Initial condition

Initial condition

• 1.
  Gujarat power engineering & research institute(104) SUB : Circuits & Networks Topic: Initial Conditions Prepared by : Tindwani Jay (131040109059) Patel Ravi (131040109051) Purohit Sarthak (131040109048) Guided by : Prof. Ashvin Patel
• 2.
  Introduction • Thereare many reasons for studying initial and final conditions. The most important reason is that the initial and final conditions evaluate the arbitrary constants that appear in the general solution of a differential equation. • Conditions existing at instant are known as initial condition or initial state. Conditions at t =  are used, these are known as final condition. • Assuming that switches act in zero time. To differentiate between the time immediately before and after the operation of the switch, we will use t = 0- and t = 0+ respectively. • The conditions existing just before the switch is operated will be designed as i(0-), v(0-), q(0-). And conditions just after will be i(0+), v(0+), q(0+).
• 3.
  Initial and finalconditions in elements The inductor  The switch is closed at t = 0. Hence t = 0- corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed.  if i(0-) = 0, we get i(0+) = 0. This means that at t = 0+, inductor will act as an open circuit, irrespective of the voltage across the terminals. If i(0-) = I0, then i(0+) = I0. In this case at t = 0+, the inductor can be thought of as a current source of I0 A. The equivalent circuits of an inductor at t= 0+ is shown in Fig.
• 4.
  • The final-conditionequivalent circuit of an inductor is derived from the basic relationship. v = L di dt • Under steady condition, di = 0. This means, v = 0 and dt hence L acts as short at t =  (final or steady state). The final- condition equivalent circuits of an inductor is Shown in fig. The initial -condition equivalent circuit of an inductor The final -condition equivalent circuit of an inductor
• 5.
  THE CAPACITOR The switch is closed at t = 0. Hence, t = 0 corresponds to the instant when the switch is just open and t = 0+ corresponds to the instant when the switch is just closed. The expression for voltage across the capacitor is given by v(0+) = v(0-).  Thus the voltage across a capacitor cannot change instantaneously.  If v(0-) = 0, then v(0+) = 0. This means that at t = 0+, capacitor q acts as short circuit. Conversely, if v(0-) = 0 then v(0+) = q 0 .These conclusions are summarized in Fig. c c
• 6.
  Initial-condition equivalent circuitsof a capacitor  The final–condition equivalent network is derived from the basic relationship v = C dv dt dv dt   Under steady state condition, = 0. This is, at t = , i = 0. This means that t =  or in steady state, capacitor C acts as an open circuit. The final condition equivalent circuits of a capacitor is shown in Fig.
• 7.
  Final-condition equivalent circuitsof a capacitor The resistor  The cause effect relation for an ideal resistor is given by v = Ri . From this equation, we find that the current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, voltage will change instantaneously if current changes instantaneously.
• 8.
  Procedure for evaluatinginitial conditions There is no unique procedure that must be followed in solving for initial conditions. We usually solve for initial values of currents and voltages and then solve for the derivatives. For finding initial values of currents and voltages, an equivalent network of the original network at t = 0+ is constructed according to the following rules: (1) Replace all inductors with open circuit or with current sources having the value of current flowing at t = 0+. (2) Replace all capacitors with short circuits or with a voltage q source of value v = if there is an initial charge. 0 c (3) Resistors are left in the network without any changes.
• 9.
  Example 1. Referthe circuit shown in Fig. 4.8. Find vc (0+). Assume that the switch was in closed state for a long time. The symbol for the switch implies that it is closed at t = 0- and then opens at t= 0+. Since the circuit is in steady state with the switch closed, the capacitor is represented as an open circuit at t = 0-. The equivalent circuit at t = 0- is as shown in Fig.4.9
• 10.
  Vc (0-) =i(0-)R2 Using the principle of voltage divider, Vc (0-) = R 2 s v R R  + 1 1 1 2 5 * 5 = 2.5  Since the voltage across a capacitor cannot change instaneously, we have Vc (0+) = Vc (0-) = 2.5V
• 11.
  That is, whenthe switch is opened at t = 0, and if the source is removed from the circuit, still Vc (0+) remains at 2.5 V. Example 1. In the network shown in Fig. , the switch is moved from position 1 to position 2 at t = 0. The steady-state has been reached before switching. Calculate , i, at t = 0+ 2 di d i , and dt dt 2
• 12.
  The symbol forswitch k implies that it is in position 1 at t = 0- and in position 2 at t = 0+. Under steady-state condition, a capacitor acts as an open circuit. Hence at t = 0-, the circuit diagram is as shown in Fig. . (a) We know that the voltage across a capacitor cannot change instantaneously. This means that Vc (0+) = Vc (0-) = 40 V.
• 13.
  At t =0- , inductor is not energized. This means that i(0-) = 0. Since current in an inductor cannot change instantaneously, i(0+) = i(0-) = 0. Hence, the circuit diagram at t = 0+ is as shown in Fig. . The circuit diagram for t > 0+ is as shown in Fig.. (b) (c)
• 14.
  KVL clockwise, weget t Ri + L  ( )  0..........(1) o  Ri + L v ( ) 0 At t = 0+ , we get   c (0 ) Ri(0+) + L v (0 ) c Apllying di i T dt dt di t dt di dt     0 (0 ) 20 * 0+ 1 40 = 0 (0 ) = -40A/sec di dt di dt    
• 15.
  Diferentiating equation (1)with respect to t, we get 2 di d i i dt dt C R + L = 0 2  putting t = 0+ in the above equation, we get (0 ) R di dt  2 d i i 2 dt C d 2 i 2 2 dt C 2 2 (0 ) (0 ) + L = 0 (0 ) 0 R * (-40) + L = 0 d i (0 ) L = 800A/sec dt      