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B
2W
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C
3W
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D
4W
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Solution
The correct option is B2W Let us use the technique of successive reduction. The resistor in arm XQ can be removed, as it represents a balanced Wheatstone bridge.
Now arms PQR and PR are in parallel, their equivalent resistance is given by, 6×36+3=2Ω
SR can be removed as it represents another balanced Wheatstone bridge and equivalent resistance between PT will be, 6×186+18=92Ω
Further, net equivalent resistance,