Two coils of self inductance `L_(1)` and `L_(2)` are connected by in parllel and the then to cell of emf `epsilon` and of resistance R throught a Key . Find the instantaneous current throught L after the key is closed .

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Correct Answer - `I_(1)=(epsiL_(2)0)/(R(L_(1)+L_(2))) I =(epsiL_(2)0)/(R(L_(1)+L_(2)))`

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