Question

a wire resistance 27 ohm is bent to form and equilateral triangle.Find the resistance acrossany two vertices of a triangle.

Answer 1

Given :

A wire of resistance 27Ω is bent to form an equilateral triangle.

To Find :

Equivalent resistance across any two vertices of a triangle.

Solution :

❖ Let length of the resistor be l. It is bent to form an equilateral triangle.

Length of each side of triangle = l/3

We know that resistance is directly proportional to the length of conductor$.$

• Therefore resistance of each side of triangle will be 27/3 = 9Ω

A] R₂ and R₃ are connected in series so their equivalent resistance will be

➙ R₂₃ = R₂ + R₃

➙ R₂₃ = 9 + 9

➙ R₂₃ = 18Ω

B] Finally R₁ and R₂₃ come in parallel so net equivalent resistance of the circuit will be

➙ 1/R = 1/R₁ + 1/R₂₃

➙ 1/R = 1/9 + 1/18

➙ 1/R = 3/18

➙ R = 18/3

➙ R = 6Ω

Answer 2

Given :

A wire of resistance 27Ω is bent to form an equilateral triangle.

To Find :

Equivalent resistance across any two vertices of a triangle.

Solution :

❖ Let length of the resistor be l. It is bent to form an equilateral triangle.

Length of each side of triangle = l/3

We know that resistance is directly proportional to the length of conductor$.$

Therefore resistance of each side of triangle will be 27/3 = 9Ω

A] R₂ and R₃ are connected in series so their equivalent resistance will be

➙ R₂₃ = R₂ + R₃

➙ R₂₃ = 9 + 9

➙ R₂₃ = 18Ω

B] Finally R₁ and R₂₃ come in parallel so net equivalent resistance of the circuit will be

➙ 1/R = 1/R₁ + 1/R₂₃

➙ 1/R = 1/9 + 1/18

➙ 1/R = 3/18

➙ R = 18/3

➙ R = 6Ω