Three resistors each of 4,Omega are connected | vedclass

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(B) Given three resistors, each with resistance $R = 4\,\Omega$.Case $1$: All three resistors are connected in series.$R_{eq} = R + R + R = 4 + 4 + 4

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$3$

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(B) Given three resistors, each with resistance $R = 4\,\Omega$.Case $1$: All three resistors are connected in series.$R_{eq} = R + R + R = 4 + 4 + 4 = 12\,\Omega$.Case $2$: All three resistors are connected in parallel.$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{eq} = \frac{4}{3} \approx 1.33\,\Omega$.Case $3$: Two resistors are in series, and the third is in parallel with the combination.$R_{series} = 4 + 4 = 8\,\Omega$.$R_{eq} = \frac{8 	imes 4}{8 + 4} = \frac{32}{12} = 2.67\,\Omega$.Case $4$: Two resistors are in parallel, and the third is in series with the combination.$R_{parallel} = \frac{4 	imes 4}{4 + 4} = 2\,\Omega$.$R_{eq} = 2 + 4 = 6\,\Omega$.Comparing these results with the given options, the value $3\,\Omega$ cannot be obtained.

Three resistors each of $4\,\Omega$ are connected together to form a network. The equivalent resistance of the network cannot be ............ $\Omega$.

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