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How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

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(a) To get total resistance 4 Ω, connect 3 Ω and 6 Ω resistors in parallel and 2 Ω resistance in series with the resultant.

⟹12=2/1=2 Ω Now, the resultant 12 and 2 Ω resistors are in series. So the net resistance =12+2 Ω=2+2=4 Ω (b). To get total resistance 1 Ω, connect 2 Ω, 3 Ω and 6 Ω resistors in parallel.
The net resistance in parallel is given by

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