A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance (in Ω) between any two vertices of the triangle is:

Question

A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance (in Ω) between any two vertices of the triangle is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/291b458a6907e565e62c9e4ae8af2578-.png" /> Resistance, R propto l so resistance of each side of the equilateral triangle = 6 Ω Resistance Req between any two vertices (1/Req) = (1/12) + (1/6) ⇒ Req. = 4Ω

Q. A uniform metallic wire has a resistance of $18 Ω$ and is bent into an equilateral triangle. Then, the resistance (in $Ω$ ) between any two vertices of the triangle is:

Resistance, $R \propto l$ so resistance of each side of the
equilateral triangle $= 6 Ω$
Resistance $R_{e q}$ between any two vertices
$\frac{1}{R _{e q}} = \frac{1}{12} + \frac{1}{6}$
$\Rightarrow R_{e q .} = 4Ω$

Q. A uniform metallic wire has a resistance of 18Ω and is bent into an equilateral triangle. Then, the resistance (in Ω) between any two vertices of the triangle is:

Resistance, R∝l so resistance of each side of the equilateral triangle =6Ω Resistance Req​ between any two vertices Req​1​=121​+61​ ⇒Req.​=4Ω

Q. A uniform metallic wire has a resistance of $18 Ω$ and is bent into an equilateral triangle. Then, the resistance (in $Ω$ ) between any two vertices of the triangle is:

Resistance, $R \propto l$ so resistance of each side of the
equilateral triangle $= 6 Ω$
Resistance $R_{e q}$ between any two vertices
$\frac{1}{R _{e q}} = \frac{1}{12} + \frac{1}{6}$
$\Rightarrow R_{e q .} = 4Ω$