lecture24 - Self-Inductance and Circuits Self-Inductance I...

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Self-Inductance and Circuits Self-Inductance and Circuits Inductors in circuits RL circuits
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Power (VI) absorbed by inductor is: 2 2 1 LI U L = 2 1 2 f 0 I f L U L I dI LI = = L dU LIdI = The total potential energy stored is: L dU dI LI dt dt = Find the potential energy of an inductor with L=400mH and a final current of 10A.
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Self-Inductance Self-Inductance dt dI L L - = ε I 2 2 1 LI U L = Potential energy stored in an inductor: Self-induced emf:
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RL circuits: current increasing RL circuits: current increasing The switch is closed at t =0; Find I (t). - = - = = - - I R L R L IR dt dI IR dt dI L ε 0 ε L R I Kirchoff’s loop rule:
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Solution Solution R L = τ Time Constant: Note that H/ = seconds Ω (show as exercise!) ( 29 ε / 1 ) ( t e R t I - - =
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0 1 τ 2 τ 3 τ 4 τ 63% ε /R I t Time Constant: Current Equilibrium Value: R L = τ ( 29 ε / 1 ) ( t e R t I - - = R I =
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Example Calculate the inductance in an RL circuit in which R=0.5 and the current increases to one fourth of its final value in 1.5 sec.
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lecture24 - Self-Inductance and Circuits Self-Inductance I...

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