The voltage across the inductor in a circuit is directly proportional to the rate of flow of current through the inductor.
For this firstly calculate the current flowing through the branch containing L and R2. Then differentiate current I with respect to time t.
Given: L = 400mH
R1 = 2Ω
R2 = 2Ω
E = 12V
In the branch containing L and R2, I = E/R2 1−e−R2t/L
Differentiating I with respect to t:
dI/dt = ER2e−R2t/L. R2/L
= E/Le−R2t/L
Therefore, VL = L dI/dt = Ee−R2t/L
= 12e−5tV
Then the potential drop across L as a function of time is 12e−5tV.
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