Three conductors draw currents of 1 A, 2 A and 3 A respectively, when connected in turn across a battery. If they are connected in series and the combination is connected across the same battery, the current drawn will be

Options

A. `(2)/(7) A`
B. `(3)/(7) A`
C. `(4)/(7) A`
D. `(5)/(7) A`

Explanation

Let `R_(1), R_(2) and R_(3)` be resistance of three conductors respectively
As per question

When they are connected in series, their equivalent resistance is
`R_(eq) = R_(1) + R_(2) + R_(3)=V + (V)/(2) + (V)/(3)`
`=V (1+(1)/(2) + (1)/(3)) = (11)/(6)V`

When they are connected in series, across the same, battery, then the current in the circuit is
`I = (V)/(R_(eq)) = (V)/((11)/(6)V) = (6)/(11)A`
None of the given options is correct