Physics-
General
Easy

Question

Energy required to establish a current of 4 A in a coil of self-inductance L =200mH is

  1. 0.16 J  
  2. 0.18 J  
  3. 0.40 J  
  4. 1.6 J  

The correct answer is: 1.6 J


    Energy stored in a self-inductor, E equals fraction numerator 1 over denominator 2 end fraction L i to the power of 2 end exponent
    equals fraction numerator 1 over denominator 2 end fraction cross times 200 cross times 10 to the power of negative 3 end exponent open square brackets 4 close square brackets to the power of 2 end exponent equals 1.6 blank J

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