JEE MAIN - Physics (2005 - No. 16)
A coil of inductance $$300$$ $$mH$$ and resistance $$2\,\Omega $$ is connected to a source of voltage $$2$$ $$V$$. The current reaches half of its steady state value in
$$0.1$$ $$s$$
$$0.05$$ $$s$$
$$0.3$$ $$s$$
$$0.15$$ $$s$$
Explanation
KEY CONCEPT : The charging of inductance given
by, $$i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)$$
$${{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}$$
Taking log on both the sides,
$$ - {{Rt} \over L} = \log 1 - \log 2$$
$$ \Rightarrow t = {L \over R}\log 2 = {{300 \times {{10}^{ - 3}}} \over 2} \times 0.69$$
$$ \Rightarrow t - 0.1\,\sec .$$
by, $$i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)$$
$${{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}$$
Taking log on both the sides,
$$ - {{Rt} \over L} = \log 1 - \log 2$$
$$ \Rightarrow t = {L \over R}\log 2 = {{300 \times {{10}^{ - 3}}} \over 2} \times 0.69$$
$$ \Rightarrow t - 0.1\,\sec .$$
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