1.

If the current flowing through a coil is reduced by 50%, then what is the energy in the coil?

Answer»

SOLUTION :`E_2/E_1 = 1/2L(I/2)^2//1/2LI^2 = 1/4`
`THEREFORE E_2 = 1/4E_1`
`therefore E_1-E_2= 3/4E_1`


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