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Question

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.

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