A resistance of 20 $Ω$ is connected to a source of an alternating potential V = 220 sin (100 $π$ t). The time taken by the current to change from the peak value to rms value, is

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0.2 s

0.25 s

2.5 × $10^{- 3}$ s

4.5 × $10^{- 3}$ s

answer is C.

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Detailed Solution

$\begin{array}{l} i = \frac{V}{R} = \frac{220 S i n 100 π t}{20} = 11 S i n 100 π t \\ \frac{1}{\sqrt{2}} \times 11 = 11 S i n 100 π t \Rightarrow \frac{π}{4} = 100 π t \Rightarrow t = 2.5 \times 10^{- 3} \sec \end{array}$

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