Three resistances each of $4 O h m$ are connected in the form of an equilateral triangle. The effective resistance between any two corners is:

Ohm

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Ohm

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8 Ohm

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12 Ohm

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Solution

The correct option is B
Ohm

Given,

Resistance $R$ = $4 O h m$

Between two corners 2 resistances are in series and in parallel with one resistance

In series Resistance = $R + R$ = $2 R$

Now $2 R$ is in parallel with $R$

Effective resistance = $\frac{R_{1} . R_{2}}{R_{1} + R_{2}}$

This 8 Ohm will be parallel to 4 Ohm which gives a resultant = $\frac{4 * 8}{4 + 8}$ = $\frac{8}{3}$ Ohm.

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Three resistances each of 4 Ohm are connected in the form of an equilateral triangle. The effective resistance between any two corners is:

Three resistances each of $4 O h m$ are connected in the form of an equilateral triangle. The effective resistance between any two corners is: