A current increases at a uniform rate from 1A to 4A in a period of 0.5ms. If this current is applied to an inductor of 1.5H, determine the voltage induced.
Now the induced voltage will be given by:
e = - L x ( rate of change of current ) = -L x ( change in current / time ) = -1.5 x ( ( 4 - 1 ) / 0.0005 ) = -9000V
Example 2:
A current of 1.5A flows in an inductor of 2H.
Determine the flux present if there are 600 turns on the inductor.
Re-arranging NΦ = L I to make Φ the subject gives:
Φ = L I / N = 2 x 1.5/600 = 5 mWb
Example 3:
Determine the energy stored in an inductor of 60mH when a current of 2A is applied to it.
Using W = ½ C V2 gives:
Using E = 0.5 L I2 gives:
E = 0.5 x 60 x 10-3 x (22) = 30 x 10-3 x 4 = 120 mJ
