If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will be

Question

If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will be

If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will be
A. `(20)/(19)Omega`
B. `(19)/(20)Omega`
C. `(10)/(20)Omega`
D. `(29)/(10)Omega`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`

5 views Answered 2023-02-05 15:29