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 a.c. = 50 Hz)

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

Answer 1

The correct option is B 200 V

Given : $I_o=\frac{2}{\pi }$ ampere $\nu =50$ Hz $L=1H$

Thus $w=2\pi \nu =2\pi \left(50\right)=100\pi$

Alternating current flowing through the coil is given by $I=I_o\sin wt$

Differentiating it wr.t. time we get $\frac{dI}{dt}=I_{o}w\cos wt$

$\therefore$ $\frac{dI}{dt}{|}_{max}=I_{o}w=\frac{2}{\pi }\times 100\pi =200$ ampere per second

Peak e.m.f induced $E=L\frac{dI}{dt}$

$⟹$ $E=1\times 200=200$ V