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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 146 Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Modeling Copyright © Orchard Publications Solutions to EndofChapter 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 14 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 . 138 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.