Ch04 Networks with L and C 1s09

Ch04 Networks with L and C 1s09 - Chapter 4 Inductors,...

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Inductors, Capacitors Inductors, Capacitors RLC Circuits RLC Circuits Chapter 4 Department of Electrical and Electronics Engineering University of the Philippines - Diliman Department of Electrical and Electronics Engineering 2 Outline Fundamental Capacitor Characteristics Fundamental Inductor Characteristics L and C Combinations RC Op-Amp Circuits Mutually Coupled Circuits Equilibrium Equations for RLC Circuits Department of Electrical and Electronics Engineering 3 Capacitor The capacitor is a circuit element that consists of two conducting surfaces separated by a non- conducting ( dielectric ) material. It is an important element as it has the ability to store energy in its electric field. + v - C i Department of Electrical and Electronics Engineering 4 Fundamental Characteristics The charge on the capacitor is q = Cv Since the current is dt dq i = Then for a capacitor dt ) Cv ( d i = or dt dv C i = + v - C i Re-arranging the equation and integrating dt i C dv 1 = - = t dt i C ) t ( v 1
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Department of Electrical and Electronics Engineering 5 - + = 0 0 1 1 t dt ) t ( i C dt ) t ( i C ) t ( v This can be expressed as two integrals + = t dt ) t ( i C ) t ( v ) t ( v 0 0 1 Power is given by dt ) t ( dv C ) t ( v ) t ( i ) t ( v ) t ( p = = Hence the energy stored in the electric field is - = t C dt dt dv ) t ( Cv ) t ( w -∞ = ) t ( v ) ( v dv ) t ( v C ) t ( v ) ( v ) t ( Cv ) t ( w C -∞ = 2 2 1 J ) t ( Cv ) t ( w C 2 2 1 = Department of Electrical and Electronics Engineering 6 5 10 15 5 10 20 15 25 30 35 t v(t) + v - 2mF i 5 , 0< t < 5 s 2t-5 , 5< t < 10 s -0.5t+20, 10< t < 30 s 5 , t > 35 s v(t) = Example: Find the current i C (t) given At t < 5 sec: i C (t) = 2mF (0) = 0 A 5 < t < 10 sec: i C (t) = 2mF (2) = 4 mA 10 < t < 30 sec: i C (t) = 2mF (-0.5) = -1 mA t > 30 sec: i C (t) = 2mF (0) = 0 A dt dv C i = Department of Electrical and Electronics Engineering 7 5 10 15 5 10 20 15 25 30 35 t v(t) 0 -1 4 5 10 30 35 t i(t) mA Capacitor is charging (absorbing power) Capacitor is discharging (delivering power) Plot of the capacitor voltage and current: Department of Electrical and Electronics Engineering 8 Example: Compute the energy stored in a 4- μ F capacitor at time t=3ms given the current i(t). The capacitor is initially uncharged. 16 1 2 3 4 current ( μ A) t (ms) -8 10 5 8x10 -3 t , 0< t < 2ms -8x10 -6 , 2< t < 4 ms i(t) = At 0< t < 2ms: + = t dt ) t ( i C ) t ( v ) t ( v 0 0 1 - - = t xdx ) ( ) ( ) t ( v 0 3 6 10 8 10 4 1 = 10 3 t 2 V
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Department of Electrical and Electronics Engineering 9 At time t =2ms: v(2ms) = 10 3 (2x10 -3 ) 2 = 4 mV In the period 2ms < t < 4ms: - + = t x dt ) t ( i C ) ms ( v ) t ( v 3 10 2 1 2 - - - - - + = t x dx ) ( ) ( x 3 10 2 6 6 3 10 8 10 4 1 10 4 = 8x10 -3 – 2 t V t x x x 3 10 2 3 2 10 4 - - - = 10 3 t 2 V , 0< t < 2ms –2 t + 8x10 -3 V, 2< t < 4 ms v(t) = Department of Electrical and Electronics Engineering 10 The energy in the electric field of the capacitor at time t = 3 ms:
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Ch04 Networks with L and C 1s09 - Chapter 4 Inductors,...

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