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If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will beA. `(20)/(19)Omega`B. `(19)/(20)Omega`C. `(10)/(20)Omega`D. `(29)/(10)Omega` |
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Answer» Correct Answer - A The resistance `2 Omega, 4 Omega` and `5 Omega` are connected in parallel combination. Therefore, resultant resistance is given by `(1)/(R )=(1)/(2)+(1)/(4)+(1)/(5)=(19)/(20) rArr R = (20)/(19) Omega` |
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