Three resistance of one ohm each are connected in parallel. Such connection is again connected with `2//3 Omega` resistor in series. The resultant resistance will be

Options

A. `(5)/(3) Omega`
B. `(3)/(2) Omega`
C. `1 Omega`
D. `(2)/(3) Omega`

Answer

C

Explanation

(c ) Resistance of `1 ohm` group `= (R )/(n) = (1)/(3) Omega`
This is in series with `(2)/(4) Omega` resistor.
`:.` Total resistance `= (2)/(3) + (1)/(3) = (3)/(3) = 1 Omega`