A wire of resistance R is bent into a triangular pyramid as shown… — Solution | JEE Physics

MCQMediumJEE 2025Combination of Resistors

A wire of resistance $R$ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $A$ and $B$ is $\frac{R}{n}$ . The value of $n$ is:

A triangular pyramid resistor network with vertices A and B at the base ends, one top vertex, and one central junction connected symmetrically by equal resistive segments.

A
$16$
B
$14$
C
$10$
D
$12$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A wire of total resistance $R$ is bent into a triangular pyramid with all segments of equal length, so each segment has resistance

r = \frac{R}{6}

Find: The equivalent resistance between $A$ and $B$ , written as $\frac{R}{n}$ .

By symmetry, the network forms a balanced Wheatstone-bridge type arrangement between $A$ and $B$ . Hence the branches can be reduced to three parallel paths: two paths of resistance $2r$ each and one direct path of resistance $r$ .

So,

\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r}

Therefore,

\frac{1}{R_{AB}} = \frac{2}{r}

which gives

R_{AB} = \frac{r}{2}

Substitution and Conclusion

Now substitute

r = \frac{R}{6}

into

R_{AB} = \frac{r}{2}

Then,

R_{AB} = \frac{1}{2}\cdot \frac{R}{6} = \frac{R}{12}

Using Symmetry Directly

Because all six edges of the triangular pyramid are equal, each edge has resistance $\frac{R}{6}$ . Symmetry about the line joining $A$ and $B$ makes the bridge balanced, so no extra current analysis is needed for the middle balancing branch. The network reduces immediately to two branches of $2r$ in parallel with one branch of $r$ , giving

R_{AB} = \left(\frac{1}{2r} + \frac{1}{2r} + \frac{1}{r}\right)^{-1} = \frac{r}{2} = \frac{R}{12}

Therefore, the correct option is D, so $n = 12$ .

Common mistakes

Assuming each segment has resistance $R$ . This is wrong because $R$ is the resistance of the entire wire, not one edge. Since the pyramid has six equal segments, each segment must be $\frac{R}{6}$ .
Ignoring symmetry and treating the network as an arbitrary resistor combination. This misses the balanced-bridge simplification. Use symmetry first to identify equal-potential points and reduce the circuit correctly.
Adding all visible paths directly in series. That is incorrect because the paths between $A$ and $B$ are alternative current routes, so they combine in parallel after the symmetry-based reduction.

Practice more Combination of Resistors questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Combination of Resistors

Answer

Step-by-step solution

Common mistakes

Practice more Combination of Resistors questions

Related questions