SSC JE Electrical Engineering 2014 Set 2 Paper

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Question : 136 of 200

Marks: +1, -0

The voltage across 5 – H inductor is

V( t)=\{\table30t^2{,},t>0;0{,},t<0

Find the energy stored at t = 5 s.
Assume zero initial current.

312.5 kJ
0.625 kJ
3.125 kJ
156.25 kJ

Solution:

Stored Energy

(E) = 1 /2 LI^ 2

Now

e(t) = L{di(t) }/ {dt}

∴

i(t) = 1 /L∫e(t)dt

=1/5∫↖{5}↙{0} 30t^2 dt

= 6 ∫↖{5}↙{0}t ^2dt

= 6 /3[t^3]^{5} _{0}= 2 × (5)^3 = 250 A

∴

E = 1 /2 LI ^2 = 1 /2 ×5× (250) ^2

= 156.25 kJ

∴ Option (d) is correct.

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