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(D) When three resistors of $4\,\Omega$ are connected in an equilateral triangle,let the corners be $A$,$B$,and $C$.To find the effective resistance b

Vedclass · Accepted Answer

$\frac{8}{3}\,\Omega$

Vedclass · Answer

(D) When three resistors of $4\,\Omega$ are connected in an equilateral triangle,let the corners be $A$,$B$,and $C$.To find the effective resistance between any two corners (e.g.,$A$ and $B$),we observe the circuit configuration.The resistor connected directly between $A$ and $B$ is in parallel with the series combination of the other two resistors (connected between $A-C$ and $C-B$).Resistance of the series branch = $4\,\Omega + 4\,\Omega = 8\,\Omega$.Now,this $8\,\Omega$ branch is in parallel with the $4\,\Omega$ resistor connected directly across $A$ and $B$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{8} = \frac{2+1}{8} = \frac{3}{8}$.Therefore,$R_{eq} = \frac{8}{3}\,\Omega$.

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

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