JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 12)
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :
2.5 ms
25 ms
2.5 s
0.25 ms
Explanation
$$I = {I_0}\sin \omega t $$
$$ \Rightarrow $$ $${{{I_M}} \over {\sqrt 2 }} = {I_M}\sin \omega t$$
$$ \Rightarrow $$ $$\omega t = {\pi \over 4}$$
$$ \Rightarrow $$ $$t = {\pi \over {4\omega }}$$ $$ = {\pi \over {4\left( {2\pi f} \right)}}$$
$$ \Rightarrow $$ t = $${1 \over {8 \times 30}} = {1 \over {400}}$$ = 2.5 ms
$$ \Rightarrow $$ $${{{I_M}} \over {\sqrt 2 }} = {I_M}\sin \omega t$$
$$ \Rightarrow $$ $$\omega t = {\pi \over 4}$$
$$ \Rightarrow $$ $$t = {\pi \over {4\omega }}$$ $$ = {\pi \over {4\left( {2\pi f} \right)}}$$
$$ \Rightarrow $$ t = $${1 \over {8 \times 30}} = {1 \over {400}}$$ = 2.5 ms
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