Q102 / 150 · MHT CET 2020 (19 Oct Shift 1) · Physics
An alternating current of frequency $50 \mathrm{~Hz}$ has the peak value as $14 \cdot 14 \mathrm{~A}$. The time
taken by the alternating current in reaching from zero to maximum value and r.m.s. value of current will be respectively
- A$0.025 \mathrm{~s}, 5 \mathrm{~A}$
- B$0 \cdot 005 \mathrm{~s}, 5 \mathrm{~A}$
- Correct answerC$0 \cdot 005 \mathrm{~s}, 10 \mathrm{~A}$
- D$0 \cdot 025 \mathrm{~s}, 10 \mathrm{~A}$
Solution
(D)
$\mathrm{f}=50 \mathrm{~Hz}, \mathrm{~T}=\frac{1}{50} \mathrm{~s}=0.02 \mathrm{~s}$
Time taken to reach from zero to maximum value is
$\begin{array}{l}
t=\frac{T}{4}=\frac{0.02}{4}=0.005 \mathrm{~s} \\
I_{m s}=\frac{I_{0}}{\sqrt{2}}=\frac{14.14}{1.414}=10 \mathrm{~A}
\end{array}$
Hence, the correct answer is option: 'C'.