JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 17)
An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 $$\Omega $$
resistor. The ratio of the currents at time t = $$\infty $$ and at t = 40 s is close to : (Take e2 = 7.389)
1.06
0.84
1.15
1.46
Explanation
i = i0(1 - $${e^{ - {{Rt} \over L}}}$$)
i$$\infty $$ = i0(1 - $${e^{ - \infty }}$$) = i0
$$ \therefore $$ $${{{i_\infty }} \over {{i_{40s}}}}$$ = $${{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}$$
= $${1 \over {\left( {1 - {e^{ - 2000}}} \right)}}$$ $$ \approx $$ 1
Then most appropriate option is 1.06
i$$\infty $$ = i0(1 - $${e^{ - \infty }}$$) = i0
$$ \therefore $$ $${{{i_\infty }} \over {{i_{40s}}}}$$ = $${{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}$$
= $${1 \over {\left( {1 - {e^{ - 2000}}} \right)}}$$ $$ \approx $$ 1
Then most appropriate option is 1.06
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