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Question

An inductor of inductance L = 400 mH and resistors of resistance R1=2Ω and R2=2Ω are connected to a battery of emf 12 V as shown in the figure. 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


  1. 6e5tV
  2. 12e5tV
  3. 6(1et/0.2)V
  4. 12te3tV


Solution

The correct option is B 12e5tV
I1=ER1=122=6AE=Ldl2dt+R2×I2I2I0(1et/t0)I0=ER2=122=6At0=LR=400×10320.2I2=6(1et/0.2)
potential drop across L=ER2l2=122×6(1ebt)=12 e5t

 

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