Physics-
General
Easy

Question

An alternating current of frequency 200 rad/sec and peak value 1A as shown in the figure, is applied to the primary of a transformer. If the coefficient of mutual induction between the primary and the secondary is 1.5 H, the voltage induced in the secondary will be

  1. 300 V    
  2. 191 V    
  3. 220 V    
  4. 471 V    

The correct answer is: 191 V


    e equals negative M fraction numerator d i over denominator d t end fraction equals negative 1.5 fraction numerator left parenthesis 1 minus 0 right parenthesis over denominator left parenthesis T divided by 4 right parenthesis end fraction equals negative fraction numerator 6 over denominator T end fraction, T equals fraction numerator 2 pi over denominator omega end fraction equals fraction numerator 2 pi over denominator 200 end fraction equals fraction numerator pi over denominator 100 end fraction
    Þvertical line e vertical line equals fraction numerator 600 over denominator pi end fraction equals 190.9 V stack – with tilde on top 191 V

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