Dear Siddharth,

Ans:- I think your ans is wrong why Let me explain

Let the current in the inductor is i1 and through the parallel resistor is i2

Then from kirchoff,s laws we get

10(i1+i2)+10(i1+i2)+10 i2=6..............(1)

L di1/dt =10i2

or 5 di1/dt=10i2...............(2)

Solving these two eq we get

3/5=2i1+3/2di1/dt

integrating from i1 varying from 0 to i1 we get the current through the inductance is

i1=0.3(1-exp-4t/3)

hence initially at t=0, i1=0 i.e no current flows through the inductance

So, from eq (1) we get i2=0.2 amp

Hence the current through the battery is = 0.2 amp (ans)

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SOUMYAJIT IIT_KHARAGPUR