A uniform metallic wire has a resistance of 18 ohms and is bent into an equilateral triangle. Then,

Q: Why do we divide the total resistance by 3?

An equilateral triangle has three sides of equal length. Since the wire is uniform, its resistance is directly proportional to its length ($R = \rho L/A$). Therefore, the total resistance is divided equally among the three sides.

Q: Why are two sides in series and one in parallel?

When you measure resistance between two vertices (let's say A and B), the current has two paths to go from A to B. One path is directly through the side AB. The other path goes through side AC and then side CB. Since the same current must flow through AC and CB in that second path, they are in series. This entire series combination is parallel to the direct path AB.

Q: Does it matter which two vertices I choose?

Not at all! Because the triangle is perfectly symmetrical and all sides have the same resistance ($6\ \Omega$), the equivalent resistance between any two vertices will always be exactly the same.

Q: What if the wire was bent into a square instead?

If it were a square, the wire would be divided into four equal parts. Each side would have a resistance of $18 / 4 = 4.5\ \Omega$. The resistance between adjacent corners would be one side in parallel with three sides in series.

Q: Can I use the formula $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$ for more than two resistors?

No, that specific product-over-sum formula only works for exactly two resistors in parallel. If you have three or more, you must use the general formula: $1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3 + \dots$

Question

A uniform metallic wire has a resistance of $18\ \Omega$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is

Options:

(A) $12\ \Omega$

(B) $8\ \Omega$

(C) $2\ \Omega$

(D) $4\ \Omega$

Sigma Percentile · Accepted Answer

### Visualizing the Geometry

Imagine a straight metallic wire with a total resistance of $18\ \Omega$. Now, visualize taking this wire and carefully bending it to form a perfect equilateral triangle. This is our physical setup.

Focus on the geometry here. Because it's an equilateral triangle, the wire is divided into three segments of exactly equal length.

**Since we know that resistance is directly proportional to length ($R \propto L$), the total resistance will also be divided equally among the three arms.**

### Calculating Individual Arm Resistance

So, let's set up the calculation for the resistance of each individual arm. We simply divide the total resistance, $18\ \Omega$, by $3$.

$$ R_{	ext{arm}} = \frac{18}{3} $$

This gives us exactly $6\ \Omega$ for each side of the triangle.

### Analyzing the Circuit Paths

The question asks for the equivalent resistance between any two vertices. Let's pick vertices A and B as our terminals. Imagine connecting a battery across these two points to see how the current would flow.

Look closely at the path from A to B that goes through vertex C. The current flowing through arm AC has nowhere else to go but through arm CB.

**This means arms AC and CB are in a perfect series combination.**

Substituting the values, we get $6\ \Omega + 6\ \Omega$, which is $12\ \Omega$ for the entire upper branch.

### The Final Parallel Combination

Now, this entire upper branch of $12\ \Omega$ is connected directly across our terminals A and B.

**This means it is in parallel with the bottom arm AB, which has a resistance of $6\ \Omega$.**

To find the final equivalent resistance of two parallel resistors, we use the classic product over sum formula.

$$ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} $$

Let's substitute our values carefully into the parallel resistance formula: $12$ multiplied by $6$, divided by $12$ plus $6$.

$$ R_{eq} = \frac{12 	imes 6}{12 + 6} $$

Let's do the math. $12 	imes 6$ is $72$, and $12 + 6$ is $18$.

$$ R_{eq} = \frac{72}{18} $$

And $72$ divided by $18$ gives us exactly $4\ \Omega$. **That is our final, elegant answer!**

### The Way Forward

Now, think about this. What if the same wire was bent into a square instead of a triangle?

How would the resistance between adjacent corners or opposite corners change? Try calculating that on your own to truly master this concept!

Animated Solution for Physics - Current Electricity: A uniform metallic wire has a resistance of $18 Ω$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is

Select Answer:

Visualized Solution

$The Equilateral Triangle$

$Resistance and Length$

$Resistance of Each Arm$

$Calculating Arm Resistance$

$Choosing the Terminals$

$Series Combination$

$Series Resistance$

$Parallel Combination$

$Substituting Values$

$Simplifying the Expression$

$Final Equivalent Resistance$

$What if it was a square?$

The Sigma Insight: Combination of Resistors

Visualizing the Geometry

Calculating Individual Arm Resistance

Analyzing the Circuit Paths

The Final Parallel Combination

The Way Forward

Similar Questions

A square shaped wire with resistance of each side $3Ω$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $Ω$ will be

A wire of resistance $R$ is bent to form a square $A B C D$ as shown in the figure. The effective resistance between $E$ and $C$ is [ $E$ is mid-point of arm $C D$ ]

Five equal resistances are connected in a network as shown in figure. The net resistance between the points $A$ and $B$ is

A current of 6 A enters one corner $P$ of an equilateral triangle $P QR$ having three wires of resistance $2 Ω$ each and leaves by the corner $R$ . The currents $i_{1}$ in ampere is ......... .

Six equal resistances are connected between points $P, Q$ and $R$ as shown in the figure. Then, the net resistance will be maximum between

The effective resistance between points $P$ and $Q$ of the electrical circuit shown in the figure is

If each of the resistances in the network shown in the figure is $R$ , what is the resistance between the terminals $A$ and $B$ ?

The equivalent resistance between points $A$ and $B$ of the circuit given below is ....... .

All resistances in the diagram are in ohm. Find the effective resistance between the points A and B.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Equilateral Triangle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance and Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance of Each Arm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Arm Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Choosing the Terminals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Series Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Series Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Equivalent Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was a square?