A resistance of $20 \Omega$ is connected to a source of an alternating potential $\mathrm{V}=200 \sin (10…

A resistance of $20 \Omega$ is connected to a source of an alternating potential $\mathrm{V}=200 \sin (10 \pi \mathrm{t})$. If $t$ is the time taken by the current to change from the peak value to rms value, then ' $t$ ' is (in seconds).
  1. $25 \times 10^{-1}$
  2. $2.5 \times 10^{-4}$
  3. $25 \times 10^{-2}$
  4. $2.5 \times 10^{-2}$

Solution

$\mathrm{R}=20 \Omega, \mathrm{~V}=200 \sin (10 \pi \mathrm{t})$ $\therefore \quad I=\frac{V}{R}=10 \sin (10 \pi t)$ ....(i) $\therefore \mathrm{I}_{\mathrm{o}}=10, \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{o}}}{\sqrt{2}}=\frac{10}{\sqrt{2}}$ $\therefore$ From equation (i), $10=10 \sin \left(10 \pi t_1\right) \Rightarrow \sin \left(10 \pi t_1\right)=\sin \frac{\pi}{2} \quad \therefore t_1=\frac{1}{20} s$ Again, $\frac{10}{\sqrt{2}}=10 \sin \left(10 \pi \mathrm{t}_2\right) \Rightarrow \sin \left(10 \pi \mathrm{t}_2\right)=\sin \frac{\pi}{4}$ $\therefore \mathrm{t}_2=\frac{1}{40} \mathrm{~s}$ $\therefore \quad \Delta t=t_1-t_2=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}=2.5 \times 10^{-2} \mathrm{~s}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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