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An inductor of inductance L equals 400 m H and resistors of resistances R subscript 1 end subscript equals 2 capital omega and R subscript 2 end subscript equals 2 capital omega are connected to a battery of e m f blank 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t equals 0. The potential drop across L as a function of time is

  1. 6 e to the power of negative 5 t end exponent V  
  2. fraction numerator 12 over denominator t end fraction e to the power of negative 3 t end exponent V  
  3. 6 open parentheses 1 minus e to the power of fraction numerator negative t over denominator 0.2 end fraction end exponent close parentheses V  
  4. 12 e to the power of negative 5 t end exponent V  

The correct answer is: 12 e to the power of negative 5 t end exponent V


    E(across B C)equals L fraction numerator d I subscript 2 end subscript over denominator d t end fraction plus R subscript 2 end subscript I subscript 2 end subscript

    I subscript 2 end subscript equals I subscript 0 end subscript left parenthesis 1 minus e to the power of negative t divided by t subscript 0 end subscript end exponent right parenthesis
    I subscript 0 end subscript equals fraction numerator E over denominator R subscript 2 end subscript end fraction equals fraction numerator 12 over denominator 2 end fraction equals 6 blank A
    tau equals t subscript 0 end subscript equals fraction numerator L over denominator R end fraction equals fraction numerator 400 cross times 10 to the power of negative 4 end exponent over denominator 2 capital omega end fraction equals 0.2 blank S
    therefore I subscript 2 end subscript equals 6 left parenthesis 1 minus e to the power of negative t divided by 0.2 end exponent right parenthesis
    Potential drop areas L equals E minus R subscript 2 end subscript I subscript 2 end subscript
    equals 12 minus 2 cross times 6 left parenthesis 1 minus e to the power of negative t divided by 0.2 end exponent right parenthesis
    equals 12 e to the power of negative t divided by 0.2 end exponent equals 12 e to the power of negative 5 t end exponent V

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