Lecture 10_v2 - A RC A V RC t A k RC t t V k RC t V k RC dt V dV RC dt V dV RC V dt dV dt dV C I I R V C R = =-= = =-=-=-=-= ⇔-=-=-= β‡’-=-= = βˆ

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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 1 Lecture 10 Inductors RC circuits: natural response
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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 2 Governing equations: Inductor ( 29 ( 29 0 1 0 I d V L I dt dI L V dt dI dt dB V t + = = τ + V - I current b magnetic field change in field b voltage Define Inductance: Henrys: H=s V(t) I(t) = = = = t LI d d dI LI W dt dI LI VI P 0 2 2
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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 3 Equivalents ( 29 ( 29 ( 29 2 1 2 2 2 1 2 2 2 1 2 1 2 1 L L L V V dt dI L L dt dI L V V dt dI L L dt dI L dt dI L V V V + = + = + = + = + = dt dI L L L dt dI dt dI L L L L dt dI L dt dI L L L L V L L L L V L V L V dt dI + = + = + = + = + = 2 1 1 2 2 1 1 2 2 2 2 1 1 2 2 1 1 2 2 1 V(t) L 1 L 2 I I 2 parallel Series L 1 L 2 I(t) Current-divider V 2 + 2 1 1 2 L L L L Voltage-divider + V - L 1 +L 2
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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 4 Parallel RC circuit ( 29 ( 29 ( 29 ( 29 ( 29 0 , exp 0 exp 0 exp exp ln 0 0 0 V V RC t V t V V
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Unformatted text preview: A RC A V RC t A k RC t t V k RC t V k RC dt V dV RC dt V dV RC V dt dV dt dV C I I R V C R = = -= = = -= -= +-= +-= ⇔ +-=-=-= β‡’-=-= = ∫ ∫ Ο„ + V-C R By KCL, governing Eq, get 1 st order Differential equation: Solution is decaying exponential set by β€œRC time constant” and initial condition Solve by sep. variables, integrating V V t 0 Ο„ V exp(-t/ Ο„ ) ( 29 ( 29 ( 29 -= = t I R t V t I exp...
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This note was uploaded on 06/10/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell University (Engineering School).

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Lecture 10_v2 - A RC A V RC t A k RC t t V k RC t V k RC dt V dV RC dt V dV RC V dt dV dt dV C I I R V C R = =-= = =-=-=-=-= ⇔-=-=-= β‡’-=-= = βˆ

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