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Prove that E= half LI•I

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Power absorbed or dissipated by an inductor P = dE/dt...'(1) If 'E' is the energy and 't' is time. Voltage drop across inductor = L * dI/dt but P = VI So, dE/dt = L*(dI/dt) *I dE = LIdI Integrating from 0 to E on LHS and 0 to I on RHS E = (1/2) * L *(I^2)
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Headstart to assured 95+ score in math

Please refer NCERT text.
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mc2
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This is an expression indicating the magneto static potential energy stored in an inductor.we can obtain it in the same way as of getting electrostatic potential energy stored in capacitor
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Power absorbed or dissipated by an inductor P = dE/dt and Voltage drop across inductor = L * dI/dt Also, P = VI So, dE/dt = L*(dI/dt) *I So, dE = LIdI Integrating both sides E = (1/2) * L *(I^2)
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Please refer NCERT text.
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