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Answer :
If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors,
\(\Rightarrow \) 6 \(\Omega \) + 6 \(\Omega \) + 6 \(\Omega \) = 18 \(\Omega \),
which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be \(\frac{6}{2} \) = 3 \(\Omega \) is also not desired.
Hence, we should either connect the two resistors in series or parallel.
(a) Two resistor in parallel
Two 6 \(\Omega \) resistors are connected in parallel. Their equivalent resistance will be
$$ \frac{1}{(\frac{1}{6} + \frac{1}{6} )}= \frac{6}{2} = 3 \Omega . $$
The third 6 \(\Omega \) resistor is in series with 3 \(\Omega \). Hence, the equivalent resistance of the circuit is 6 \(\Omega \)+ 3 \(\Omega \) = 9 \(\Omega \).
(b) Two resistor in parallel
Two 6 \(\Omega \) resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 \(\Omega \).
The third 6 \(\Omega \) resistor is in parallel with 12 \(\Omega \). Hence, equivalent resistance will be
$$ 12 x \frac{6}{12} + 6 = 4 \Omega . $$
Therefore, the total resistance is 4\(\Omega \)