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Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of 9 Ω.

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#### Solution

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., 6 Ω + 6 Ω + 6 Ω = 18 Ω, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be `6/2` = 3 Ω is also not desired. Hence, we should either connect the two resistors in series or parallel.**Two resistors in parallel:**

Two 6 Ω resistors are connected in parallel. Their c will be

`1/(1/6 + 1/6) = (6 xx 6)/(6 + 6) = 3 Omega`

The third 6 Ω resistor is in series with 3 Ω. Hence, the equivalent resistance of the circuit is 6 Ω + 3 Ω = 9 Ω.

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