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MHT CET202015 Oct 2020Evening ShiftPhysicsAlternating CurrentActual

Alternating current of peak value ( 2 ) A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is 1 H . The peak e.m.f. induced in secondary coil is (Frequency of a.c. =50 ~Hz )

Options

  1. A400 ~V
  2. B200 ~V
  3. C300 ~V
  4. D100 ~V

Correct answer

B. 200 ~V

Step-by-step solution

Given (: I ₀= 2 ) ampere (v=50 ~Hz ~L =1 H ) Thus (w=2 v=2 (50)=100 ) Alternating current flowing through the coil is given by (I=I₀ w t ) Differentiating it wr.t. time we get ( dI dt = I ₀ w w t ) ( . dI dt |_ = I ₀ ~W = 2 100 =200 ) ampere per second Peak e.m.f induced ( E = L dI dt ) ( E =1 200=200 ~V

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