R_eq = 2k + 4k = 6 kΩ
I = V/R_eq = 24 / 6000 = 4 mA (same through both)
V₁ = I × R₁ = 4 mA × 2 kΩ = 8 V
V₂ = I × R₂ = 4 mA × 4 kΩ = 16 V
Check: 8 + 16 = 24 V ✔ (KVL satisfied)
By divider: V₂ = 24 × 4/(2+4) = 16 V ✔
Worked numerical 2 — parallel with divider
A 12 V source feeds 6 Ω and 3 Ω in parallel. Find R_eq, total current, and each branch current.
R_eq = (6 × 3)/(6 + 3) = 18/9 = 2 Ω (< 3 Ω ✔)
I_total = 12 / 2 = 6 A
I₆ = 12/6 = 2 A I₃ = 12/3 = 4 A
Check: 2 + 4 = 6 A ✔
By divider: I₆ = 6 × 3/(6+3) = 2 A ✔ (other resistor on top)
Worked numerical 3 — series-parallel combination
A 10 Ω resistor is in series with a parallel pair of 20 Ω and 20 Ω, across a 30 V supply. Find total current and the voltage across the parallel section.
Step 1 — collapse the parallel pair:
R_p = (20 × 20)/(20 + 20) = 10 Ω
(equal resistors in parallel: just R/n = 20/2 = 10 Ω)
Step 2 — now it's a simple series circuit:
R_eq = 10 + 10 = 20 Ω
I_total = 30 / 20 = 1.5 A
Step 3 — voltages:
V across 10 Ω = 1.5 × 10 = 15 V
V across parallel = 1.5 × 10 = 15 V
Each 20 Ω branch carries 15/20 = 0.75 A ✔ (sums to 1.5 A)
Always work from the innermost combination outwards. Redraw the simplified circuit after each step — it costs ten seconds and prevents the mistake of applying a divider to the wrong pair.
💡 Exam angle: n equal resistors in parallel give R/n — instant answer, use it. For combination networks, label every node before you start; two points joined by a plain wire are the same node, and spotting that often collapses a scary-looking network into two resistors.
Syllabus points
Equivalent resistance
Voltage/current division (numerical)
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