Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of…
Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of transformer. The coefficient of mutual inductance between primary and secondary coil is $1 \mathrm{H}$. The peak e.m.f. induced in secondary coil is (Frequency of a.c. $=50 \mathrm{~Hz}$ )
$400 \mathrm{~V}$
$200 \mathrm{~V}$
$300 \mathrm{~V}$
$100 \mathrm{~V}$
Solution
Given \(: \mathrm{I}_{0}=\frac{2}{\pi}\) ampere \(v=50 \mathrm{~Hz} \mathrm{~L}=1 \mathrm{H}\)
Thus \(w=2 \pi v=2 \pi(50)=100 \pi\)
Alternating current flowing through the coil is given by \(I=I_{0} \sin w t\)
Differentiating it wr.t. time we get \(\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{I}_{0} \mathrm{w} \cos w \mathrm{t}\) \(\left.\therefore \frac{\mathrm{dI}}{\mathrm{dt}}\right|_{\max }=\mathrm{I}_{0} \mathrm{~W}=\frac{2}{\pi} \times 100 \pi=200\) ampere per second
Peak e.m.f induced \(\mathcal{E}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\)
\(\Longrightarrow \mathcal{E}=1 \times 200=200 \mathrm{~V}\)