Equivalent Resistance of Tetrahedron

 

❓ Question

A wire of total resistance R is bent to form a triangular pyramid (tetrahedron) as shown in the figure, with each segment having the same length (and hence same resistance).

The equivalent resistance between points A and B is given as

$$\frac{R}{n}\mathrm{.}$$

Find the value of $n$.

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🖼️ Question Image

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✍️ Short Solution

This is a classic JEE symmetry-based resistance problem.
No Kirchhoff, no long equations — sirf symmetry + equivalent paths 🔥

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🔹 Step 1 — Count number of equal segments

A triangular pyramid (tetrahedron) has:

• 4 vertices

• 6 edges

The wire of total resistance $R$ is bent uniformly into these 6 equal segments.

So resistance of each edge:

$$r=\frac{R}{\mathrm{6}}$$

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🔹 Step 2 — Identify symmetry in the network

We are asked resistance between A and B.

Important symmetry observation 👇

• Points C and D are symmetrically placed with respect to A and B

• So their potentials will be equal

📌 Hence:

• No current flows in the wire CD

• We can safely remove CD from the circuit (it doesn’t affect A–B resistance)

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🔹 Step 3 — Redraw the simplified circuit

After removing CD, the network becomes:

• Direct path: A → B (one edge)

• Two identical indirect paths:

  • A → C → B

  • A → D → B

Each indirect path has two edges in series.

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🔹 Step 4 — Calculate resistances of paths

• Direct AB:

$$r = \frac{R}{6}$$

• Path ACB:

$$r + r = \frac{R}{6} + \frac{R}{6} = \frac{R}{3}$$

• Path ADB:

$$\frac{R}{3}$$

So between A and B, we have three parallel branches:

$$\frac{R}{6},\quad \frac{R}{3},\quad \frac{R}{3}$$

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🔹 Step 5 — Find equivalent resistance

Parallel combination:

$$\frac{1}{R_{AB}} = \frac{1}{R/6} + \frac{1}{R/3} + \frac{1}{R/3}$$
$$= \frac{6}{R} + \frac{3}{R} + \frac{3}{R} = \frac{12}{R}$$

So:

$$R_{AB} = \frac{R}{12}$$

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✅ Final Answer

$$\boxed{n = 12}$$

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⭐ Golden JEE Insight

• Symmetry ⇒ equal potential ⇒ zero current

• Such branches can be removed instantly

• Tetrahedron, cube, hexagon —
  👉 Symmetry is the fastest shortcut

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