In the circuit arrangement shown in figure, the switch `S` is closed at `t = 0`. Find the current in the inductance as a function of time? Does the current through `10 Omega` resistor vary with time or remains constant.

Answer

A, C

Explanation

At `t=0`, equivalent resistance of an inductor in infinite and at `t = oo`, equivalenet resistance is zero.

`:.` Initial current thorugh inducto `=0`
Final current through inducto `=36/10=3.6A`
To find equivalent time constant we will have to short circuit the battery and find net resistasnce across inductor.
`R_("net")=(10xx5)/(10+5)=10/3Omega`
`tau_L=L/R_("net")=3/10ms`
Current through inductor will increase exponentially from `0` to `3.6 A`.
`:. i=36 (1-e^(-t//tau)L)`, where `tau_L=3/10ms=300mus`
current through `10 Omega` will vary with time.