Resistors in Series and Parallel — Equivalent Resistance
Two formulas, one for series and one for parallel, cover almost every circuit-reduction question you will meet before university. The physics is short; the marks are usually lost either by forgetting to take a reciprocal or by mixing the series rule into a parallel question. This guide gives both formulas, four fully worked circuits, and the checks that confirm an answer is right before you move on.
Series resistors
In series, the same current flows through every resistor (there is only one path), while the voltage splits across them in proportion to each resistance. Adding a resistor in series always increases the total resistance.
Parallel resistors
In parallel, the same voltage appears across every branch, while the current splits between them in proportion to each branch's conductance (1/R). Adding a resistor in parallel always decreases the total resistance — a useful sanity check: Req in parallel must always be smaller than the smallest individual resistor.
Worked example 1 — three resistors in series
R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω, connected to a 20 V supply.
Req = 2 + 3 + 5 = 10 Ω
Current (same through all three): I = V/Req = 20/10 = 2 A
Voltage across each: V₁ = IR₁ = 2×2 = 4 V, V₂ = IR₂ = 2×3 = 6 V, V₃ = IR₃ = 2×5 = 10 V. Check: 4 + 6 + 10 = 20 V ✓ — the individual drops must always add back to the supply voltage.
Worked example 2 — two resistors in parallel
R₁ = 6 Ω, R₂ = 3 Ω, connected across a 12 V supply.
1/Req = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so Req = 2 Ω — smaller than either individual resistor, as expected.
Total current: I = V/Req = 12/2 = 6 A
Branch currents (same 12 V across both): I₁ = 12/6 = 2 A, I₂ = 12/3 = 4 A. Check: 2 + 4 = 6 A ✓ — the branch currents must add back to the total current.
Worked example 3 — a mixed series-parallel circuit
R₁ = 4 Ω is in series with a parallel combination of R₂ = 6 Ω and R₃ = 3 Ω. Find the total resistance.
Step 1 — reduce the parallel part first: 1/Rp = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so Rp = 2 Ω
Step 2 — now add the series part: Rtotal = R₁ + Rp = 4 + 2 = 6 Ω
Always simplify the innermost parallel or series group first, then work outward — trying to combine everything in one step is where most mixed-circuit errors happen.
Worked example 4 — n equal resistors in parallel (the shortcut)
Three 10 Ω resistors are connected in parallel. Find Req.
1/Req = 1/10 + 1/10 + 1/10 = 3/10, so Req = 10/3 = 3.33 Ω
For n equal resistors R in parallel, this always simplifies to Req = R/n — here R/n = 10/3 = 3.33 Ω, matching the long method exactly. This shortcut is worth memorising for objective-type questions where speed matters.
Common mistakes that cost marks
- Adding resistors directly in parallel the way you would in series — this is the single most common error in the whole chapter. Parallel always needs the reciprocal formula.
- Forgetting to flip the reciprocal at the end. Computing 1/Req = 1/2 and writing Req = 1/2 instead of Req = 2 — the final answer must be inverted back.
- Mixing up which quantity is "same" in each case. Series: same current. Parallel: same voltage. Swapping these leads to using the wrong resistor's value when finding an individual current or voltage drop.
- Simplifying a mixed circuit outside-in instead of inside-out. Reduce the innermost group first; you cannot combine R₁ with the parallel pair until the pair has already been reduced to a single value.
- Getting a parallel answer larger than the smallest resistor. If that happens, an arithmetic slip has occurred somewhere — parallel resistance is always smaller than every individual branch.
Where series and parallel resistance appears in exams
| Exam | Typical use |
|---|---|
| CBSE Class 10 | Electricity chapter — series vs parallel comparison, household wiring reasoning |
| CBSE Class 12 | Current electricity — Kirchhoff's laws built on top of series/parallel reduction |
| JEE Main & Advanced | Multi-loop resistor networks reduced step by step before applying Ohm's law |
| GATE (Engineering) | Circuit reduction as a first step in almost every network-analysis question |
Check a reduced circuit's current or voltage instantly. Once you have an equivalent resistance, plug it straight into the Ohm's Law calculator along with the supply voltage or current to verify your working.
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