SSC JE Electrical Engineering 2014 Set 1 Paper

Question : 125 of 200

Marks: +1, -0

The flux through each turn of a 100-turn coil is

(t^ 3 -2t)

mWh, where ‘t’ is in seconds. Find the magnitude of the induced emf at t = 2 s.

Solution:

Magnitude of the induced EMF =

{+Ndϕ} /{dt}

∴ Flux

( ϕ ) = t^3 -2

Em Wh,
N is Number of turn in coil
Now

{dϕ}/ {dt}=(3t^ 2 -2)

Now

({dϕ}/ {dt})_{t=25}=3×2^2-2

=12 -2 = 10

EMF = 100 × 10 × 10

^{–3 }

Volt = 1 Volt

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