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Three resistor 2 Omega , 4 Omega and 6 Omega are combined In parallel. What is the total resistance of the combination ? |
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Answer» Solution :If EQUIVALENT resistance of parallel CONNECTION of given resistances is `R_(p)`, then according to law, `(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))` = `(1)/(2) + (1)/(4) + (1)/(5)` `= (10 + 5 + 4)/(20)` `THEREFORE (1)/(R_(p)) = (19)/(20)` `therefore R_(p) = (20)/(19) Omega` |
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