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- Electrical Circuit Analysis - ECA
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**Visvesvaraya Technological University Regional Center - VTU**- Electrical and Electronics Engineering
- 9 Topics
**14436 Views**- 188 Offline Downloads
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- Initial Conditions - ( 1 - 11 )
- Laplace Transformation And Applications - ( 12 - 17 )
- Wave form Synthesis - ( 18 - 22 )
- Ramp Function - ( 23 - 46 )
- Mesh Analysis - ( 47 - 54 )
- Node Voltage Analysis - ( 55 - 68 )
- Two Port Network - ( 69 - 81 )
- Network function - ( 82 - 85 )
- Unbalanced Three Phase Systems - ( 86 - 101 )

Topic:

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 d2i/dt2 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 d2i(0+)/dt2 =k >0 2. i(0+)=0, di(0+)/dt = K>0 and d2i(0+)/dt2 =>0

3. i(0+)=K>0, di(0+)/dt =0 and d2i(0+)/dt2 =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 Voltage source Io Finally short + - 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 d2i(0+)/dt2 9. Apply initial conditions like i(0+) , di/dt at t=0+ find out d2i(0+)/dt2 . 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 d2i/dt2 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: d2i(0+)/dt2 = - R/L di/dt , substituting the known values we get -1000A/s2 . Ans: 0, 100 A/sec, 1000 A/sec2

Problem 2. RL (parallel) If K is opened at t=0, Find v, dv/dt and d2v/dt2 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 iL(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: d2v(0+)/dt2 = - R/L dv/dt , substituting the known values we get 10*106 V/s2 . Ans: 100, -104 v/sec , 106 v/s2

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