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 $AC = 50\, Hz$)

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

Answer 1

Answer: (B)

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