Transients Problems

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    Circuits Transient Problems 1 M H Miller

    TRANSIENTS

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    Problem 7.30The problem is to determine io(t) for t 0 in the circuit drawn below, right. The first step would seem to be

    to determine the initial state of the circuit. But sometimes a moments thought might be useful. Forexample the circuit geometry is not significant for the analysis; it is the topology that is significant. This issomething to keep in mind. We can redraw the circuit diagram as is shown to the right,

    In the new drawing it is clear that the two capacitors in parallel are equivalent to a single 200F capacitor,and since this is in series with the 200F capacitor the net equivalent to all three capacitors is a single100F capacitor.

    Now let us calculate the initial state of the circuit, i.e., the state at t= 0- just before the switch is closed. Anymethod of circuit analysis may be used; the branch voltages and currents do not , of course, depend on themanner in which they are calculated. The problem assumes the circuit has come to a steady state before theswitch is closed, so that the result of the differentiation operation d/dt on any variable is 0. There istherefore no current through the capacitor(s) and so they are effectively open-circuit for the initial statecalculation. The switch branch also is open-circuit. Both are removed from the circuit diagram (see below)to simplify the drawing.

    Different methods of calculating io(t=0) are illustrated next; any one will provide all the neededinformation..

    Method #1: Convert the source branches to Norton equivalents, i.e., a current source in parallel with aresistor

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    Method #2: Superposition

    Method #3: Node-to-datum analysis

    Method #4: Mesh analysis

    Method #5: Apply Thevenin theorem to circuit 'seen' by the 6K.

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    Now that the initial conditions are known we close the switch. The complete solution consists of thegeneral (transient) solution to which is added theparticular solution. Since the sources are DC theparticular solution simply is a constant, say K. Thegeneral solution is that which satisfies the circuitequation with all source strengths set to zero; thecircuit diagram is shown to the right with the tripletof capacitors replaced by the one equivalent capacitorfor simplicity. The circuit can be simplified further,

    but this will be done 'on the fly'. The particularsolution has the general form est, and we use thisknowledge to simplify use of the capacitor volt-ampere relation. For example replace all the resistorsby a single equivalent 2/3 K and write a nodal equation

    e (0.5K + 100s) =0

    from which we find

    and s =-15. Hence

    where the particular solution has been added to provide the complete solution. We have the initial condition

    io(t=0) = 1/6 ma. In addition we can ascertain the steady-state value of io(t), i.e. as t-> . This involves

    analysis of the circuit with the sources in place but with the capacitors effectively open-circuit. It is left as

    an exercise to verify (using any analysis method) that the current is 1/9 ma as t-> .

    It follows that (with current units of ma) K= 1/9, and for the initial current to be 1/6 A = 1/18. Hence

    The other circuit currents and voltage follow directly, e.g., e = 6io(t). The voltage at the node common to

    the three capacitors in the original circuit is half of this, etc.

    Problem 7.321) Observe that the circuit topology is different before and after the switch opens. The circuit is to beanalyzed for t 0, i.e., after the switch opens.

    2) The circuit behavior depends on the initial energy stored in the circuit, i.e., at t = 0. That initial energy isprovided by the history of the circuit prior to t = 0.

    3) The connection between the circuit state immediately before and immediately after the switch is openedis provided by the inductor. As discussed the inductor current cannot change instantaneously, i.e.,

    iL(t=0) = iL(t=0+). Hence we may start by computing iL(t=0-).

    4) For this computation we suppose the circuit had been assembledfar enough in the past for all transients to become negligible. (All thetransients decay exponentially; far enough into the past means about

    5 time constants or so since e-5 < 0.01.) There is no further changeof current or voltage with time. This means zero voltage drop acrossan inductor, since diL/dt = 0. The circuit is redrawn to the right for

    clarity. The inductor is replaced by a short-circuit (zero voltage drop),and it is the current through this short-circuit that is to be calculated.Note the switch is closed at this point. By inspection note that

    iL(t=0-) = 1A. After all 6||3 = 2, 2 in series with 2 = 4, and

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    12v across 4 is 3A. (even though a 4 resistor bridges the short-circuit it has no effect on thecalculation; Ohm's law requires zero current through this resistor and it is effectively an open-circuit.) The3A source current divides between the 6 and 3, with 1A passing through the short-circuit into the 6.

    5) At this point we have calculated the desired current, and we turn to the

    circuit, as it exists for t 0. In this circuit we have determined that iL(t=0+)

    = 1A. That current divides between the resistors such that the current

    through the 6 is 4/(4+6+3) = 4/13, and hence we have vo(0+) = 24/13 v.The general solution (the particular solution here is zero there is no fixed

    source and the voltage must eventually decay to zero as the initial energystored in the magnetic field of the inductor is dissipated in the resistors.

    There are several ways to solve for the timeconstant, e.g., combine the resistors. However toillustrate the convenience of knowing in advance that the solution has the

    exponential form est we write a loop equation. The voltage drop across theresistors is readily expressed in terms of the current by Ohm's law. Theimpedance equivalent to the inductor in parallel with the 4 resistor is (4)(s/2)/(4+s/2)). This follows the general form for combining two resistors in parallel,although we cannot reduce the equivalent impedance to a simple number.

    Still I I {3 + (4)(s/2)/( 4+s/2)).+ 6} = 0

    and to avoid the nonphysical solution I=0 we set the factor in brackets to zero. This gives a value for s of -72/13. Hence

    Vo(t) = Ae-72t/13

    and since vo(0+) = 24/13

    Vo(t) = (24/13)e-72t/13

    Problem 7.52While the basic procedure to solve this problem is nodifferent than for the preceding problem it may beconsidered 'tricky' if you substitute a vague intuition orfeeling for KVL, KCL, and the v-a relations.1) The connection bridging the circuit topology before theswitch is closed to the different circuit topology after theswitch is closed is the inductor current; it cannot changeinstantaneously. Hence we calculate the current in theinductor just before the switch close.

    2) Assuming (as usual) the circuit transients have died down (exponentially) by the time we are prepared toclose the switch the voltage across the inductor is zero. Then we may note from KVL, for example, that

    12 = -(2+4)ia + 2ia, and so ia = -3A. Then the current through the inductor is iL(0-) = 2ia/4 = -1.5A.

    3) Now close the switch. We then applysuperposition, i.e., recognize that the completesolution consists of a general (source-free) partialsolution, and a particular (source-dependent) partialsolution. We know that the general solution has the

    form Aest, and the particular solution for fixedsources is a constant. Consider the general solutionfirst.

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    4) The general solution is for the source-free circuit, so turn off thesources, i.e., set the independent source amplitudes to zero as illustrated to theleft. Note that this short-circuits the 2 resistor, so that the resistor iseffectively an open-circuit (Ohm's law). Then one might observe that thevoltage across the 4 resistor is 2ia, so Ohm's law requires ia = 2ia/4, and thisrequires ia=0. Then a loop equation provides 2si +4i=0, or s = -2.

    Therefore the general solution is Ae-2t, where the constant of integration

    remains to be evaluated for a specific variable (which will be vo(t) )..5) To compute the particular solution first restore the sources; see figure above. Then use the condition

    that the transient dies down as t-> , i.e., no further changes with time. Hence the voltage across theinductor is zero (but note this is a different circuit than that used to calculate the initial condition. This time

    we observe -4 = -4ia + 2ia (this is as t-> ). Note that the 12v source does not enter into this observation,

    neither does the inductor branch. This is simply an application of KVL. Hence ia(t-> ) = 2, and

    vo(t> ) is 4v.

    6) The complete solution (for t0) is vo(t)= Ae-2t + K, where the integration constants have values as

    required to meet the boundary conditions at t=0 ( vo(0)= -4iL(0) =-6v), and t-> (vo(t-> )= -4 Thus

    vo(t) = -10e-2t + 4.

    7) Just for fun solve using Thevenin's theorem.

    The Thevenin voltage is the open-circuit voltage across the terminals. As before we may note that2ia = 4ia -4, and ia= 2A. Hence VT = 4v. The short-circuit current is 2ia/4 = ia/2, and even in this

    calculation ia=2A. Hence RT =4. Replace the inductor, observe that R/L = 2 and apply the boundary

    conditions as before.

    Problem 7.531) Observe that the circuit topology is different before andafter the switch opens. The circuit is to be analyzed for t 0,i.e., after the switch opens.

    2) The circuit behavior depends on the initial energy stored inthe circuit, i.e., at t = 0. That initial energy is provided by thehistory of the circuit prior to t = 0.

    3) The connection between the circuit state immediatelybefore and immediately after the switch is opened is providedby the inductor. As discussed the inductor current cannot change

    Instantaneously, i.e., iL(t=0-) = iL(t=0+). Hence we may start by computing iL(t=0-).

    4) For this computation we suppose the circuit had been assembled far enough in the past for all transientsto become negligible. (All the transients decay exponentially; far enough into the past means about 5 time

    constants or so since e-5 < 0.01.) There is no further change of current or voltage with time. This meanszero voltage drop across an inductor, since diL/dt = 0. The circuit is redrawn below for clarity. The

    inductor is replaced by a short-circuit (zero voltage drop), and it is the current through this short-circuit that

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    is to be calculated. Note the switch is closed at this point. By inspection note that iL(t=0-) = 1A. After all

    6||3 = 2, 2 in series with 2 = 4, and 12v across 4 is 3A (even though a 4 resistor bridges theshort-circuit it has no effect on the calculation; Ohm's law requires zero current through this resistor and itis effectively an open-circuit.) The 3A source current divides between the 6 and 3, with 1A passingthrough the short-circuit into the 6.

    5) At this point we have calculated the desired current, and we turn to the

    circuit, as it exists for t 0. In this circuit we have determined that iL(t=0+) =

    1A. That current divides between the resistors such that the current throughthe 6 is 4/(4+6+3) = 4/13, and hence we have vo(0+) = 24/13 v. Thegeneral solution (the particular solution here is zero there is no fixed sourceand the voltage must eventually decay to zero as the initial energy stored in themagnetic field of the inductor is dissipated in the resistors.

    There are several ways to solve for the time constant,e.g., combine the resistors. However to illustrate the

    convenience of knowing in advance that the solution has the exponential form est

    we write a loop equation. The voltage drop across the resistors is readilyexpressed in terms of the current by Ohm's law. The impedance equivalent to theinductor in parallel with the 4 resistor is (4)(s/2)/( 4+s/2)). This follows thegeneral form for combining two resistors in parallel, although we cannot reducethe equivalent impedance to a simple number.

    StillI{ 3 + (4)(s/2)/( 4+s/2)).+ 6} = 0

    and to avoid the nonphysical solution I=0 we set the factor in {} to zero. This gives a value for s of -72/13Hence

    Vo(t) = Ae-72t/13

    and since vo(0+) = 24/13

    Vo(t) = (24/13)e-72t/13

    Problem 7.53Assuming the circuit is in a steady-state condition justbefore the switch is closed the inductors are effectivelyshort-circuit. There is then no current through either the6 or 12 resistors, since they are short-circuit, and soio(t=0) = 0. More importantly (since it provides the

    connection between the circuit state just before and justafter the switch closes) the current through the inductors(combined) is 3A. Note: Redraw the circuit with theinductors replaced by short-circuits and use, say,superposition.) After the switch is closed, and after the

    transients die down (i.e., t-> ), the inductors again areeffectively short-circuit. The inductor current then approaches 10/3A. Actually it is easiest (probably) justto recognize that the inductors are in parallel (replace by an equivalent 6/5H) and determine the Theveninequivalent of the circuit 'seen' by the inductor. Or just follow the (arrow) step by step reduction illustrated.Start by converting the 12v & 4 into a Norton equivalent.; combine the two current sources as indicated.Then replace the 6A current source and the 12 resistor by a Thevenin equivalent, and combine theresistors to reduce the circuit to a voltage source, a resistor, and an inductor as shown. The solution for theinductor current then follows directly. For io(t) use the resistive current divider expression.

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    Problem 8.25With the switch in the position shown a steady-statecurrent of 4A flows in the inductor; the capacitor is

    uncharged. At t=0 the switch position changes toform an RLC loop. Anticipating a solution of the

    form est (this is a source-free loop) write a loopequation

    and factor the quadratic to (s+1)(s+8)=0.

    Then express the current as

    i(t) = Ae-t + Be-8t

    Since the initial current is 4A, A + B =4. Note that i(t) is the current through the capacitor, and the voltage

    across the capacitor is determined by integrating :vc(t) = (8/3)(-Ae-t - (B/8)e-8t). The initial voltage is zero,

    so that -A -B/8 = 0. Solve to obtain

    i(t) = (-4e-t + 32e-8t)/7

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    Problem 8.27The initial charge (voltage) on the capacitor is zero. The initialcurrent in the inductor also is zero, and hence (KCL) the initialcapacitor current is zero. After the switch is thrown the circuit

    approaches (t->) a steady-state capacitor voltage of 8 volts. Thetransient solution has the form est, and writing a node equationobtain

    From the roots of the quadratic, and noting that the particular solution is a constant, write

    v(t) = Ae-5t + Be-t + K

    From the condition as t-> determine K = 8. At t=0 we have v(t=0) = A + B +8 =0. Also at t=0 we havedv/dt = 0 (zero current condition), or -5A -B = 0. Solve to obtain

    v(t) = 2e-5t -10e-t + 8

    Problem/Comment:

    Continually using initial conditions and conditions as t-> tends to leave an erroneous impression thatthere is nothing else. This example is intended to dispel that impression. In this example we suppose thatby means not shown the capacitor is charged to 5 volts and connected at t=0 to the series combination of a2 and 3 resistors.. The switch shown remains closed until t=1second, at which time it opens. Thecircuit may be analyzed in two parts, 0 t 1, and t 1. For 0 t 1 the circuit is simply a series

    combination of a 1F capacitor and a 2 resistor, and it is easy to determine i(t) = 2.5e-2t, and v(t) = 5e-2t.

    For t 1the form of the solution also is not difficult. The switch isopen, and the circuit becomes equivalent to a series combination of a 1F

    resistor and a 5 resistor. Hence i(t) = Ae-5t, v(t) = Be-5t, where theconstants of integration A and b are to be determined. What isneeded is some information which relates the circuit state just before theswitch opens to the state just after the switch opens. There isnothing really new herethe voltage across the capacitor cannot

    change instantaneously, and so v(t=1-) = v(t=1+). Now v(t=1-) = 5e-2

    and v(=1+) = Be-5.

    . Hence v(t) = 5e-2

    e-5(t-1)

    for t

    1, and from Ohm's law we have i(t) = e-2

    e-5t

    for t

    1. It is convenient to use t-1 in the exponent both as a reminder that this part of the solution is valid for t 1 and because t-1=0 at this starting point. In this way it is easier to see that that the solution is the expectedexponential decay starting from t' = t - 1 = 0.

    Incidentally note that i(t=1-) i(t=1+), i.e., the current changes abruptly. Since there is no inductor in thiscircuit there is no problem.