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5A 50 mH inductor is in series with a 10 2 resistor and a battery with an emf oj 25 V. At t = 0 the switch is closed. Find Q(c) the rate at which energy is stored in the inductor, |
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Answer» SOLUTION :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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