A battery of emf `E` and of negligible internal resistance is connected in a `L-R` circuit as shown in figure. The inductor has a piece of soft iron inside it. When steady state is reached in the piece of soft iron is abruptly pulled out suddenly so that the inductance of the inductor decreases to `nL` with `nlt1` with battery remaining connected. Calculate.

a. Current as a function of time assuming `t=0` at the instant when piece is pulled.
b. the work done to pull out the piece.
c. thermal power generated in the circuit as as function of time.
d. power supplied by the battery as a function of time.
HOW TO PROCEED When the inductance of an inductor is abruptly changed, the flux passing through it remains constant.
`phi =` constant
`:. Li =` constant `(L = (phi)/(i))`

Explanation

a. At time `t=0` steady state current in the circuit is `i_0=E//R`. Suddenly ,`L` reduces to `nL(nlt1)`, so current in the circuit at time `t=0` will increase to `i_0/n=E/(nR)`. Let i be the current at time `t`.

Applying Kirchoff's loop rule, we have
`E-nL((di)/(dt))-iR=0`
`:. (di)/(E-iR)=1/(nL)dt`
`:. int_(i_o//n)^i(di)/(E=iR)=1/(nL)int_0^tdt`
solving this equation we get
`i=i_0-(i_0-i_0/n)e^(-t//tauL)`
Here, `i_0=E/R`
`and tau_L=(nL)/R`

Form the i-t equation we get `i=i_0/n` at` t=0` and `i=i_0` at `t=oo`
The `i-t` graph is as known in figure.
At `=0`, current in the circuit is `(i_0)/(n)`. Current in the circuit in steady state will be again `i_(0)`. So, it will decrease exponentially from `(i_(0))/(n)` to `i_(0)`. From the `i-t` graph, the equation can be fomed without doing any calculation.