Question 82: Current Electricity

Question

A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Accepted Answer

$$ 	ext{Each side of the triangle has a resistance of } R_{	ext{side}} = \frac{9\,\Omega}{3} = 3\,\Omega. $$
When measuring the equivalent resistance between any two vertices, there are two paths:
A direct path along one side with resistance:
$$    R_1 = 3\,\Omega.    $$
An indirect path passing through the other two sides in series:
$$    R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega.    $$
These two paths are in parallel, so the equivalent resistance is calculated by:
$$ \frac{1}{R_{	ext{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{6\,\Omega}. $$
Thus,
$$ R_{	ext{eq}} = \frac{6\,\Omega}{3} = 2\,\Omega. $$
The equivalent resistance across any two vertices of the equilateral triangle is therefore:
$$ 2\,\Omega. $$

Current Electricity

A wire of resistance 9 \Omega is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.

Integer / Numeric Answer Type