NDA Physics · Electricity and Magnetism

Combination of Resistors

Resistors in series add (R = R₁ + R₂ + …); resistors in parallel combine by reciprocals (1/R = 1/R₁ + 1/R₂ + …). Every network reduces by collapsing the innermost series/parallel groups one step at a time.

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Why this matters

This is the chapter's marquee subtopic — 16 PYQs at 38% HARD, the bank's single biggest HARD pool. Master five shapes and you own these marks: pure series, pure parallel, mixed series-parallel reduction, the 'cut a wire then reconnect' trick, and minimum-vs-maximum resistance. Series always gives the LARGEST equivalent, parallel the SMALLEST — that one fact answers a surprising number of questions outright.

Concept 1 of 5

Resistors in series

Intuition

In series the same current flows through every resistor (one path), and the voltages across them add up. So the resistances simply add — the combination is always BIGGER than the largest single resistor.

Definition

Resistors in series carry the same current; their voltages add. The equivalent resistance is the sum: ** $R_\text{series} = R_1 + R_2 + \cdots$ ** — always larger than the biggest individual resistor.

Series equivalent

R_\text{series} = R_1 + R_2 + \cdots + R_n

Worked example

A 2 Ω, a 3 Ω, and a 5 Ω resistor are connected in series. What is the equivalent resistance?

Series resistances add directly.
$R = 2 + 3 + 5 = 10\,\Omega$ .

Answer:10 Ω.

Practice this concept3 quick reps

Series = same current, voltages add

Don't confuse the two combinations. Series: one current path, add the resistances. The combination can never be smaller than any single resistor in it.

Concept 2 of 5

Resistors in parallel

Intuition

In parallel the same voltage sits across every resistor, and the currents through them add. Giving the charge extra paths can only make it easier to flow, so the equivalent resistance is always SMALLER than the smallest single resistor. For n EQUAL resistors R, the parallel value is just R/n.

Definition

Resistors in parallel have the same voltage across them; their currents add. The reciprocal of the equivalent equals the sum of reciprocals: ** $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$ ** — always smaller than the smallest branch. Two resistors: $R = \dfrac{R_1 R_2}{R_1 + R_2}$ . ** $n$ equal resistors** $R$ : equivalent $= R/n$ .

Parallel equivalent

\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots \qquad (\text{two: } R = \tfrac{R_1R_2}{R_1+R_2})

Worked example

A 6 Ω and a 3 Ω resistor are connected in parallel. What is the equivalent resistance?

Two-resistor shortcut: $R = \dfrac{R_1 R_2}{R_1+R_2}$ .
$R = \dfrac{6\times 3}{6+3} = \dfrac{18}{9} = 2\,\Omega$ .
Note it's smaller than 3 Ω, the smaller branch — as parallel always is.

Answer:2 Ω.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 2Electricity and MagnetismEASY

Three equal resistors are connected in parallel configuration in a closed electrical circuit. Then the total resistance in the circuit becomes

[Q119 · Sep · 2021]

Parallel value is SMALLER than the smallest branch

If you compute a parallel combination and get something bigger than one of the branches, you've made an error. Adding paths reduces resistance. For two equal R the answer is R/2, never 2R.

Drill 3 more on resistors in parallel

Concept 3 of 5

Reducing mixed series-parallel networks

Intuition

Most circuit questions are nested: a parallel cluster sitting inside a series chain, or vice versa. The method is always the same — find the innermost pure-series or pure-parallel group, collapse it to one resistor, redraw, and repeat until a single number is left.

Definition

Reduction algorithm: (1) spot a sub-group that is purely series OR purely parallel; (2) replace it with its equivalent; (3) redraw and repeat. A short-circuit (0 Ω) branch across two points forces those points to the same potential — current takes the zero-resistance path. For self-similar infinite ladders, set the whole network equal to $R_\infty$ and solve the resulting equation (the rest of the ladder is identical to the whole).

Series adds (always larger); parallel combines by reciprocals (always smaller than the smallest branch). Reduce a network innermost-group first.

Worked example

Three resistors 2 Ω, 4 Ω and 4 Ω: the two 4 Ω are in parallel, and that pair is in series with the 2 Ω. Find the equivalent resistance.

Innermost group: 4 Ω ∥ 4 Ω = $4/2 = 2\,\Omega$ (equal pair).
Now in series with the 2 Ω resistor: $2 + 2 = 4\,\Omega$ .

Answer:4 Ω.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 3Electricity and MagnetismMODERATE

Two resistances of 5.0

\Omega

and 7.0

\Omega

are connected in series and the combination is connected in parallel with a resistance of 36.0

\Omega

. The equivalent resistance of the combination of three resistors is

[Q63 · Sep · 2024]

Collapse innermost first — don't add everything blindly

You can't add a series and a parallel resistor in one step. Identify a sub-group that is PURELY one kind, reduce it, redraw, and only then look at the next group. Mixing the two rules in a single step is the most common network error.

Drill 7 more on reducing mixed series-parallel networks

Concept 4 of 5

Cutting a wire and reconnecting it

Intuition

Cut a wire of resistance R into n equal pieces and each piece has resistance R/n (shorter = less resistance). Reconnect those n pieces in parallel and you divide again by n — so the final resistance is R/n². A ring measured across a diameter is just two equal half-rings in parallel.

Definition

Cut a wire of resistance $R$ into $n$ equal pieces ⟹ each piece is $R/n$ . Connect all $n$ pieces in parallel ⟹ equivalent $= \dfrac{R/n}{n} = \dfrac{R}{n^2}$ . A uniform ring of total resistance $R$ , measured across any diameter, behaves as two $R/2$ arcs in parallel = $R/4$ .

Cut into n, reconnect in parallel

R_\text{final} = \dfrac{R}{n^2}

Worked example

A 12 Ω wire is cut into three equal pieces, and the three pieces are connected in parallel. What is the equivalent resistance?

Each piece: $12/3 = 4\,\Omega$ .
Three 4 Ω in parallel (equal): $4/3\,\Omega$ .
Shortcut check: $R/n^2 = 12/9 = 4/3\,\Omega$ . ✓

Answer:4/3 Ω ≈ 1.33 Ω.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 4Electricity and MagnetismMODERATE

An electric wire of resistance 50 ohm is cut into five equal wires. These wires are then connected in parallel. What is the equivalent resistance of this combination?

[Q55 · Apr · 2022]

Cut + parallel = R/n², not R/n

Two effects stack: cutting into n pieces makes each R/n, AND paralleling n of them divides by another n. The combined result is R/n². Stopping at R/n is the dominant wrong answer.

Drill 2 more on cutting a wire and reconnecting it

Concept 5 of 5

Minimum and maximum resistance

Intuition

Given a fixed set of resistors, you get the LARGEST possible resistance by wiring them all in series, and the SMALLEST by wiring them all in parallel. So 'which arrangement gives minimum resistance?' is really 'which is the most-parallel of the cheapest resistors?'

Definition

For any fixed collection of resistors: all in series ⟹ maximum equivalent (the sum); all in parallel ⟹ minimum equivalent (below the smallest branch). To minimise resistance, parallel the smallest-valued resistors; to maximise, put the largest in series.

Worked example

You have two 6 Ω resistors. What are the maximum and minimum resistances you can make with both of them?

Maximum — series: $6 + 6 = 12\,\Omega$ .
Minimum — parallel: $6/2 = 3\,\Omega$ .

Answer:Maximum 12 Ω (series), minimum 3 Ω (parallel).

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 5Electricity and MagnetismMODERATE

Which of the following arrangement of resistors offers minimum effective resistance between points X and Y?

[Q109 · Sep · 2023]

Minimum ≠ fewest resistors

Minimum resistance means MOST parallel paths of the SMALLEST resistors — not the smallest count of components. Three 3 Ω in parallel (1 Ω) beats two 1 Ω in series (2 Ω): more parallelism wins even with larger individual values.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Resistors in series
Series equivalent
$R_\text{series} = R_1 + R_2 + \cdots + R_n$
Resistors in parallel
Parallel equivalent
$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots \qquad (\text{two: } R = \tfrac{R_1R_2}{R_1+R_2})$
Cutting a wire and reconnecting it
Cut into n, reconnect in parallel
$R_\text{final} = \dfrac{R}{n^2}$

Watch out for (5)

Series = same current, voltages add
→ Resistors in series
Parallel value is SMALLER than the smallest branch
→ Resistors in parallel
Collapse innermost first — don't add everything blindly
→ Reducing mixed series-parallel networks
Cut + parallel = R/n², not R/n
→ Cutting a wire and reconnecting it
Minimum ≠ fewest resistors
→ Minimum and maximum resistance

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