A resistance of 20 Ω is connected to a source of alternating potential V = 220 sin (100π t). The time taken by current to change from its peak value to r.m.s. value is (sin π/4 = 1/√(2)) (a) 0.2 s (b) 0.25 s (c) 25 × 10⁻³ s (d) 2.5 × 10⁻³ s

Answer: (d). ω = 100π ⇒ T = 1/50 s = 20 ms. From peak ( θ = π/2) to rms ( sin θ = 1/√(2), θ = 3π/4): Δθ = π/4, i.e. T/8. t = 20/8 ms = 2.5 × 10⁻³ s So the answer is (d) 2.5 × 10⁻³ s.

A resistance of 20 Ω is connected to a source of alternating… | MHT CET 2026 (14 May, Shift 1)

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AC Basics: Peak, RMS & Average Values

A resistance of $20\ \Omega$ is connected to a source of alternating potential $V = 220 \sin(100\pi t)$ . The time taken by current to change from its peak value to r.m.s. value is $\left(\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\right)$

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Answer: (d)

$\omega = 100\pi \Rightarrow T = \frac{1}{50}$ s $= 20$ ms.
From peak ( $\theta = \frac{\pi}{2}$ ) to rms ( $\sin\theta = \frac{1}{\sqrt{2}}$ , $\theta = \frac{3\pi}{4}$ ): $\Delta\theta = \frac{\pi}{4}$ , i.e. $\frac{T}{8}$ .
$t = \frac{20}{8}$ ms $= 2.5 \times 10^{-3}$ s
So the answer is (d) $2.5 \times 10^{-3}$ s.

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