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A 20 H inductor is placed in series with...

A 20 H inductor is placed in series with 10 W resistor and an emf of 100 V is suddenly applied to the combination. At t = 1 s from t = 1 s from the start, find the rate at which energy is being stored in the magnetic field around the inductor (givne `e^(-0.5)=0.61`).

Text Solution

Verified by Experts

The rate at which energy is supplied to the inductor is
(dU^(L))/(dt)=+ Li(di0/(dt)Rightarrow(di)/(dt)=+epsilon/Le^(-Rt/L)`
Therefore `P_(L)=(dU_(L)/(dt)=iepsilone^(-t/zeta)`
We now substitute for i to obtain `P_(L)=(zeta^(2))/(R)[e^(-t/zeta]`
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