JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 6)

A 10 $$\Omega$$ resistance is connected across 220V $$-$$ 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :
2.5 ms
1.5 ms
3.0 ms
4.5 ms

Explanation



$$\Rightarrow$$ i = i0sin$$\omega$$t

when i = i0

i0 = i0sin$$\omega$$t1 $$\Rightarrow$$ $$\omega$$t1 = $${\pi \over 2}$$ ..... (i)

When i = $${{{i_1}} \over {\sqrt 2 }}$$

$${{{i_1}} \over {\sqrt 2 }}$$ = i0sin$$\omega$$t2 $$\Rightarrow$$ $$\omega$$t2 = $${\pi \over 4}$$ ...... (ii)

Time taken by current from maximum value to rms value

$$ \Rightarrow ({t_1} - {t_2}) = {\pi \over {2\omega }} - {\pi \over {4\omega }} = {\pi \over {4\omega }} = {\pi \over {4 \times 2\pi f}}$$

$$ = {1 \over {8 \times 50}}$$

$${1 \over {400}}$$ sec

= 2.5 ms

Comments (0)

Advertisement