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Numerical
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of 4 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 series:
Two 6 Ω resistors are in series. Their equivalent resistance will be the sum 6 + 6 = 12 Ω.
The third 6 Ω resistor is in parallel with 12 Ω. Hence, equivalent resistance will be
`1/(1/12+1/6)=(12xx6)/(12+6)=4Omega.`
Therefore, the total resistance is 4 Ω.
Is there an error in this question or solution?