×
Every great story on the planet happened when someone decided not to give up, but kept going no matter what.

# Note for Electrical Circuit Analysis - ECA By vtu rangers

• Electrical Circuit Analysis - ECA
• Note
• Visvesvaraya Technological University Regional Center - VTU
• Electrical and Electronics Engineering
• 9 Topics
• 14436 Views
Page-1

#### Note for Electrical Circuit Analysis - ECA By vtu rangers

Topic: / 101

0 User(s)

#### Text from page-1

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

#### Text from page-2

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.

#### Text from page-3

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

#### Text from page-4

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