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Question
Six resistances are connected together as shown in the figure. Calculate the equivalent resistance between points A and B.
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Solution
The resistors R2, R3 and R4 in series.
∴ R' = R2 + R3 + R4
= 2 + 3 + 5 = 10 Ω
Now R' and R5 are in parallel.
∴ `1/("R"'') = 1/("R"') + 1/"R"_5`
`= 1/10 + 1/10 = 2/10 = 1/5`
∴ R'' = 5 Ω
Now R1, R” and R6 in series between the points A and B. The equivalent resistance between A and B is
R = R1 + R” + R6 = 2 + 5 + 5 = 12 Ω
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