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

Answer

8

Explanation

`L_(1) = 1 `mH, `r_(1) = 3 Omega`
`L_(2) = 2 `mH, ` r_(2) = 4 Omega`
R = 12 `Omega`
Current through R = `(5V)/(12 Omega) = (5)/(12) A `
At t = 0 current through both the inductors = 0 after a sufficiently long time, current in the inductors in stabilised For t `rarr intfy `
`i_(1) ("through"L_(1)) = (5V)/(3 Omega) = (5)/(3)` A
`I_(2) ("through"L_(2)) = (5V)/(4 Omega) = (5)/(4) A `
`(I_(max))/(I_(min)) = (((5)/(12) + (5)/(3) + (5)/(4)))/(((5)/(12)))` = 8