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?

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

= 12^e−5tV

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

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An inductor of inductance L= 400mH and resistors R1=2\(\Omega\) and R2=2\(\Omega\) 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 tim