SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Question : 198 of 200

Marks: +1, -0

Two coupled coils with

\L_1 = 0.5 \H

and

\L_2 = 4.0 \H

have a co-efficient of coupling 0.8. Find maximum value of the inductance EMF in the coil 2 if a current of

i _{1}=20 \sin 314 t\A

is passed in coil 1.

Solution:

{ \L }_{1}=0.5 \H

\L _{2}=4 \H

\K =0.8

\I =20 \sin 314 t

The induced emf in coil 2 due to current in coil 1 is given by

\V _{ 2 }= \M ( \di _{ 1 } / \dt )

\M = k √{{\L}_{1} {\L}_{2}}

\M =0.8 ×√{0.5 ×4}

\M = 1 . 1 3 H

\V _{2}=1.13 \d / \dt (20 \sin 314 t )

\V _{2}=1.13 ×20 ×314 ×\cos 314 t

The maximum value of

\V_2

= 1.13 × 20 × 314 = 7.1 kV

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