Initial conditions Necessity and advantages: Initial conditions assist To evaluate the arbitrary constants of differential equations Knowledge of the behavior of the elements at the time of switching Knowledge of the initial value of one or more devivatives of a response is helpful in anticipating the form of the response It is useful in getting to know the elements individually and in combined networks Gives a better understanding of non linear switching circuits The general solution of a first order differential equation contains an arbitrary constant, the order of differential equation increases the no of arbitrary constants increase Solution of differential equation= CF + PI Geometrical interpretation of Derivatives Normally i(0+), di/dt, and d 2 i/dt 2 at t=0+ will be evaluated i(0+) indicates where the current starts di/dt at t=0+ gives the slope fo the curve and the second derivative gives the rate of change of slope Ex: 1. i(0+)=0, di(0+)/dt = 0 and d 2 i(0+)/dt 2 =k >0 2. i(0+)=0, di(0+)/dt = K>0 and d 2 i(0+)/dt 2 =>0
3. i(0+)=k>0, di(0+)/dt =0 and d 2 i(0+)/dt 2 =0 Behaviour of the elements at t=0+ and at t= Element Equivalent circuit at t=0+ At t= Resistor Resistor Resistor Inductor Open circuit Short circuit Capacitor Short circuit Open circuit Inductor with initial current Capacitor with initial charge Current source Io Finally short Voltage source + - Finally open circuit Procedure to find initial conditions: 1. History of the network, at t=0- find i(0-), v(0-), preferably current through inductor and voltage across the capacitor before switching, 2. Write the circuit at t=0+ by looking at the table given above 3. Find i(0+), and v(0+) 4. Write general circuit after the switching operation 5. Write general integro differential equation 6. Obtain an expression for di/dt 7. Apply initial conditions like i(0+) find di/dt at t=0+ 8. obtain an expression for d 2 i(0+)/dt 2 9. Apply initial conditions like i(0+), di/dt at t=0+ find out d 2 i(0+)/dt 2. and the process is repeated.
Problems1. R-L Circuit (series) If K is closed at t=0, find the values of i, di/dt and d 2 i/dt 2 at t=0+ if R=10 Ω, L=1H and V=100V As per the steps indicated : 1. Previous history before the switch is closed i(0-) =0. 2. Write the general circuit after switching : Inductor acts as open, as inductor wont allow current to change instantaneously Hence i(0+) = 0 3. Write general network after switching : V= Ri + L di/dt 4. Obtain an expression for the first derivative: di/dt = (V R(i) )/L, substituting the values we get di/dt = 100 A/s 5. Obtain an expression for the second derivative: d 2 i(0+)/dt 2 = - R/L di/dt, substituting the known values we get -1000A/s 2. Ans: 0, 100 A/sec, 1000 A/sec 2
Problem 2. RL (parallel) If K is opened at t=0, Find v, dv/dt and d 2 v/dt 2 at t=0+, If I=1 amp, R=100 Ω, and L=1H As per the steps indicated : 6. Previous history before the switch is opened indicates that the switch Is closed i L (0-) =0. 7. Write the general circuit after switching : Inductor acts as open, as inductor will not allow current to change instantaneously Hencev(0+) =i*r=100 V 8. Write general network after switching : I= v/r + 1/L 9. Obtain an expression for the first derivative: Diff we get dv/dt at to0+ = -10000V/sec 10. Obtain an expression for the second derivative: d 2 v(0+)/dt 2 = - R/L dv/dt, substituting the known values we get 10*10 6 V/s 2. Ans: 100, -10 4 v/sec, 10 6 v/s 2
Problem 3: R-C Circuit(series) If K is closed at t=0, find the values of i, di/dt and d 2 i/dt 2 at t=0+ if R=1000 ohms, C= 1μf and V=100V History of the network: Vc(0-) = 0, i(0-)=0 Capacitor acts as a short Hence network after switching, Voltage source, switch, resistor, and capacitor as s/c makes a closed path Therefore i(0+)=v/r = 0.1 A. Write the general network after the switch is closed: V= Ri + 1/c Differentiate : 0 = R di/dt + i/c di(0+)/dt= - i(0+)/rc, substituting the values we get 100 A/s differentiate again to obtain second derivative: d 2 i(0+)/dt 2 = - (1/RC) di(0+)/dt, substitutiting we get + 10 5 A/s 2. Ans: 0.1, -100 A/s, 100,000 A/s 2
Problem 4: RC circuit(parallel) If K is opened at t=0, Find v, dv/dt and d 2 v/dt 2 at t=0+, if I=10A,R=1000 Ω and C=1μf History of the network, before the switch is opened: Vc(0-) = 0, Network after the switch is opened. Cap acts as a short circuit hence v(0+) = 0; General network after the switch is opened: I= V/R + C dv/dt Obtain an expression for dv(0+)/dt = (I V(0+)/R )/c, substituting we get dv(0+) / dt = 10 5 V/s Expression for second derivative: d 2 v(0+)/dt 2 = (-1/RC)*dv(0+)/dt=-10 8 V/s 2 Ans: 0, 10 5 v/s, - 10 8 v/s 2
Problem5: In the network shown, switch K is changed from position a to b at t=0, steady state being established at position a. V=100, R= 1000Ω, L=1H, C=1μf. Find : i, di/dt and d 2 i/dt 2 at t=0+ When the switch is at position a, SS being established, L acts as a short circuit, (t= ) state: Hence : i(0-)= V/R = 0.1A, Vc(0-)=0V. When the switch is closed to position b, Cap acts as short, R behaves as R and L acts as a current source, there fore i(0+)=0.1a Write general network after switching: Write integro diff equation: 0= Ri+Ldi/dt + 1/c,-------------(a) Rearranging the above, we get the expression for di/dt = (-Ri/L) V c /L substituting the values di(0+)/dt = - 100 A/s differentiating equation (a), we get d 2 i(0+)/dt 2 = -R/L (di/dt) + i/rc substituting the values we get d 2 i(0+)/dt 2 = 0.
Problem6: In the network shown, switch K is changed from position a to b at t=0, steady state being established at position a. Show that i 1 (0+)=i 2 (0+)= -V/(R 1 +R 2 +R 3 ), i 3 (0+)=0. Solution: Steady state being established at position a, Means inductors act as short circuited, capacitors act as open circuit Switch being at position a, we can see that i 1 (0-)=0, i 2 (0-)=0 i 3 (0-)=0 V c1 (0-)=0, V c2 (0-)=0 and V c3 (0-)=V. When switch is moved to position b, Only C3 will act as a voltage source, No current i 1, i 2 and i 3 hence inductor L 1 and L 2 acts as open And Vc3 = V We get i 3 (0+)=0 and i 1 =i 2 = -V/ R 1 +R 2 +R 3
Problem7: Steady state reached with switch k open, Switch is closed at t=0. V=100V, R1=10 Ω, R2=R3=20Ω, L=1H, C=1μ f Find: v 0 across C and indicate polarities,initial values of i 1, i 2, di 1 /dt, di 2 /dt at t=0+ find di/dt at t= Solution: Switch k is opened : SS being established : Inductor acts as short circuit: capacitor acts as open : I 1 (0-)= V/R1+R2 = 3.33 A I 2 (0-)=0; Vc(0-)= i1*r2 = 66.67V At t=0 switch is closed : hence R1 becomes redundant< current will not flow through R1 as switch is acting as short across the resistor R1. Inductor acts as a current source of value 3.33 A, Vc(0+)= 66.67 V I 2 (0+) = (100 66.67)/20 = 1.67A I 1 (0+)= 3.33 A General network after switching : V=i 1 R 2 + L di 1 /dt-------------------(a) V=i 2 R 3 + 1/c -----------------(b) di 1 (0+)/dt = (V-i 1 R 2 ) /L = 33.3A/s diff eqn (b) we get di 2 (0+)/dt=- i 2 (0+)/R 3 C = -83500 A/s. Ans: 66.7 volts 3.33A, 1.67A 33.4 A/s, -83500 A/s 0
Problem8: Switch K is closed at t=0, with zero capacitor voltage and zero inductor curent. Solve for: v 1 and v 2 at t=0+, v 1 and v 2 at t=, dv 1 /dt, dv 2 /dt at t=0+, d 2 v 2 /dt 2 at t=0+ History: at t=0- V c (0-)=0, i L (0-)=0, Network at t=0+: Capacitor acts as short circuit, inductor acts as open circuit. Hence v 1 (0+) =0, V 2 (0+)=0, i 1 (0+)= V/R 1, i 2 (0+)=0, V c (0+)=0 General network after switching: V=Ri+ 1/c ----------------(a) 0= 1/c + L di 2 /dt + R 2 i 2. (b) 2 diff eqn (a) we get di 1 (0+)/dt = - V/CR 1 di 2 (0+)/dt=0, d 2 i 2 /dt 2 = V/R 1 LC dv 1 /dt= L (d 2 i 2 /dt 2 )= V/R 1 C dv 2 (0+)/dt=0 Solution: 0,0 0, vr 2 / (R 1 +R 2 ) L d 2 i 2 /dt 2 = L V / (LCR 1 ), 0 Rv/(R 1 CL)
Problem 9: In the network shown, switch K is opened at t=0 after the network has attained a steady state with switch closed. Find: Expression for voltage across the switch at t=0+ If the parameters are adjusted such that i(0+)=1 and di/dt at t=(0+) = -1, What is the value of the derivative of the voltage across the switch dv k /dt at t = 0+ Solution: History: switch is closed indicates that the voltage across R1 and C is zero, inductor acts as short ckt as steady state having been reached, There fore i(0-) = V/R 2, voltage across cap =0, Switch is opened: Inductor acts as a current source of value V/R 2, capacitor acts as short Hence: i(0+)= V/R 2 only as inductor doesnot allow any sudden change in current. Now general network after switching: V= R 1 i+ 1/c + R 2 i + L di/dt but R 1 i+ 1/c is V k V k (0+) = R 1 V / R 2 dv k /dt = R 1 di 1 /dt + i1/c = -R1 + 1/C Solution vr 1 /R 2 1/c R 1