Q145 / 180 · NEET 2026 (Cancelled) · Physics
The peak value of an alternating current is $5$ A and frequency is $60$ Hz. How long will the current, starting from zero, take to reach the peak value ?
- A$\dfrac{1}{120}$ s
- B$\dfrac{1}{60}$ s
- C$\dfrac{1}{30}$ s
- Correct answerD$\dfrac{1}{240}$ s
Solution
The alternating current is given by $I = I_0 \sin(\omega t)$.
The time taken to reach the peak value from zero corresponds to a phase angle of $\dfrac{\pi}{2}$.
$\omega t = \dfrac{\pi}{2}$
$2 \pi f t = \dfrac{\pi}{2}$
$t = \dfrac{1}{4f}$
Given $f = 60$ Hz, we get:
$t = \dfrac{1}{4 \times 60} = \dfrac{1}{240}$ s.
Answer: $\dfrac{1}{240}$ s
The time taken to reach the peak value from zero corresponds to a phase angle of $\dfrac{\pi}{2}$.
$\omega t = \dfrac{\pi}{2}$
$2 \pi f t = \dfrac{\pi}{2}$
$t = \dfrac{1}{4f}$
Given $f = 60$ Hz, we get:
$t = \dfrac{1}{4 \times 60} = \dfrac{1}{240}$ s.
Answer: $\dfrac{1}{240}$ s
Hence, the correct answer is option: 'D'.