A wire of resistance $9\text{ }\Omega$ is bent into the shape of an equilateral triangle. What is the equivalent resistance between any two vertices of the triangle? (Shift I Memory Based)

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YESPYQ · Accepted Answer

Correct answer: b) $2\text{ }\Omega$. Total resistance of the wire: ${R}_{\text{total}}=9\text{ }\Omega$ Resistance of each side: ${R}_{\text{side}}=\frac{{R}_{\text{total}}}{3}=3\text{ }\Omega$ Between any two vertices, there are two paths: Direct resistance ${R}_{1}=3\text{ }\Omega$. The other two sides (${R}_{2}$) add up: ${R}_{2}=6\text{ }\Omega$ These are in parallel: ${R}_{\text{eq}}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}=\frac{(3)(6)}{3+6}=2\text{ }\Omega \mathrm{.}$ Thus, the equivalent resistance is $2\text{ }\Omega$.

A wire of resistance \(9\text{ }\Omega\) is bent into the shape of an equilateral triangle. What is the equivalent resis…

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