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)
Solution
i = i<sub>0</sub>(1 - ${e^{ - {{Rt} \over L}}}$)
<br><br>i<sub>$\infty$</sub> = i<sub>0</sub>(1 - ${e^{ - \infty }}$) = i<sub>0</sub>
<br><br>$\therefore$ ${{{i_\infty }} \over {{i_{40s}}}}$ = $${{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}$$
<br><br> = ${1 \over {\left( {1 - {e^{ - 2000}}} \right)}}$ $\approx$ 1
<br><br>Then most appropriate option is 1.06
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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