View attachment 2370
can someone explain to me how to find the mathematical expression for iL.

An inductor that is charging with a DC source has a current running through it as

\(\displaystyle i_L = \dfrac{E}{R} \left ( 1 - e^{-tR/L} \right ) \)

where E is the supplied voltage and R is the resistance in the rest of the circuit. (I'm ignoring the impedance of the inductor here. If you need to account for this then you need to add in the impedance of the inductor. In that case replace R with \(\displaystyle X_L + R\).)

The potential drop across the inductor is then given by

\(\displaystyle v_L = L \dfrac{di}{dt} = E \left ( e^{-tR/L} \right )\)

You might find

**this video** to be helpful.

Can you finish from here?

-Dan