# An inductor of inductance L= 400mH and resistors R1=2Ω and R2=2Ω are connected to a battery of emf 12V . The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of time is:

oltage across the inductor in the circuit is proportional to the rate of current flow through the inductor. For this first calculate the current flowing through the branch containing L and R2. Then differentiate current i with respect to time t.

Given that,

L=400mH

R1=2Ω

R2=2Ω

E= 12V

In the branch containing L and R2, i=E/R1−e−R2t/L

Differentiating current, i with respect to time t we get,

⇒ di/d= ER2e−R2t/L. R2/L = E/Le−R2t/L

Therefore, VL = L di/d=Ee−R2t/L = 12e−5tV

Then the potential drop across L as a function of time is 12e−5tV.

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