Second Order Transient Analysis

Second Order Transient Analysis

How do recognize it?

A transient problem is one that asks you to find voltage (or current) vs. time. It also includes some instant event: a switch opens or closes, the power supply turns on, a part is suddenly pulled out the circuit, a fuse blows. Your answer can either be a formula for v(t) or i(t) or it might be a graph of v or i on the vertical axis and time on the horizontal axis.
Second-order problems have two independent energy storage elements (capacitors and/or inductors). Note that two capacitors in parallel or series only count as one capacitor because you can find the equivalent capacitance (same with inductors).

How do you do it?

1. For first-order, the only form possible was an exponential response. For second-order there are three possible forms. You can compute which one based on the following three constants that are coefficients of the differential equation that describes these types of circuits:

   Constant	Series RLC	Parallel RLC
   a	1	1
   b	R/L	1/(RC)
   c	1/(LC)	1/(LC)

   Note that in each case, R is the Thevenin resistance seen by the LC pair. If the LC pair is in parallel but the Thevenin R is in series, you may need to convert to a Norton so that all the elements are in parallel.
2. Now compare three constants to determine which of the three response forms you have:
   • if b^2 > 4*a*c then you have overdamped.
   • if b^2 = 4*a*c then you have critically damped.
   • if b^2 < 4*a*c then you have underdamped.

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First-Order Transient Analysis

First-Order Transient Analysis

How do recognize it?

A transient problem is one that asks you to find voltage (or current) vs. time. It also includes some instant event: a switch opens or closes, the power supply turns on, a part is suddenly pulled out the circuit, a fuse blows. Your answer can either be a formula for v(t) or i(t) or it might be a graph of v or i on the vertical axis and time on the horizontal axis.
First-order problems have only one energy storage element (either a capacitor or inductor, but not both). Note that two capacitors in parallel or series only count as one capacitor because you can find the equivalent capacitance (same with inductors).

How do you do it?

1. Find the inital condition (t=0+)
   1. Find the current for each inductor just before the switch is flipped (t=0-). Current through an inductor will remain the same for the instant just after the switch is flipped.
   2. Find the voltage over each capacitor just before the switch is flipped (t=0-). Voltage over a capacitor will remain the same the instant just after the switch is flipped.
   3. In order to find the currents and voltages suggested above, first determine if you are given the values in the problem or you have a formula for the current/voltage from a previous step. If you are not given any other method for finding the current/voltage, then you can usually assume that steady-state conditions exist up to the time the switch is flipped. This will simply the circuit considerably, since inductors look like shorts and capacitors look like opens in steady-state.
   4. Once you have the currents through each inductor and the voltage over any capacitors for the time just before the switch is flipped, then draw the circuit for just after the switch is flipped (t=0+). Each inductor will look like a current source (at its initial current), and each capacitor will look like a voltage source (at its initial current). Note that if the initial current on an inductor is 0 amps, then it will look like an open for that instant, and if the initial voltage on a capacitor is 0 volts, then it will look like a short for that instant. If your variable of interest is not the cap voltage or the inductor current, then use your normal circuit analysis techniques to find the variable of interest at the initial instant after the switch is flipped.
2. Find the final condition (t=infinity)
   1. A long time after the switch is flipped, the inductors will look like shorts and the capacitors will look like opens. Find the variable of interest (voltage/current) for that simplified circuit.
3. Find the time constant for the time between initial and final conditions ( 0 <= t <= infinity)
   1. The time constant for a circuit with a single capacitor is RC, where R is the Thevenin Resistance seen by the capacitor after the switch has flipped. The time constant for a circuit with a single inductor is L/R where R is the Thevenin Resistance seen by the inductor after the switch has flipped.
   2. For a circuit with more than one (equivalent) capacitor and/or inductor does not have a simple exponential response, but the time constant is still part of the generalized second order response. For a circuit with energy storage elements in series, the time constant is L/R, and for a parallel combination the time constant is RC. In both cases, R is the Thevenin Resistance seen by the energy storage elements.
4. Write down the final answer.
   1. For a circuit with a single energy storage element (first order transient), the formula is:
      
      where the "unknown" is the desired voltage or current (as a function of time), "final" is the final value of the unknown, "initial" is the initial value of the unknown, "t" is the time variable, and tau is the time constant. Note that for t=0, unknown = intial and for t=infinity, unknown = final.

What are some difficulties I might run into?

1. If the desired value is not the current through an inductor or the voltage over a cap, then you can still use this procedure, but the initial and final values must be derived based on what the inductor and capacitors do under instantaneous change and under steady-state. That is, you must "back up" to your desired value after seeing what the caps and inductors do in your circuit.
2. If there is more than one switch, treat each switch one at a time in order of the time they change. You can probably assume steady-state conditions before the first switch flip (unless you are told otherwise). Ignore the following switch flips when computing the final value (since the circuit doesn't know another switch flip is coming). For all the other switch flips, you must use your formula to compute what the initial conditions will be. The time constant will be different for each segment of time in between switch flips, since the cap and/or inductor will see a different Thevenin resistance depending on the switch positions.

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Continuity condition of Inductors

Continuity condition of Inductors
The current that flows through a linear inductor must always be a continuous. , the voltage across the inductor is not proportional to the current flowing through it but to the rate of change of the current with respect to time,()ditdt. The voltage across the inductor () is zero when the current flowing through an inductor does not change with time. This observation implies that the inductor acts as a short circuit under steady state dc current. In other words, under the steady state condition, the inductor terminals are shorted through a conducting wire. Alternating current (ac), on the other hand, is constantly changing; therefore, an inductor will create an opposition voltage polarity that tends to limit the changing current. If current changes very rapidly with time, then inductor causes a large opposition voltage across its terminals. If current changes through the inductor from one level to another level instantaneously i.e. in sec., then the voltage across it would become infinite and this would require infinite power at the terminals of the inductor. Thus, instantaneous changes in the current through an inductor are not possible at all in practice. Lv0dt􀀑
Remark-2: (i) The current flowing through the inductor cannot change instantaneously (i.e. just right before the change of current = (0)i−(0)i+ just right after the change of current). However, the voltage across an inductor can change abruptly. (ii) The inductor acts as a short circuit (i.e. inductor terminals are shorted with a conducting wire) when the current flowing through the inductor does not change (constant). (iii) These properties of inductor are important since they will be used to determine “boundary conditions”.

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Introduction to Transient Analysis

So far we have considered dc resistive network in which currents and voltages were independent of time. More specifically, Voltage (cause input) and current (effect output) responses displayed simultaneously except for a constant multiplicative factor (VR). Two basic passive elements namely, inductor →→RI=×()L and capacitor () are introduced in the dc network. Automatically, the question will arise whether or not the methods developed in lesson-3 to lesson-8 for resistive circuit analysis are still valid. The voltage/current relationship for these two passive elements are defined by the derivative (voltage across the inductor C()()LLditvtLdt=, where =current flowing through the inductor ; current through the capacitor ()Lit()()CCdvtitCdt=, = voltage across the capacitor) or in integral form as ()Cvt0011()()(0)()()(0)ttLLLCitvtdtiorvtitdtvLC=+=+∫∫ rather than the algebraic equation (VIR=) for all resistors. One can still apply the KCL, KVL, Mesh-current method, Node-voltage method and all network theorems but they result in differential equations rather than the algebraic equations that we have considered in resistive networks
An electric switch is turned on or off in some circuit (for example in a circuit consisting of resistance and inductance), transient currents or voltages (quickly changing current or voltage) will occur for a short period after these switching actions. After the transient has ended, the current or voltage in question returns to its steady state situation (or normal steady value). Duration of transient phenomena are over after only a few micro or milliseconds, or few seconds or more depending on the values of circuit parameters (like ,,RLandC).The situation relating to the sudden application of dc voltage to circuits possessing resistance (R), inductance (L), and capacitance (C) will

now be investigated in this lesson. We will continue our discussion on transients occurring in a dc circuit. It is needless to mention that transients also occur in ac circuit but they are not included in this lesson.

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Network Theorms to AC

SUPERPOSITION THEOREM

A circuit is linear when superposition theorem can be used to obtain its currents and voltages. When this theorem is applied to an ac circuit, it has to be remembered that the voltage and current sources are in the phasor form and the passive elements are impedances.

WORKED EXAMPLE 1:

A circuit is presented in Fig. 30. The task is to find the current through the load current, using superposition theorem.

The values of components are specified by equation (99). We can find the contribution due to one voltage source, by replacing the other voltage by a short circuit. Only the ideal part of the source is to be replaced by a short-circuit, and its internal impedance should be left in the circuit. For example, we can view Z₁ and Z₂ as the internal impedance of sources, V_A and V_B respectively. The application of superposition theorem is illustrated by the sketch in Fig. 31.

The equations obtained are shown below. The current supplied source can be obtained after determining the impedance seen by the source. The impedance seen by source V_A is the sum of impedance Z₁ and the parallel value of Z₂ and Z_L. Once the source current is determined, current I_L1 through impedance Z_L can be found out using the current division rule. In the same way, we can find current I_L2 through impedance Z_L due to source V_B acting alone. Using superposition theorem, we get the load current as the phasor sum of I_L1 and I_L2.

The answer can be obtained much more easily by using nodal analysis. The voltage across the load Z_L can be found out by solving a single KCL equation. The purpose here is to illustrate the application of superposition theorem. Its direct application is not always handy, bit its importance lies in that the entire analysis of linear circuits rests on it.

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MAXIMUM POWER TRANSFER THEOREM

In some low power circuits, the need for maximum power transfer tends to be more important than the need for efficiency. For example, extraction of maximum signal from a weak source such as a transducer may be necessary. Another example is the case of impedance matching, where the output impedance of an amplifier is to be matched with that of the load such as a loudspeaker. On the other hand, efficiency is the important criterion for a voltage source with low output impedance. For example, efficiency is of utmost importance in electrical power generation and transmission, and it is over 80% for electrical power generation units. Under the maximum power transfer condition, efficiency is no more than 50 %.

When maximum power transfer is to be brought about in ac circuits, the result depends upon type of load. At first, a circuit with resistive load is taken up for study. In the circuit in Fig 32a, the load is purely resistive, designated as R . The source has source impedance containing source resistance, R_G, and positive source reactance X_G. A typical ac voltage source has this type of internal impedance.

When the load is purely resistive, maximum power transfer to the load resistance occurs when the load resistance equals the magnitude of the source impedance. This point is proved as follows.

We consider the case when the load is purely resistive. From Fig. 32a, current supplied by the source can be obtained as shown by equation (105). Then power delivered to the load is obtained as shown by equation (106). In equation (106), the current is expressed as the ratio of source voltage over impedance. A factor of half is present, because the source voltage is represented by its peak value. To find the value of load resistance at which maximum power transfer occurs, equation (106) is differentiated with respect to load resistance R_L, and the derivative is equated to zero. Then the value of load resistance R_L at maximum power transfer can be obtained.

From equation (106), the derivative obtained is shown by equation (107). The formula used for obtaining the derivative is shown below equation (108). On setting the derivative to zero, we obtain the value of load resistance R_L as shown by equation (108). When the load resistance equals the magnitude of source impedance, the power transferred to the load is maximum. That the power transferred is the maximum and not the minimum can be verified by evaluating the second derivative with respect to load resistanceR_L and finding it to be negative. Equation (108) states the condition for maximum power transfer when the load is purely resistive.

When the load impedance is complex, the condition for maximum power transfer is obtained as follows.

From Fig. 32b, the load current can be obtained as shown by equation (109). Then power delivered to the load is obtained as shown by equation (1110). First let us consider the case when the load resistance R_L is fixed and the load reactance X_L is variable.

When the load resistance R_L is fixed and the load reactance X_L is variable, maximum power transfer occurs when equation (111) is satisfied. In this case, the maximum power transferred is obtained from equation (112). If the source impedance has inductive reactance, then the load impedance should contain capacitive reactance. In this case, suffix L is a pointer to load reactance. It is not a pointer to inductive reactance.

If both the load resistance R_L and the load reactance X_L are variable, one of the conditions for maximum power transfer is specified by equation (113). Then the power delivered to the load is obtained as shown by equation (114). When load resistance R_L is variable, equation (114) is differentiated with respect to load resistance R_L. The derivative obtained is shown next.

The derivative obtained is expressed by equation (115). As shown by equation (116), the derivative is zero, when the load resistance equals the source resistance. The maximum power transfer to load occurs when equation (116) is satisfied. In this case, the maximum power transferred to the load is obtained from equation (117).

The condition for maximum power transfer is specified by equation (118), when both the load resistance R_L and the load reactanceX_L are variable . That is, the load impedance is the conjugate of the source impedance and the power factor of the circuit is unity. Since the source impedance normally has inductive reactance, the load impedance should contain capacitive reactance. In this case, suffix L is a pointer to load reactance. It is not a pointer to inductive reactance.

WORKED EXAMPLE 2:

An example is presented now to illustrate the application of maximum power transfer theorem.

For the circuit in Fig. 33, find the value of load resistance at which maximum power transfer occurs and find the maximum power transferred to the load. The values of components used are specified below.

The solution is as follows:

At first, the circuit containing the source, the resistors, R₁ and R₂ and the capacitance should be replaced by its Thevenin's equivalent circuit. Then the results of maximum power transfer theorem can be applied. In short, the procedure is as follows. Get Thevenin's Equivalent Circuit. Then load resistance R_L equals the magnitude of Thevenin's impedance.

From the circuit in Fig. 33, the open-circuit voltage is the voltage across the capacitor, as shown by equation (120). Next we find the Thevenin's impedance by replacing source voltage V_S by a short-circuit. Then the Thevenin's impedance is obtained as shown by equation (121). The load resistance equals the magnitude of Thevenin's impedance and its value is shown by equation (122).

The Thevenin's equivalent circuit is shown by the circuit in Fig. 34. Note that Thevenin's equivalent circuit does not contain the load resistance. Some worked examples are presented next, to illustrate the use of network theorems and the use of mesh and nodal analysis.

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reciprocity theorem.

In many electrical network it is found that if positions ofvoltage source and ammeter are interchanged, the reading of ammeter remains same. It is not clear to you. Let’s explain in details. Suppose avoltage source is connected to a passive network and an ammeter is connected to other part of the network to indicate the response. Now any one interchanges the positions of ammeter and voltage source that means he or she connects the voltage source at the part of the network where the ammeter was connected and connects ammeter to that part of the network where the voltage source was connected. The response of the ammeter means current through the ammeter would be same in both cases. This is where the property of reciprocity comes in circuit. The particular circuit which has this reciprocal property is called reciprocal circuit. This type of circuit perfectly obeys reciprocity theorem.

The voltage source and the ammeter used in this theorem must be ideal. That means the internal resistance of both voltage source and ammeter must be zero. The reciprocal circuit may be a simple or complex network. But every complex reciprocal passive network can be simplified to a simple network. As per reciprocity theorem in a linear passive network, supply voltage V and output current I are mutually transferable. The ratio of V and I is called the transfer resistance. The theorem can easily be understood by this following example

Reciprocity Theorem

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Tellegen theorem

This theorem has been introduced in the year of 1952 by Dutch Electrical Engineer Bernard D.H. Tellegen. This is very useful theorem in network analysis. According to Tellegen theorem the summation of instantaneous powers for the n number of branches in an electrical network is zero. Are you confused ? Let’s explain. Suppose n number of branches in an electrical network have i1, i2, i3, ………….inrespective instantaneous currents through them. These currents satisfyKirchhoff current law. Again, suppose these branches have instantaneous voltages across them are v1, v2, v3, ……….. vn respectively. If these voltages across these elements satisfy Kirchhoff Voltage law then,

Where vk is the instantaneous voltage across the kth branch and ik is the instantaneous current flowing through this branch. Tellegen theorem is applicable mainly general class of lumped networks consists of linear, non-linear, active, passive, time variant and time variant elements. This theorem can easily be explained by the following example.

In the network shown, arbitrary reference directions have been selected for all of the branch currents, and the corresponding branch voltages have been indicated, with positive reference direction at the tail of the current arrow. For this network, we will assume a set of branch voltages satisfy the Kirchhoff voltage law and a set of branch current satisfy Kirchhoff current law at each node. We will then show that these arbitrary assumed voltage and currents satisfy the equation

and it is the condition of Tellegen theorem,

In the network shown in the figure, let v1, v2 and v3 be 7, 2 and 3 volts respectively. Applying Kirchhoff voltage law around loop ABCDEA. We see that v4 = 2 volt is required. Around loop CDFC, v5 is required to be 3 volt and around loop DFED, v6 is required to be 2. We next apply Kirchhoff current law successively to nodes B, C and D.

At node B let ii = 5 A, then it is required that i2 = − 5 A. At node C let i3 = 3 A and then i5 is required to be − 8. At node D assume i4 to be 4 then i6 is required to be − 9. Carrying out the operation of equation,

we get, 7 X 5 + 2 X ( − 5) + 3 X 3 + 2 X 4 + 3 X ( − 8) + 2 X ( − 9) = 0
Hence Tellegen theorem is verified.

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