Circuits & Networks-II Lecture-I Transients PDF

Circuits & Networks-II Unit-II Lecture-I Transients K V V P Chari Assosiate Professor DEPT OF EEE SNIST  Syllabus 1. RL,RC and RLC series in time domain with one operation of switch with DC...

Circuits & Networks-II Unit-II Lecture-I Transients K V V P Chari Assosiate Professor DEPT OF EEE SNIST  Syllabus 1. RL,RC and RLC series in time domain with one operation of switch with DC supply. 2. RL,RC and RLC series in time domain with two operations of switch with DC supply. 3. RL,RC and RLC series in time domain with one operation of switch with AC supply. 4. RL,RC and RLC series in time domain with two operations of switch with AC supply.  Syllabus 5. RL,RC and RLC series in Laplace domain with one operation of switch with DC supply. 6. RL,RC and RLC series in Laplace domain with two operations of switch with DC supply. 7. RL,RC and RLC series in Laplace domain with one operation of switch with AC supply. 8. RL,RC and RLC series in Laplace domain with two operations of switch with AC supply. Transients A sudden change of state is called Transient. In this transient period the response is highly depends on time. R, L and C are the electrical elements. The resistor, R is time invariant element. The L and C are time variant elements. They cannot consume any power bur can store energy The behavior of R , L and C with Switching action are explained below.  Resistor, R When a DC supply is given to a resistor, through a switch as shown in the figure As per Ohm’s law, V=I *R or I=V/R ,As the resistor is a time independent element It responds instantaneously ( no need of any transient period) i.e. when the voltage is applied, the current can be passed.immediately  Inductor, L When a DC supply is given to an inductor, through a switch as shown in the figure The above integration is considered without initial conditions… which can be discussed later. Suppose the inductor allows current at t=0, V is L di/0 which is infinite, which is a conflict for a given voltage. For example, a 10V is applied to a 2Henry inductor, a 0.5Ampere is flowing at swith is closed at t=0, It results 10 = ∞, which is a conflict. So the inductor does not allow any flow of current as the switch is closed and it takes some time to allow the current. So at t=0+, L is open circuit.  Just switch is closed ( t= 0+)for an inductor with supply it cannot be changed the current immediately. As sufficient time is given, it allows the flow of current ie at t = ∞, the L is short circuit. At t=0+, the inductor behaves like ---open circuit which is callaed Transient state of inductor At t= ∞, the inductor behaves like ---short circuit which is callaed Study state of inductor.  Capacitor, C When a DC supply is given to a Capacitor, through a switch as shown in the figure The above integration is considered without initial conditions… which can be discussed later. Suppose the capacitor allows voltage at t=0, I is C dv/0 which is infinite, which is a conflict for a given current. For example, a 10A is flowing to a 2Farad capacitor, a 0.5V is flowing at switch is closed at t=0, It results 10 = ∞, which is a conflict. So the Capacitor does not charge any voltage as the switch is closed at t=0 and it takes sufficient time to charge the voltage. So at t=0+, C is short circuit.  Just switch is closed ( t= 0+)for a capacitor with supply it cannot be changed the voltage immediately. As sufficient time is given, it charges voltage ie at t = ∞, the C is open circuit.( potential difference between C and supply is same) At t=0+, the capacitor behaves like ---shortcircuit which is callaed Transient state of capacitor At t= ∞, the capacitor behaves like ---open circuit which is callaed Study state of capacitor.  Transient, Study state and initial conditions t=0+ t=∞ t = 0- initial Energy Element transient study state conditions stored state Inductor, L OPEN SHORT IL (0) ½ LI2 Capacitor, SHORT OPEN VC(0) ½ CV2 C  Switching operation w.r.t. time for L and C  t = 0+ means the switch is just closed (immediately after the switch is closed) is called transient state  t = ∞ means the sufficient time to reach the steady state. (long back the switch is closed)  t=0- means before the switch is operated, the once excited L and C have initial current and initial voltage respectively called initial conditions. Inductor with initial current Capacitor with initial voltage  Conclusion of L and C behavior Just switch is closed for an inductor with supply it cannot change the current immediately or instantaneously Similarly just switch is closed the voltage across or instantaneously capacitor cannot be changed immediately  Circuits & Networks-II Unit-II Lecture-II Transients K V V P Chari Assosiate Professor DEPT OF EEE SNIST  Transient, Study state and initial conditions t=0+ t=∞ t = 0- initial Energy Element transient study state conditions stored state Inductor, L OPEN SHORT IL (0) ½ LI2 Capacitor, SHORT OPEN VC(0) ½ CV2 C RL series circuit with DC excitation, V In series circuit, current is same and voltages are different. The complete solution of this differential equation of first orde is 1. Complementary function and 2. Perticular Integral  the complete solution, I(t) = ICF + IPI Where ICF is transient and IPI is steady state response RL series circuit with DC excitation, V the complete solution, I(t) = ICF + IPI For 1) ICF consider the complementary equation, Where D is the differential operator d/dt. ICF = is transient response C1 is a constant and 2) IPI = V/R is steady state response. from the circuit diagram at t=∞, L is shorted. RL series circuit with DC excitation, V Therefore the complete solution, I(t) = V/R + To calculate C1 use the initial conditions at t =0, I(t) = I(0). At t= 0, the inductor is opened. I.e. I(0) = 0 Therefore the complete solution at t=0 is 0 = V/R + C1 C1 = -V/R and substitute C1 in above equation, I(t) = V/R –V/R Or RL series circuit with DC excitation, V Also VR = R*I(t) = And VL = V – VR = The I(t) graph is Also the time constant for RL seris circuit is, τ = L/R. Exampl e1 1.Write equations for complementary equation for roots 2.Write ICF with unknown constants 3.Calculate IPF at t=∞ from circuit diagram. 4.Now I(t) = ICF + IPI 5.For constants use initial conditions at t=0 from circuit 1. Write equations diagram. for 6.Write the complete 2. Write ICF with unknown constants 3.Calculate IPF at t=∞ from circuit diagram. 4.Now I(t) = ICF + IPI Answer ca be verified by this formula directly. 5. For constants use initial conditions at t=0 6.Write the complete equation Example2 on RL with DC voltage Standard formula with initial conditions Example3. of RL with two switch operation Procedure for two operations of switch 1. First steady state is considered i.e. switch was in closed position, so at t=∞, calculate …….old circuit 2. Now consider at t=0 switch condition ie just opened. Write equations for complementary function for roots 3.Write ICF with unknown constants 4.Calculate IPF at t=∞ from circuit diagram. …….new circuit 5.Now I(t) = ICF + IPI 6.For constants use initial conditions at t=0 from circuit diagram. Solution: 1) When the switch was in closed position for a long time, the circuit reached in steady state condition i.e. a) the current through inductor is constant and hence, the voltage across the inductor terminals a and b is zero or in other words, inductor acts as short circuit It can be seen that the no current is flowing through resistor. The following are the currents through different branches just before the switch ‘S ’ is opened i.e., at t=0− is The algebraic sum of these two currents is flowing through the inductor (Step 1) 2) when the switch is in opened position now Write equations for new cicuit.  RC series circuit with DC excitation, V I(t) In series circuit, current is same and voltages are different. The complete solution of this Differentiating both sides w.r.t. time differential equation of first order is 1. Complementary function and 2. Perticular Integral RC series circuit with DC excitation, V the complete solution, I(t) = ICF + IPI For 1) ICF :consider the complementary equation, Where D is the differential operator d/dt. ICF = C1 is transient response and C1 is a constant Since right side of is zero, there is no IPI. IPI = 0 RC series circuit with DC excitation, V Therefore the complete solution, I(t) = ICF = C1 To calculate C1 use the initial conditions at t =0, I(t) = I(0). At t= 0, the Capacitor is shorted I.e. I(0) = V/R Therefore the complete solution at t=0 is V/R = C1 substitute C1 in the above equation, I(t) = V/R RC series circuit with DC excitation, V Also VR = R*I(t) = V And VC = V – VR = v(1- ) The I(t) or VR graph is Also the time constant for RC seris circuit is, τ = RC VR  Example for R C series circuit. Solve current when the switch is closed at t=0 Procedure for two operations of switch 1. Write equations for complementary function for roots 2.Write ICF with unknown constants 3.Calculate IPF at t=∞ from circuit diagram. 4.Now I(t) = ICF + IPI 5.For constants use initial conditions at t=0 from circuit diagram. 6.Write the complete equation 1. Write the equations for complementary function for roots (IPI = 0)  Circuits & Networks-II Unit-II Lecture-III Transients K V V P Chari Assosiate Professor DEPT OF EEE SNIST  RLC series circuit with DC excitation, V In series circuit, current is same and voltages are different. Randomly circuit values are The complete solution of this assigned in above diagram. differential equation of second order is 1. Complementary function and 2. Perticular Integral Differentiating both sides w.r.t. time Since VS is constant RLC series circuit with DC excitation, V the complete solution, I(t) = ICF + IPI For ICF :consider the complementary equation, which is differential equation with second order Where D is the differential operator d/dt.  Since the right side of is zero, there is no IPI. IPI = 0  RLC series circuit with DC excitation, V Now solving for D from the equation, Based on the value of discriminant ∆, the roots are of three types. Where ∆ = Which can vary based on the values of R, L and C. RLC series circuit with DC excitation, V Case1: If ∆ is positive, the roots are real and different, D = -α1 and –α2, the system is called overdamped and the I(t) = ICF = Case2: If ∆ is zero, the roots are real and equal, D = -α and –α, the system is called critically damped and the I(t) = ICF = RLC series circuit with DC excitation, V Case3: If ∆ is negative, the roots are complex conjugate., D = -α ± j β, the system is called under damped and the I(t) = ICF = The second order Under damped system can reach the steady state fast and used in control systems and power systems  Example1 for case3: Solve current, I(t) when the switch is closed at t=0 R = 20 ohm for the given L= 0.15Henry Since ∆= values is C = 100 μF and negative V = 100V i.e. case 3. therefore the roots are Procedure for two operations of switch 1. Write equations for complementary function for roots 2.Write ICF with unknown constants 3.Calculate IPF at t=∞ from circuit diagram. 4.Now I(t) = ICF + IPI 5.For constants use initial conditions at t=0 from circuit diagram. 6.Write the complete equation  Example1 for case3 on RLC series circuit 1. Write equations for complementary function for roots Above is case3, complex conjugate roots. 2. Write ICF with unknown constants  Example1 for case3 on RLC series circuit Since right side of complementay equation is zero, there is no IPI. IPI = 0 3. Calculate IPF at t=∞ I(t) = t =0 5. For constants I(0 ) use initial conditions C To calculate at t=0 1 and C2, use the initial conditions at t =0, I(t) = I(0). At t= 0, the inductor is opened  Example1 for case3 on RLC series circuit Therefore the complete solution at t=0 , the equation VL(0) I(t) = ICF = I(0 ) t=o, V Ie C1 = 0, Now for C2, another equatation is required at t = 0 From the figure, VL(0) = V(0) = L di(0)/dt or di(0)/dt = V(0)/ L = 100/0.15 = 666.67  Example1 for case3 on RLC series circuit Substitute C1 value in above equation, Now differentiate both sides di/dt = At t = 0, di(0)/dt = 666.67A/sec = 249.5 C2 = C2 2.672 Therfore the 6. Write the complete solution is I(t) = complete equation  Solve current, I(t) when the switch is closed Exampl at t=0 e2 for case1: RL series circuit with AC supply 1.For the calculation of C1, use initial conditions at t = 0, I(0) 2.For Im and φ, use the steady state formulas for RL series circuit with sinusoidal ac response Particular Integral from steady state analysys IPI = Im sin(ωt- φ) Where And tan φ = ωL/R Particular Integral from steady state analysys Impedan ce in ohm Examp le In series, Initial conditions with two operations of a switch. Find i(t ) when the switch is moved from postion1 to position2 position1 was already in steady state and position2 is transient i.e. equations to be wriiten.  1) First steady state is considered i.e. position1 was in closed position, so at t=∞, calculate IL(∞) of position1 = IL(0) to position2.  Transient, Study state and initial conditions t=0+ t=∞ t = 0- initial Energy Element transient study state conditions stored state Inductor, L OPEN SHORT IL (0) ½ LI2 Capacitor, SHORT OPEN VC(0) ½ CV2 C procedure 1. First steady state is considered i.e. position1 was in closed position, so at t=∞, calculate IL(0) …….old circuit 2. Now consider at t=0 switch condition ie just opened. Write equations for complementary function for roots 3.Write ICF with unknown constants 4.Calculate IPF at t=∞ from circuit diagram. …….new circuit 5.Now I(t) = ICF + IPI 6.For constants use initial conditions at t=0 from circuit diagram. At t=0, I(0) = 5A, substitute in Therefore I(t) = 5 Example2. Find i(t ) when the switch is moved from postion1 to position2 position1 was already in steady state and position2 is transient i.e. equations to be wriiten.  1) First steady state is considered i.e. position1 was in closed position, so at t=∞, calculate IL(∞) of position1 = ----------------------------------------------------------------------- ---- Therefore C1 = 8 and i(t) = Example 3 Find i(t ) when the switch is moved from postion1 to position2  Find i(t ) when the switch is moved from postion1 to position2 i.e RC with AC RLC with AC supply Lapalce transform Laplace transform can be used for DC or AC supply with initial conditions for RLC circuits at once as 1.Find Laplace function i(s) to differential equation 2.Partial fractions mehod to i(s)and then 3.Calaculate Inverse Laplace transform for i(t). Example1 with Laplace transform 1. Find Laplace function i(s) to differential equation 3. Calaculate Inverse Laplace transform for i(t). Example2 with Laplace transform Find i(t ) when the switch is moved from postion1 to position2 Solve Time Example3 with time domain and domain and Laplace then s- transform domain For C1, use initial conditions at t=0 Find i(t ) when the switch is moved from postion1 to position2 in s-domain. L with initial current in s- domain. C with initial voltage in s- domain. Solve the network using lapalace transform Find i(t ) when the switch is moved from postion1 to position2 using s-domain Thank you RC with two swiching operations  Solution: As we know the voltage across the capacitor cannot change instantaneously due to the principle of conservation of charge. Therefore, the voltage across the capacitor just before the switch is closed voltage across the capacitor just after the switch is closed (note the terminal ‘ a ’ is positively charged. It may be noted that the capacitor current before the switch ‘ S’ is closed is. On the other hand, at t=0, the current through 10Ω resistor is zero but the current through capacitor can be computed as ( voltage across the capacitor cannot change instantaneously at instant of switching)  The rate of change of capacitor voltage at time ‘t = 0 ’ is expressed as