Q185 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
If the r.m.s. current in an a.c. circuit is $5$ A, having frequency $50$ Hz. The value of current $\dfrac{1}{300}$ s after its value becomes zero (in A) is
$\{\sin(\pi/3) = (3)^{0.5}/2\}$
$\{\sin(\pi/3) = (3)^{0.5}/2\}$
- A$5\sqrt{2}$
- Correct answerB$5\sqrt{\dfrac{3}{2}}$
- C$\dfrac{5}{6}$
- D$\dfrac{5}{\sqrt{2}}$
Solution
The equation for alternating current is given by $I = I_0 \sin(\omega t)$.
Given $I_{rms} = 5$ A, the peak current is:
$I_0 = I_{rms}\sqrt{2} = 5\sqrt{2}$ A
The angular frequency is:
$\omega = 2\pi f = 2\pi \times 50 = 100\pi$ rad/s
Substituting $t = \dfrac{1}{300}$ s into the current equation:
$I = 5\sqrt{2} \sin\left(100\pi \times \dfrac{1}{300}\right)$
$I = 5\sqrt{2} \sin\left(\dfrac{\pi}{3}\right)$
$I = 5\sqrt{2} \times \dfrac{\sqrt{3}}{2}$
$I = 5\sqrt{\dfrac{3}{2}}$ A
Answer: $5\sqrt{\dfrac{3}{2}}$
Given $I_{rms} = 5$ A, the peak current is:
$I_0 = I_{rms}\sqrt{2} = 5\sqrt{2}$ A
The angular frequency is:
$\omega = 2\pi f = 2\pi \times 50 = 100\pi$ rad/s
Substituting $t = \dfrac{1}{300}$ s into the current equation:
$I = 5\sqrt{2} \sin\left(100\pi \times \dfrac{1}{300}\right)$
$I = 5\sqrt{2} \sin\left(\dfrac{\pi}{3}\right)$
$I = 5\sqrt{2} \times \dfrac{\sqrt{3}}{2}$
$I = 5\sqrt{\dfrac{3}{2}}$ A
Answer: $5\sqrt{\dfrac{3}{2}}$
Hence, the correct answer is option: 'B'.