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