In the circuit shown, switch `S` is closed at time `t = 0`. Find the current through the inductor as a function of time `t`.

Answer

A, B, C

Explanation

At `t=0`, Current through inductor will be zero.
At `t=oo` net emf `=(2//2+4//1)/(1//2+1//1)=10/3V`
Net resistance =`(2xx1)/(2+1)=2/3Omega`
`:. i=(10//3)/(2//3)=5A`

To find equivalent time constant short circuit, both the batteries and find net resistance across inductor.
`R_("net")=(2xx1)/(2+1)=2/3Omega`
`:. tau_L=L/R_("net")=(1xx10^-3)/(2/3)=3/2000s`
Current through inductor will increase exponentially from 0 to 5 A.
`:. i=5(1-e^(-(2000t)/3))`