# The current through r and l does not instantaneously

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). The current through R and L does not instantaneously go to zero but decays smoothly. Apply Kirchhoff’s loop equation: E iR L di dt = 0 (0) iR L di dt = 0 di i = R L dt di ' i ' I 0 i = R L dt ' 0 t ln i I 0 = R L t i = I 0 e ( R / L ) t Time constant τ = L R is the time for current to decrease to 1/ e of its original value Subscribe to view the full document.

31 Energy considerations: E = iR + L di dt (0) = iR + L di dt ⇓ × i 0 = i 2 R + Li di dt Li di dt < 0 ! = i 2 R (Energy stored in the inductor decreases at a rate equal to the rate of dissipation of energy in resistor) 7 32 Subscribe to view the full document.

33 30.5 The L-C Circuit i i i i 34 Because of induced emf in inductor, current cannot change instantaneously. Although capacitor is completely discharged, current cannot change instantaneously. As current decreases, magnetic field also decreases, inducing emf in inductor in the same direction as current, slowing down decrease of current. Subscribe to view the full document.

35 Apply Kirchhoff’s loop rule: Analogy: L di dt q C = 0 d 2 q dt 2 + 1 LC q = 0 i = dq dt , di dt = d 2 q dt 2 cf . d 2 x dt 2 + k m x = 0 SHM q x , i v x , L m , 1 C k Electrical Oscillation in an L-C Circuit 36 Charge and Current: x = A cos( ω t + φ ) ! q = Q cos( ω t + φ ) i = dq dt = ω Q sin( ω t + φ ) 1 2 mv x 2 + 1 2 kx 2 = 1 2 kA 2 ! 1 2 Li 2 + q 2 2 C = Q 2 2 C Energy: Magnetic-field energy Electric-field energy Total energy Angular frequency: ω = k m ω = 1/ C L q = Q cos( ω t + φ ) i = dq dt = ω Q sin( ω t + φ ) If t = 0, q = Q and i = 0, then φ = 0 If t = 0, q = 0, then φ = ± π 2 Oscillating with time Subscribe to view the full document.

37 8 38 Subscribe to view the full document.

9 39 40 First close the switch S (upward) to fully charge the capacitor. Then, at time t = 0, flip the switch S downward. Note: initial current < 0. Apply Kirchhoff’s loop rule: iR L di dt q C = 0 d 2 q dt 2 + R L dq dt + 1 LC q = 0 30.6 The L-R-C Circuit Analyzing an L-R-C Circuit Subscribe to view the full document.

Solutions of Differential Equation: 41 a d 2 y dx 2 + b dy dx + cy = 0 b 2 4 ac < 0 : y = e α x ( A cos β x + B sin β x ) b 2 4 ac = 0 : y = ( A + Bx ) e α x b 2 4 ac > 0 : y = Ae α 1 x + Be α 2 x cf . d 2 q dt 2 + R L dq dt + 1 LC q = 0 a = 1, b = R L , c = 1 LC b 2 4 ac = R L 2 4(1) 1 LC = 1 L 2 R 2 4 L C Optional: 42 d 2 q dt 2 + R L dq dt + 1 Subscribe to view the full document. • Fall '09
• Magnetic Field, Inductor, DI

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