Three unequal resistor in parallel are equivalent to a resistance `1Omega` If two of them are in the ratio 1:2 and if no resistance value is fractional the largest of the three resistance in ohm is

Options

A. `10 Omega`
B. `8 Omega`
C. `15 Omega`
D. `6 Omega`

Answer

d

Explanation

Let the resistances be `R_(1), R_(2)` and `R_(3)`, respectively. According to the question, equivalent resistance in parallel `= 1 Omega`
In parallel combination, the equivalent resistance is given by
`(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))`
Here, `R_(eq) = 1`
`implies (1)/(1) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))`
Let `R_(3) = 2R_(2)`
If we check the option given in the question
`(1)/(1) =(1)/(r_(1)) + (1)/(R_2)) + (1)/(2R_(2)) = (1)/(R_(1)) + (3)/(2R_(2))`
If `R_(1) 1 Omeag, R_(2) = 2 Omega`
`(1)/(1) = ((1)/(1) + (3)/(2 xx 2)) = (1)/(1) + (3)/(4) = (7)/(4)`
will not hold the conditon
similarly, if we check the orther options only
`R_(1) = 2 Omega, R_(2) = 3 Omega`
`(1)/(1) = (1)/(2) + (1)/(3) = (1)/(R_(3))`
`= (5)/(6) + (1)/(R_(3)) implies - (5)/(6) = (1)/(R_(3))`
or, `(1)/(6) = (1)/(R_(3))`
So, it will satisfy the equation,
`(1)/(R_(eq)) = (1)/(r_(1)) + (1)/(R_(2) ) + (1)/(R_(3))`
Or, `(1)/(R_(eq)) = (1)/(2) + (1)/(3) + (1)/(6) = (6)/(6) = (1)/(1)`
or, `R_(eq) = 1 Omega`
`R_(3) = 6 Omega` will be heighest resistance.