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C H A P T E R 5 Transient Analysis.

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Presentation on theme: "C H A P T E R 5 Transient Analysis."— Presentation transcript:

1 C H A P T E R 5 Transient Analysis

2 Figure 5.1 Examples of transient response
0.8 0.6 0.4 0.2 1.0 1.2 1.4 1.6 1.8 2.0 t (s) (a) Transient DC voltage 0.5 (b) Transient sinusoidal voltage

3 Figure 5.2 Circuit with switched DC excitation
Complex load C L R Switch t = 0 12 V

4 Figure 5.3 A general model of the transient analysis problem
V s Circuit containing L / RC combinations Switch t = 0

5 Figure 5.5 Circuit containing energy-storage element
+ _ v R i C ( t ) S di dt 1 RC = dv A circuit containing energy-storage elements is described by a differential equation. The differential equation describing the series circuit shown is

6 Figure 5.9 Differential equations of first-order circuits
dv C dt v 1 S = 0 circuit: + _ R ( t ) di L i RL

7 Figure 5.10 Decay through a resistor of energy stored in a capacitor
= 0 t Switch t = 0 R i ( ) v C V B 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0.9 Time, s Exponential decay of capacitor current Capacitor voltage, V

8 Figure 5.15 t = 0 i L I S R

9 Figure 5.16 10 mA i L ( t ) v R (0 +

10 Figure 5.22 A more involved RC circuit
1 t = 0 V v C + _ 2 3

11 Figure 5.23 The circuit of Figure 6.45 for t > 0
+ R R 1 2 R v C 3 C V V 1 2 _

12 Figure 5. 24 Reduction of the circuit of Figure 5
Figure 5.24 Reduction of the circuit of Figure 5.23 to Thevenin equivalent form V 1 R 2 / 3 T + _ =

13 Figure 5.25 The circuit of Figure 5.22 in equivalent form for t > 0
+ _ R T C V

14 Figure 5.39 Second-order circuits
(a) + _ v T (t) R C L Parallel case Series case (b) ( t )

15 Figure 5.43 Response of overdamped second-order circuit
2 1.5 1 0.5 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0.9 t (s) x N ( ) e

16 Figure 5.44 Response of critically damped second-order circuit
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t (s) x N ( ) e te

17 Figure 5.45 Response of underdamped second-order circuit

18 Figure 5.46 + _ t = 0 v C ( ) V S R = 5000 L = 1 H = 1 F = 25 V i

19 Figure 5.48 t = 0 C i ( ) L = 2 H = 2 F R = 500 I S = 5 A v + _

20 Figure 5.52 i V B N 1 2 L P , R spark plug switch closed C = 100 +


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