×

Close

- Electrical Circuit Analysis - ECA
- Note
**Visvesvaraya Technological University Regional Center - VTU**- Electrical and Electronics Engineering
- 9 Topics
**14762 Views**- 189 Offline Downloads
- Uploaded 1 year ago

Touch here to read

Page-2

- 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:

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

Problem 3: R-C Circuit(series) If K is closed at t=0, find the values of i, di/dt and d2i/dt2 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: d2i(0+)/dt2 = - (1/RC) di(0+)/dt , substitutiting we get + 105 A/s2 . Ans: 0.1, -100 A/s, 100,000 A/s2

## Leave your Comments