JEE MAIN - Physics (2007 - No. 9)
An ideal coil of $$10H$$ is connected in series with a resistance of $$5\Omega $$ and a battery of $$5V$$. $$2$$ second after the connection is made, the current flowing in ampere in the circuit is
$$\left( {1 - {e^{ - 1}}} \right)$$
$$\left( {1 - e} \right)$$
$$e$$
$${{e^{ - 1}}}$$
Explanation
KEY CONCEPT : $$I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
(When current is in growth in $$LR$$ circuit)
$$ = {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
$$ = {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)$$
$$ = \left( {1 - {e^{ - 1}}} \right)$$
(When current is in growth in $$LR$$ circuit)
$$ = {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
$$ = {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)$$
$$ = \left( {1 - {e^{ - 1}}} \right)$$
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