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Alternating current of peak value `((2)/(pi))` ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is (Frequency of AC= 50 Hz)A. 100VB. 200VC. 300VD. 400V |
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Answer» Correct Answer - B According to question, peak value of current `l_0 =sqrt2xxl_(rms)=2/piA` Coefficient of mutual inductance 1H As we know, Induced emf in secondary coil is given by `E_s=M.(dl)/dt` [`Where,l=l_0sinomegat`] `E_s=Momegal_0cos(omegat)` =`1xx2pixx50xx2/picos(2pixx50xxt)` `because omega=2pin` For t= 0 we have `E_s 4xx50 =200V` |
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