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Note of Initial conditions by Manjunatha P

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Chapter 1 Initial-Conditions 1.0.1 Introduction: Most of the transmission lines, electrical circuits and communication networks are made up of network elements like resistor R,inductor, L and Capacitor C. These networks are connected by voltage and current sources. It is most useful to understand the behavior of the network when we switched on the network by supplying voltage source. It is most important to determine the transient response of R-L, R-C, R-L-C series circuits for d.c and a.c excitations. Assuming that at reference time t = 0, the switch in the circuit is closed and also assuming that switch act in zero time. To differentiate between the time immediately before and immediately after the operation of a switch, t = 0− and t = 0− signs are used. The condition existing just before the switch is operated will be designated as i(0−),v(0−), q(0−) and the conditions existing after closing of a switch is designated as as i(0+),v(0+), q(0+). Also initial conditions of a network depend on the past history of the network prior the closing of the network at to t = 0− and the network structure at t = 0+, after switching. The evaluation of voltages and currents and their derivatives at t = 0+, are known as initial conditions and evaluation of condition at t = ∞ are known as final conditions. The following are the objectives of studying the behavior of the circuit for Initial-Conditions: • The most important reason is that the initial and final conditions must be known evaluate the arbitrary constants that appear in the general solution of a differential equation. • The initial conditions give knowledge of the behavior of the circuit elements at the instant of switching • The final conditions give knowledge of the behavior of the circuit elements after the settling of circuit at t=∞ ———————————————————————————————————————- 1.1 initial Conditions K V iR R Resistor Figure 1.1: Series resonance circuit When the switch K is closed at t=0 the current I is flowing in a circuit and is given by Consider a circuit which consists of resistor R connected as shown in Figure 1.1. The circuit resistor V R is connected by a voltage source V in series with I= switch K as shown in Figure. R 1

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1.1. INITIAL CONDITIONS Chapter 1. Initial-Conditions If the capacitor is initially charged with charge q0 coulombs at t=0-, then at t=0+ the capacitor is Consider a circuit which consists of inductor L equivalent to voltage source v0 = qc0 which is as shown connected as shown in Figure 1.2. The inductor L is in Figure 1.6 connected by a voltage source V in series with switch K K t  0 K as shown in Figure. When the switch K is closed at t=0 the current flowing in a inductor at t = 0+ is zero + + iC C - q V iC V q the inductor acts as a open circuit at t = 0+ which is v0  o v0  o C C as shown in Figure 1.3. Inductor K Figure 1.6: Capacitor circuit at t = 0+ iL V L Figure 1.2: Inductor circuit K t  0 K iL V iL LV OC Figure 1.3: Inductor circuit at t = 0+ The final-condition equivalent circuit of an inductor is derived from the basic relationship. v=L The final condition of capacitor circuit is derived from the following relationship. The voltage across capacitor is dv v=C dt Under steady state condition, rate of change of voltage capacitor is dv dt = 0. This means, v = 0 and hence C acts as open circuit at t = ∞. The equivalent circuits of a capacitor at t = ∞ is as shown in Figure 1.7 K t  K iC V C iC V OC di dt Figure 1.7: Capacitor circuit at t = ∞ Under steady state condition, rate of change of current If the capacitor is initially charged with voltage v0 di flowing in inductor is dt = 0. This means, v = 0 and then the final condition at t = ∞ of a capacitor circuit hence L acts as short at t = ∞. The equivalent circuits is replaced with voltage source v0 with open circuit of an inductor at t = ∞ is as shown in Figure 1.4 which is as shown in Figure 1.8 K iL V K t  t 0 LV iL v=C SC Figure 1.4: Inductor circuit at t = 0+ Capacitor K t  K V dv dt iC C + - q V v0  o C iC q v0  o C + OC Figure 1.8: Capacitor circuit at t = ∞ Consider a circuit which consists of capacitor C connected as shown in Figure 1.5. The capacitor is connected by a voltage source V in series with switch K as shown in Figure 1.5. When the switch K is closed at t=0 capacitor C acts as short circuit and current flows in a capacitor instantaneously. K t 0 K V iC C V iC SC Figure 1.5: Capacitor circuit at t = 0+ Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 2

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1.1. INITIAL CONDITIONS Chapter 1. Initial-Conditions Table 1.1: Initial and Final Conditions at t=0- L at t = ∞ at t=0+ R R R L O. C S.C IO IO IO O. C S.C C - C+ -+ q V= 0 C q0 -+ q V 0 C O. C Procedure for Evaluating Initial Conditions: 1. Before closing or opening the switch at t=0- find the history of the network, at t=0- find i(0-), v(0-), i.e., current through inductor and voltage across the capacitor before switching 2. Draw the circuit after switching operation at t=0+. 3. Replace inductor with open circuit or by current source having source 4. Replace capacitor with short circuit or with a voltage source vc = q0 c if it has an initial charge q0 . 5. Find i(0+), and v(0+) at t=0+ 6. Obtain an expression for di dt 7. Obtain an expression for d2 i dt2 and find and find di dt at t=0+ d2 i dt2 at t=0+ 8. Similarly determine voltages across circuit elements and its derivatives. Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 3

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1.1. INITIAL CONDITIONS Chapter 1. Initial-Conditions Q 1) In the circuit shown in Figure 1.9 the switch K s closed at t=0, with capacitor uncharged. Find the di d2 i values i, dt and dt 2 at t=0+, for element values as follows V=100 V R = 1000 Ω and C = 1 µF . K Q 2) In the circuit shown in Figure 1.11 the switch K s closed at t=0, with zero current in the conductor. di d2 i Find the values i, dt and dt 2 at t=0+, for element values as follows V=100 V R = 10 Ω and L = 1 H. K R L i V C i V R Figure 1.11: Example Figure 1.9: Example Solution: Solution: KVL for the given circuit is V = Ri + L 1 V = Ri + C Z idt (1.1) i(0+) = V 100 = = 0.1A R 1000 i(0+) = 0 (1.2) K (1.3) i (1.6) R i (0+ ) V R t 0 V (1.5) at t=0+ the inductor acts as open circuit which is as shown in Figure 1.12 At t=0+ the capacitor acts as short circuit which is as shown in Figure 1.10 V = Ri(0+) di dt Figure 1.12: Example SC From equation conditions (1) and substituting initial Figure 1.10: Example Differentiating equation (1) 0=R di i + dt C L (1.4) Substituting initial conditions di i(0+) (0+) + dt C i(0+) − RC 0.1 − − 100A/sec 1000 × 1 × 10−6 1 di (0+) − RC dt .1 − (−100) 1000 × 1 × 10−6 .1 − (−100) 1000 × 1 × 10−6 1 × 105 A/sec2 0 = R di (0+) = dt di (0+) = dt d2 i (0+) = dt2 = = = di dt = V − Ri di (0+) = V − Ri(0+) = 100 − 0 dt di 100 (0+) = = 100A/sec dt 1 d2 i R di 10 (0+) = − (0+) = − × 100 dt2 L dt 1 2 = −1000A/sec L Q 3) In the circuit shown in Figure 1.13 V=10 v R = 10 Ω L = 1 H and C = 10 µF and vc (0) = 0, find di d2 i i(0+) = 0, dt (0+) and dt 2 (0+). K R L i V C Figure 1.13: Example Dr. Manjunatha P Professor Dept of E&CE, JNN College of Engineering, Shivamogga 4

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