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(A) To find the effective resistance between points $A$ and $B$,we observe the circuit configuration.$1$. The resistors $BC$ $(60\,\Omega)$ and $CA$ $

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$32$

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(A) To find the effective resistance between points $A$ and $B$,we observe the circuit configuration.$1$. The resistors $BC$ $(60\,\Omega)$ and $CA$ $(100\,\Omega)$ are connected in series because the same current flows through them when a potential difference is applied across $A$ and $B$.$2$. The equivalent resistance of this series branch is $R_{series} = 60\,\Omega + 100\,\Omega = 160\,\Omega$.$3$. This series combination is now in parallel with the resistor $AB$ $(40\,\Omega)$.$4$. The effective resistance $R_{eq}$ between $A$ and $B$ is given by the parallel combination formula: $\frac{1}{R_{eq}} = \frac{1}{R_{AB}} + \frac{1}{R_{series}} = \frac{1}{40} + \frac{1}{160}$.$5$. Calculating this: $R_{eq} = \frac{40 	imes 160}{40 + 160} = \frac{6400}{200} = 32\,\Omega$.

Three resistors are connected to form the sides of a triangle $ABC$. The resistances of the sides $AB$,$BC$,and $CA$ are $40\,\Omega$,$60\,\Omega$,and $100\,\Omega$ respectively. The effective resistance between the points $A$ and $B$ in $\Omega$ will be:

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