Question 11 (ii) Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (ii)4Ω?
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Solution
(b) Two resistors in series
Two 6Ω resistors are in series. Their equivalent resistance will be the sum 6Ω+6Ω=12Ω. The third 6Ω resistor is in parallel with 12Ω. Hence, equivalent resistance will be R=1112+16=12×612+6=4Ω Therefore, the total resistance is 4Ω.