Question

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 1

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At `t=0 I_9min)=5/12`
At `t=(oo)I_(max)=(5)/(R_(eq))=(5)/(3//2)=10/3`
[(1)/(R_(eq))=1/3+1/4+1/12=8/12]`
`:. (I_(max)/(I_(min))=10/3xx12/5=8`.