A uniform wire of resistance 9Ω is bent in the form of an equilateral triangle. Find the effective resistance across a side of the triangle.

Question

A uniform wire of resistance 9&Omega; is bent in the form of an equilateral triangle. Find the effective resistance across a side of the triangle.

Prepp Team · Accepted Answer

Calculating Effective Resistance in an Equilateral Triangle Wire

This problem asks us to find the effective resistance across one side of a uniform wire that has been bent into the shape of an equilateral triangle. The total resistance of the wire is given as 9 $\Omega$.

Understanding the Problem Setup

When a uniform wire with total resistance $R_{total}$ is bent into an equilateral triangle, the resistance is distributed equally along each side. Since an equilateral triangle has three equal sides, the resistance of each side will be one-third of the total resistance.

Total resistance of the wire = 9 $\Omega$.
Number of equal sides in an equilateral triangle = 3.
Resistance of each side $R_{side}$ = $\frac{R_{total}}{3}$.

Resistance of Each Side

Let $R_{side}$ be the resistance of one side of the triangle.

$R_{side} = \frac{9 Ω}{3} = 3 Ω$

So, each side of the equilateral triangle has a resistance of 3 $\Omega$.

Analyzing the Circuit Across a Side

To find the effective resistance across one side, let's consider finding the resistance across side AB of the triangle ABC. The current enters at point A and leaves at point B. The path through side AB is one direct path with resistance 3 $\Omega$. The other path is through the remaining two sides, AC and CB, which are connected in series between points A and B.

Path 1: Directly through side AB. Resistance = 3 $\Omega$.
Path 2: Through sides AC and CB in series. Resistance = $R_{AC} + R_{CB}$.

Calculating Resistance of the Series Path

The resistance of side AC is 3 $\Omega$, and the resistance of side CB is also 3 $\Omega$. When resistors are in series, their resistances add up.

Resistance of series path (AC + CB) = $3 \Omega + 3 \Omega = 6 \Omega$.

Calculating Effective Resistance of the Parallel Combination

Now we have two paths between points A and B:

A 3 $\Omega$ resistance (side AB).
A 6 $\Omega$ resistance (series combination of sides AC and CB).

These two paths are in parallel. The formula for the effective resistance ($R_{eff}$) of two resistors ($R_1$ and $R_2$) in parallel is:

$R_{eff} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$

In this case, $R_1 = 3 \Omega$ and $R_2 = 6 \Omega$. Plugging these values into the formula:

$R_{eff} = \frac{3 Ω \times 6 Ω}{3 Ω + 6 Ω}$

$R_{eff} = \frac{18 Ω^{2}}{9 Ω}$

$R_{eff} = 2 Ω$

The effective resistance across a side of the equilateral triangle is 2 $\Omega$.

Summary of Calculation Steps

Step	Description	Calculation	Result
1	Resistance per side	Total Resistance / 3	$9 Ω / 3 = 3 Ω$
2	Resistance of series part (2 sides)	Resistance of side 1 + Resistance of side 2	$3 Ω + 3 Ω = 6 Ω$
3	Effective resistance (Parallel combination)	(R_parallel * R_series) / (R_parallel + R_series)	$(3 Ω \times 6 Ω) / (3 Ω + 6 Ω) = 18 / 9 Ω = 2 Ω$

Revision Table: Resistance Calculations

Concept	Description	Formula
Resistance of uniform wire segment	Proportional to length	$R \propto L$
Resistors in Series	Total resistance is the sum of individual resistances	$R_{total} = R_{1} + R_{2} + ...$
Resistors in Parallel	Reciprocal of total resistance is the sum of reciprocals of individual resistances	$\frac{1}{R_{total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ...$

Additional Information: Resistance Networks

Problems involving wires bent into shapes often require identifying series and parallel combinations. A uniform wire means resistance is directly proportional to length. When a wire is bent into a polygon, and the resistance is measured across one side (or between two vertices), the remaining parts of the wire form a network, typically involving series and parallel connections. For symmetrical shapes like equilateral triangles, squares, or regular polygons, the resistance distribution is even across equal segments.

For an equilateral triangle, measuring across a vertex would give a different result.
For a square made of a uniform wire, measuring across a side involves a series combination of three sides in parallel with one side. Measuring across a diagonal involves two paths, each consisting of two sides in series, with these two paths in parallel.
Always identify the two points between which the effective resistance is required and trace the different current paths.

Prepp Team · Answer

$\frac{1}{3}{\rm{\Omega }}$

Prepp Team · Answer

27 &Omega;

Prepp Team · Answer

3 &Omega;