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

We should know that ideal inductor behaves like an infinite resistance at t = 0 , when circuit is switched on. And it behaves like zero resistance after current is established in the circuit , after a long time. But here inductors are said to have some internal resistance, and hence initially they will behave like an infinite resistance and after a long time they will behave simply like resistance of given value. In one parallel branch, there is one pure resistance .
At t = 0 there would not be any current through the parallel branches containing inductor and current will flow only through branch of pure resistance . it will be the case of minimum current drawn from the battery .
Minimum current drawn from the battery can be written as follows?
` I_(min) = (E)/(R) = (5)/(12)` A
After a long time, when current sets in branches containing inductors, then resistance of magnitude 3 `Omega, 4 Omega and 12 Omega` will be considered to be connected in parallel to each other. Eqivalent resistance of the circuit will be written as follows :
`(1)/(R) = (1)/(3) + (1)/(4) + (1)/(12) = (4 + 3 + 1)/(12) = (8)/(12) = (2)/(3)`
`rArr " " R = 3//2 Omega`
In this case current drawn from the battery will be maximum Hence maximum current drawn from the battery can be written as follows :
`I_(max) = (E)/(R) = (5)/(3) = (10)/(3) ` A
Ratio of the maximum to minimum current can be written as follows :
`(I_(max))/(I_(min)) = (10//3)/((5)/(12)) = (10 xx 12)/(5 xx 3) = 2 xx 4 = 8`