Two inductors `L_(1)`(inductors 1 mH, internal resistance 3 `Omega`) and `L_(2)` (inductance 2mH, internal resistance 4`Omega`),and a resistor R(resistance`12 omega`) are all connected in parallelacross a 5 V battery. The circuit is switched on at time t=0. The ratio of the maximum to the minimum current `(I_(max)//I_(min))` drawn from the battery is

Options

A. 8
B. 7
C. 5
D. 6

Answer

A

Explanation

At `t = 0, I_(min) = (E)/(R ) = (5)/(12) A`
Finally, in the steady state
`I_(max) = (E)/(R_(eq)) = E ((1)/(r_(1)) + (1)/(r_(2)) + (1)/(R ))`
`= 5 ((1)/(3) + (1)/(4) + (1)/(12)) = (10)/(3) A`
`:. (I_(max))/(I_(min)) = 8`