Central Electronics 600L Broadband Amplifier

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6  The plots for v C  t  and i L  t  are shown below. 6e Time (sec) 3.  At t = 0 the circuit is as shown below. 100  20 H  iL  0  + + 100 V  400  1  120 F  v 0   C At this time the inductor behaves as a short and the capacitor as an open. 2 A . 2 = V 0 = 80 V 146 Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling Copyright © Orchard Publications Solutions to EndofChapter Exercises and this establishes the second initial condition as V 0 = 80 V .

2e  2t + 1  We find i L  t  from i R  t  + i C  t  + i L  t  = 0 or i L  t  = – i C  t  – i R  t  where i C  t  = C  dv C  dt  and i R  t  = v R  t    1 + 2  = v C  t   3 . 4e  t + 1  12 3 6.    At t = 0 the circuit is as shown below where i L  0  = 12  2 = 6 A , v C  0  = 12 V , and thus the initial conditions have been established. +  12 V 2 A  iL  0  4 B +  2H vC  0  14 F  For t  0 the circuit is as shown below.

83 mF +  vC  t  3. In the circuit below the switch S has been closed for a very long time and opens at t = 0 . Compute v C  t  for t  0 . 100  20 H +  100 V S 400  + t = 0 vC  t  1  120 F  4. In the circuit below, the switch S has been closed for a very long time and opens at t = 0 . Compute v C  t  for t  0 . 138 Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling Copyright © Orchard Publications Exercises 100  20 H vS + + t = 0 S 1  120 F 400   vC  t  v S =  100 cos t u 0  t  V 5.

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