SSC JE Electrical Engineering 2013 Paper

Question : 150 of 200

Marks: +1, -0

A solenoid of inductance 250 mH and resistance 10 Ω is connected to a battery. The time taken for the magnetic energy to reach14th of its maximum value is

$\log_e (2)$
$10^{–3} \log_e (2)$
$25 \log_e (2)$
$1/40 \log_e (2)$

Validate

Solution:

let at time t, magnetic energy reaches to

1/4

th of its maximum

1/4=e^{-tR\/L}

\ln1/4=-{tR}/L

{tR}/L=\ln 4

t=L/R\ln4={2L}/R\ln e ^2

={2×250×10^{-3}}/10\ln^2

=1/40\log_e2

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