Q144 / 150 · MHT CET 2024 (15 May Shift 1) · Physics
Alternating current of peak value $\left(\frac{2}{\pi}\right)$ A flows through the primary coil of a transformer. The coefficient of mutual inductance between primary and secondary coils is 1 H . The peak e.m.f. induced in secondary coil (Frequency of a.c. $=50 \mathrm{~Hz}$ )
- A50 V
- B150 V
- C100 V
- Correct answerD200 V
Solution
$\begin{aligned}
& \omega=2 \pi v=2 \pi(50)=100 \pi \\
& I=I_0 \sin \omega t \\
& \frac{d I}{d t}=I_0 \omega \cos \omega t
\end{aligned}$
Maximum value of $\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{I}_0 \omega$ $\begin{gathered} =\frac{2}{\pi}(100 \pi)=200 \mathrm{~A} / \mathrm{S} \\ \mathrm{E}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}=1 \times 200=200 \mathrm{~V} \end{gathered}$
Maximum value of $\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{I}_0 \omega$ $\begin{gathered} =\frac{2}{\pi}(100 \pi)=200 \mathrm{~A} / \mathrm{S} \\ \mathrm{E}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}=1 \times 200=200 \mathrm{~V} \end{gathered}$
Hence, the correct answer is option: 'D'.