Question

Alternating current of peak value $\left(\frac{2}{\pi }\right)$ 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 $AC=50Hz$)

• A. $100V$
• B. $200V$
• C. $300V$
• D. $400V$

Answer 1

Answer: (B)

Let $I_p=\frac{2}{\pi }sin\omega tA$
Emf induced in the secondary coil
$\left|e\right|=M\frac{d{I}_p}{dt}=M\frac{d}{dt}\left(\frac{2}{\pi }sin\omega t\right)=\frac{2M\omega }{\pi }cos\omega t$
$\therefore$ Peak emf induced in secondary coil is
${\left|e\right|}_{max}=\frac{2M\omega }{\pi }$
$=\frac{2\times 1\times 2\pi \times 50}{\pi }$
$=200V$