An alternating voltage v(t) = 220 sin 100 π t volt is applied to a purely resistance load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is: (a) 5 ms (b) 2.2 ms (c) 3.3 ms (d) 7.2 ms

Answer: (c). i = i₀sin (100π t). Half peak: 100π t₁ = π/6 ⇒ t₁ = 1/600 s. Peak: 100π t₂ = π/2 ⇒ t₂ = 1/200 s. Δ t = 1/200 - 1/600 = 1/300 s ≈ 3.3 ms So the answer is (c) 3.3 ms.

An alternating voltage v(t) = 220 sin 100 π t volt is applied to a… | JEE Main 2019 (8 Apr, Shift 1)

Q40JEE Main 2019 · 8 APR · SHIFT 1Medium

AC Basics: Peak, RMS & Average Values

An alternating voltage v(t) = 220 sin 100 $\pi$ t volt is applied to a purely resistance load of 50 $\Omega$ . The time taken for the current to rise from half of the peak value to the peak value is:

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Answer: (c)

$i = i_0\sin(100\pi t)$ . Half peak: $100\pi t_1 = \frac{\pi}{6} \Rightarrow t_1 = \frac{1}{600}$ s.
Peak: $100\pi t_2 = \frac{\pi}{2} \Rightarrow t_2 = \frac{1}{200}$ s.
$\Delta t = \frac{1}{200} - \frac{1}{600} = \frac{1}{300}$ s $\approx 3.3$ ms
So the answer is (c) 3.3 ms.

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