JEE Mains Physics · Current Electricity

Equivalent Resistance: Series, Parallel and Symmetry

Resistances in series add and in parallel add as reciprocals; a network that is neither is first redrawn, by merging points joined by plain wire and by using symmetry to find points at the same potential.

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Why this matters

Thirty-one PYQs, nineteen of them multiple choice, and four from 2026. Fifteen use series and parallel rules directly: the largest and smallest values from equal resistors, a wire cut into pieces and regrouped, choosing a combination that gives a stated value, and wires bent into triangles, squares, polygons and circles. Sixteen give a network that must be redrawn first, by merging points joined by plain wire or by using symmetry; fifteen of those come with a figure.

Concept 1 of 2: Series, parallel and bent wires

In series the same current passes through every resistor, so their voltages add and so do their resistances. In parallel every resistor has the same voltage, so their currents add and so do their conductances 1/R. A wire bent into a loop and tapped at two points is just two arcs in parallel, each with resistance in proportion to its length.

Definition

Series: $R = R_1 + R_2 + \dots$ . Parallel: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ ; for two, $R = \dfrac{R_1R_2}{R_1 + R_2}$ .
n equal resistors R: nR in series, R/n in parallel. The largest value over the smallest is $n^{2}$ .
A wire of resistance R cut into n equal pieces: each piece is R/n; all n in parallel give $R/n^{2}$ .
A uniform loop of total resistance R tapped at two points that split its length in the fraction x and 1 − x: $R_{AB} = R\,x(1 - x)$ . It is largest when the points are opposite (x = 1/2).
A regular polygon of n sides is a loop: adjacent corners split it as 1 side and n − 1 sides.
To find which combination gives a stated value, work out each option's value; do not guess from the shape.
A resistance that depends on a variable (say m) is least where its derivative is zero.

Series, parallel, loops

R_s = \sum R_i, \qquad \frac{1}{R_p} = \sum \frac{1}{R_i}, \qquad R_{\text{loop}} = R\,x(1 - x)

Worked example

A 12 Ω wire is bent into a regular hexagon. Find the resistance between two adjacent corners and between two opposite corners.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q97Moderate

Example 1 · Current Electricity · Equivalent Resistance: Series, Parallel and Symmetry

Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance

R

would be:

A cut wire divides twice

Cutting into n pieces makes each R/n, and putting them in parallel divides by n again. The answer is R/n², not R/n.

Arcs go by length

On a bent wire the resistance of an arc is in proportion to its length. Equal angles at the centre of a circle mean equal resistances; unequal sides of a shape do not.

Check every option numerically

Combination questions offer several plausible drawings. Compute each one; the right one is often not the most symmetric.

Drill 14 more on series, parallel and bent wires

Concept 2 of 2: Redrawing networks: shorts and symmetry

Many networks look complicated only because of how they are drawn. Points joined by a plain wire are one point. Points that are mirror images across the line from input to output sit at the same potential, so a resistor between them carries nothing and can be removed. Once those moves are made, the network usually falls into series and parallel pieces.

Definition

Label every junction. Points joined by a wire with no resistor get the same label.
A resistor whose two ends have the same label carries no current: remove it.
Mirror symmetry: if the network looks the same reflected across the line joining the input and output, mirror-image points are at the same potential. They may be joined, or a resistor between them removed.
Points midway between input and output (on the perpendicular bisector of a symmetric network) are all at half the applied potential.
Ladders and trees: start at the far end and work back, each step "add in series, then put in parallel".
If nothing simplifies, set the input at potential 1 and the output at 0, write the junction law at each unknown node, find the total current I, and then $R = 1/I$ .
Two wires that cross without a dot are not joined.

The two moves

V_P = V_Q \ \Rightarrow\ I_{PQ} = 0, \qquad R_{\text{eq}} = \frac{V_{\text{in}} - V_{\text{out}}}{I_{\text{total}}}

Worked example

Between A and B: a 4 Ω from A to P, a 4 Ω from P to B, an 8 Ω from A to B, and a plain wire from P to B. Find the resistance between A and B.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q4Moderate

Example 2 · Current Electricity · Equivalent Resistance: Series, Parallel and Symmetry

A regular hexagon is formed by six wires each of resistance

r\Omega

and the corners are joined to centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

A missed short

A plain wire drawn along the edge of a figure is easy to overlook. It merges two points and can remove a whole resistor. Label the nodes before writing any formula.

A crossing is not a junction

Wires that cross without a dot, or with a hump, are separate. Joining them changes the network and the answer.

Symmetry must be about the input–output line

A network can look symmetric in many ways, but only symmetry with respect to where the current enters and leaves gives equal potentials.

Drill 15 more on redrawing networks: shorts and symmetry

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 (2)

Series, parallel and bent wires
Series, parallel, loops
$R_s = \sum R_i, \qquad \frac{1}{R_p} = \sum \frac{1}{R_i}, \qquad R_{\text{loop}} = R\,x(1 - x)$
Redrawing networks: shorts and symmetry
The two moves
$V_P = V_Q \ \Rightarrow\ I_{PQ} = 0, \qquad R_{\text{eq}} = \frac{V_{\text{in}} - V_{\text{out}}}{I_{\text{total}}}$

Watch out for (6)

A cut wire divides twice
→ Series, parallel and bent wires
Arcs go by length
→ Series, parallel and bent wires
Check every option numerically
→ Series, parallel and bent wires
A missed short
→ Redrawing networks: shorts and symmetry
A crossing is not a junction
→ Redrawing networks: shorts and symmetry
Symmetry must be about the input–output line
→ Redrawing networks: shorts and symmetry

Test yourself on Current Electricity

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