TRANSIENTS
Circuits Transient Problems
1
M H Miller
Circuits Transient Problems
2
M H Miller
Problem 7.30
The 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 moment’s thought might be useful. For
example the circuit geometry is not significant for the analysis; it is the topology that is significant. This is
something 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 200µF capacitor,
and since this is in series with the 200µF capacitor the net equivalent to all three capacitors is a single
100µF capacitor.
Now let us calculate the initial state of the circuit, i.e., the state at t= 0- just before the switch is closed. Any
method of circuit analysis may be used; the branch voltages and currents do not , of course, depend on the
manner in which they are calculated. The problem assumes the circuit has come to a steady state before the
switch is closed, so that the result of the differentiation operation d/dt on any variable is 0. There is
therefore no current through the capacitor(s) and so they are effectively open-circuit for the initial state
calculation. 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 needed
information..
Method #1: Convert the source branches to Norton equivalents, i.e., a current source in parallel with a
resistor
Circuits Transient Problems
3
M H Miller
Method #2: Superposition
Method #3: Node-to-datum analysis
Method #4: Mesh analysis
Method #5: Apply Thevenin theorem to circuit 'seen' by the 6KΩ.
Circuits Transient Problems
4
M H Miller
Now that the initial conditions are known we close the switch. The complete solution consists of the
general (transient) solution to which is added the
particular solution. Since the sources are DC the
particular solution simply is a constant, say K. The
general solution is that which satisfies the circuit
equation with all source strengths set to zero; the
circuit diagram is shown to the right with the triplet
of capacitors replaced by the one equivalent capacitor
for simplicity. The circuit can be simplified further,
but this will be done 'on the fly'. The particular
solution has the general form est, and we use this
knowledge to simplify use of the capacitor voltampere relation. For example replace all the resistors
by a single equivalent 2/3 KΩ and write a nodal equation
e (0.5K + 100µs) =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 7.32
1) Observe that the circuit topology is different before and after 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 in the circuit, i.e., at t = 0. That initial energy is
provided 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 opened
is 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 assembled
far enough in the past for all transients to 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 means zero voltage drop across
an 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 = 2Ω, 2Ω in series with 2Ω = 4Ω, and
Circuits Transient Problems
5
M H Miller
12v across 4Ω is 3A. (even though a 4Ω resistor bridges the short-circuit it has no effect on the
calculation; Ohm's law requires zero current through this resistor and it is effectively an open-circuit.) The
3A source current divides between the 6Ω and 3Ω, with 1A passing through the short-circuit into the 6Ω.
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 6Ω 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 energy
stored in the magnetic 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 readily expressed in terms of the current by Ohm's law. The
impedance equivalent to the inductor in parallel with the 4Ω 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 7.52
While the basic procedure to solve this problem is no
different than for the preceding problem it may be
considered 'tricky' if you substitute a vague intuition or
feeling for KVL, KCL, and the v-a relations.
1) The connection bridging the circuit topology before the
switch is closed to the different circuit topology after the
switch is closed is the inductor current; it cannot change
instantaneously. Hence we calculate the current in the
inductor just before the switch close.
2) Assuming (as usual) the circuit transients have died down (exponentially) by the time we are prepared to
close 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 apply
superposition, i.e., recognize that the complete
solution consists of a general (source-free) partial
solution, and a particular (source-dependent) partial
solution. We know that the general solution has the
form Aest, and the particular solution for fixed
sources is a constant. Consider the general solution
first.
Circuits Transient Problems
6
M H Miller
4) The general solution is for the source-free circuit, so turn off the
sources, i.e., set the independent source amplitudes to zero as illustrated
to the
left. Note that this short-circuits the 2Ω resistor, so that the resistor is
effectively an open-circuit (Ohm's law). Then one might observe that the
voltage across the 4Ω resistor is 2ia, so Ohm's law requires ia = 2ia/4, and
this
requires 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 the
inductor 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 t≥0) 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 that
2ia = 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 =4Ω. Replace the inductor, observe that R/L = 2 and apply the boundary
conditions as before.
Problem 7.53
1) Observe that the circuit topology is different before and
after 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 in
the circuit, i.e., at t = 0. That initial energy is provided 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 opened is 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 assembled far enough in the past for all transients
to 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 means
zero 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
Circuits Transient Problems
7
M H Miller
is to be calculated. Note the switch is closed at this point. By inspection note that iL(t=0-) = 1A. After all
6||3 = 2Ω, 2Ω in series with 2Ω = 4Ω, and 12v across 4Ω is 3A (even though a 4Ω resistor bridges the
short-circuit it has no effect on the calculation; Ohm's law requires zero current through this resistor and it
is effectively an open-circuit.) The 3A source current divides between the 6Ω and 3Ω, with 1A passing
through the short-circuit into the 6Ω.
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 6Ω 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 energy stored in the
magnetic 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 readily
expressed in terms of the current by Ohm's law. The impedance equivalent to the
inductor in parallel with the 4Ω 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{ 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 -72/13.
Hence
Vo(t) = Ae-72t/13
and since vo(0+ ) = 24/13
Vo(t) = (24/13)e-72t/13
Problem 7.53
Assuming the circuit is in a steady-state condition just
before the switch is closed the inductors are effectively
short-circuit. There is then no current through either the
6Ω or 12Ω resistors, since they are short-circuit, and so
io(t=0) = 0. More importantly (since it provides the
connection between the circuit state just before and just
after the switch closes) the current through the inductors
(combined) is 3A. Note: Redraw the circuit with the
inductors 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 are
effectively short-circuit. The inductor current then approaches 10/3A. Actually it is easiest (probably) just
to recognize that the inductors are in parallel (replace by an equivalent 6/5H) and determine the Thevenin
equivalent of the circuit 'seen' by the inductor. Or just follow the (arrow) step by step reduction illustrated.
Start by converting the 12v & 4Ω into a Norton equivalent.; combine the two current sources as indicated.
Then replace the 6A current source and the 12Ω resistor by a Thevenin equivalent, and combine the
resistors to reduce the circuit to a voltage source, a resistor, and an inductor as shown. The solution for the
inductor current then follows directly. For io(t) use the resistive current divider expression.
Circuits Transient Problems
8
M H Miller
Problem 8.25
With the switch in the position shown a steady-state
current of 4A flows in the inductor; the capacitor is
uncharged. At t=0 the switch position changes to
form an RLC loop. Anticipating a solution of the
form est (this is a source-free loop) write a loop
equation
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
Circuits Transient Problems
9
M H Miller
Problem 8.27
The initial charge (voltage) on the capacitor is zero. The initial
current in the inductor also is zero, and hence (KCL) the initial
capacitor current is zero. After the switch is thrown the circuit
approaches (t->∞) a steady-state capacitor voltage of 8 volts. The
transient solution has the form est, and writing a node equation
obtain
From the roots of the quadratic, and noting that the particular solution is a constant, write
v(t) = Ae-5t + B e-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 have
dv/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 that
there is nothing else. This example is intended to dispel that impression. In this example we suppose that
by means not shown the capacitor is charged to 5 volts and connected at t=0 to the series combination of a
2Ω and 3Ω resistors.. The switch shown remains closed until t=1second, at which time it opens. The
circuit 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 2Ω 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 is
open, and the circuit becomes equivalent to a series combination of a
1F
resistor and a 5Ω resistor. Hence i(t) = Ae-5t, v(t) = Be-5t, where
the
constants of integration A and b are to be determined. What is
needed is some information which relates the circuit state just before
the
switch opens to the state just after the switch opens. There is
nothing really new here—the 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-2e-5(t-1) for t ≥ 1, and from Ohm's law we have i(t) = e-2e-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 expected
exponential 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 this
circuit there is no problem.
Circuits Transient Problems
10
M H Miller