Alternating current of peak value 2/(π) ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary…

Question

Alternating current of peak value 2/(π) ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

MARKS Content Team · Accepted Answer

Correct answer: (B) 200 V. Peak value of current I_0=I_rms×√2=2/(π) Amp Co-efficient of mutual inductance is M = 1 henry Induced emf in secondary is given by e_2=M⋅di/dt where i=i_0 .sin⁡ωt+ϕ Here ω=2πn=100 π. ∴ e_2=1×d/dt(i_0 .sin⁡ ωt) i_0 .ω .cos⁡ ωt=2/(π)×2π×50 .cos⁡ (100 πt) For, t=0 , we have e_2=4×50=200V .

Alternating current of peak value $(\frac{2}{π})$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

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Relevant Questions

Alternating current of peak value 2π ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)

Solution

Relevant Questions

Alternating current of peak value $(\frac{2}{π})$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak emf induced in secondary coil is (Frequency of AC = 50 Hz)