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Answer 1

Answer:8ΩStarting from inner loop the between point F and point E resistances 10Ω and 2.5Ω are in parallel connection, their equivalent resistance will be R eq1 = R 1 +R 2 R 1 R 2 = 10+2.510×2.5 Ω=2Ω resistance between F and E can be replaced with 2Ω Now resistance between point F and point G and R eq1 are in series their equivalent resistance will be R eq2 =10+2Ω=12ΩIn next step resistance between point F and point G and R eq2 are in series their equivalent resistance will be R eq3 =10+2Ω=12Ω resistance between G and D can be replaced with 12Ω The resistance between point G and point E and R eq3 are in parallel, their equivalent resistance will be R eq4 = R 1 +R 2 R 1 R 2 = 12+1212×12 Ω=6Ω resistance between G and E can be replaced with 6Ω for resistances between point C and point G and R eq4 are in series their equivalent resistance will be R eq5 =10+6Ω=16Ω resistance between C and D can be replaced with 16Ω Finally resistance between point B and point E and R eq5 are in parallel, their equivalent resistance will be R eq6 = R 1 +R 2 R 1 R 2 = 16+1616×16 Ω=8Ω resistance between A and B can be replaced with 8Ω Thus eqvivalent resistance between A and B is 8Ω .

Explanation:8Ω

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