An inductor of inductance L = 400 mH and resistors of resistances R1 = 2 Ω and R2 = 2 Ω is 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

6^e–5t V

$12 over straight t straight e to the power of negative 3 straight t end exponent$
$6 space open parentheses 1 minus straight e to the power of fraction numerator negative straight t over denominator 0.2 end fraction end exponent close parentheses space straight V$

12^e–5t V

Solution

12^e–5t V

straight e subscript straight L space equals space Ee to the power of negative tR subscript 2 over straight L end exponent straight e subscript straight L space equals space 12 straight e to the power of negative 2 over 4 straight t end exponent straight e subscript straight L space equals space 12 straight e to the power of negative 5 straight t end exponent

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