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Using Kirchhoff's rules for the instantaneous currents and voltages in the two-loop circuit, find the current in the inductor as a function of time.

I used Kirchhoff's rule on the left loop consisting of the battery and the right loop which consist of the inductor and gotten the following equations.

E-I

L dI

Combining both the equations give me

E-I

So from here how do we express I

I used Kirchhoff's rule on the left loop consisting of the battery and the right loop which consist of the inductor and gotten the following equations.

E-I

_{1}R_{1}- I_{3}R_{2}= 0L dI

_{2}/dt - I_{3}R_{2}= 0Combining both the equations give me

E-I

_{1}R_{1}- L dI_{2}/dt =0So from here how do we express I

_{1}into I_{2}and R to get the differentiate equation form?