Note 06
AC Circuits III: Transient Signals
In any circuit in which energy can be stored, the sudden application or removal of a source of emf causes
a momentary response until a new steady state is reached. These temporary responses of current and
voltage signals are called transients. We consider here transients in RC, RL and RCL circuits.
RC Circuits
Charging a Capacitor
A circuit for the charging and discharging of a
capacitor is drawn in Figure 6-1. We can regard the
capacitor as an ideal capacitor in series with a resistor
R or as a real capacitor with some internal resistance
R. The switch S enables us to completely discharge the
capacitor as an initial condition before charging it.
B
A
S
q(0) = 0 .
From eq[6-1] the initial condition on i(t) at t = 0 (since
q(0) = 0) is
€
R
i
Figure 6-1. Circuit for charging a capacitor. The switch S is
shown in the charging position at some instant t when an
instantaneous DC current i(t) is flowing.
Suppose at time t = 0 S is moved from position A to
position B to begin the charging. Applying KVR
around the loop at some instant of clocktime t when
the capacitor’s instantaneous charge is q(t) and an
instantaneous current i(t) is flowing we have
V−
q(t)
− Ri(t) = 0 .
C
…[6-1]
Substituting i(t) = dq(t)/dt into eq[6-1] and rearranging
we get an equation in q(t):
€
dq(t) q(t) V
+
= .
dt
RC R
di(t) i(t)
+
= 0.
dt
RC
i(0) =
V
= I,
R
…[6-3b]
…[6-2a]
…[6-2b]
Eqs[6-2a and b] are first-order differential equations in
q(t) and i(t), respectively. Eq[6-2a] is an inhomogeneous equation,
eq[6-2b] is homogeneous.
€
q(∞) = Q = VC ,
…[6-4a]
i(∞) = 0 .
…[6-4b]
and
Eqs[6-2]€through [6-4] show that when the capacitor is
charging, the charge starts at zero and rises to the
maximum €
value Q, while the current starts at a maximum I and decreases to zero. Once the capacitor is
fully charged, no more charge is transferred and the
current flow is zero.
The variations in q(t) and i(t) between the initial and
final states are what are referred to as the transient
signals. Solving eq[6-2a] the charge transient is seen to
be
q(t) = Q 1− e−t / RC .
…[6-5a]
(
)
Solving eq[6-2b] the current transient is seen to be
€
Differentiating eq[6-2a] we get an equation in i(t):
€
…[6-3a]
where I is the maximum current that flows during the
charging process. The final conditions, or long term or
steady state
€ responses, are the solutions when dq(t)/dt
= i(t) = 0 at t = ∞. These are
C
V
Before solving these equations it is useful to state
explicitly what the initial and final conditions are.
Clearly at t = 0 the capacitor is fully discharged, i.e.,
i(t) =
Q −t / RC
e
= Ie−t / RC .
RC
…[6-5b]
Eqs[6-5] are sketched in Figures 6-2. Eq[6-5a] shows
that q(t) rises from zero at t = 0 to the maximum value
Q€
at t = ∞. And it rises exponentially. The quantity τ =
RC (with units of s) is called the time constant. This is
the time required for the capacitor to charge to within
the factor 1/e of Q .
Also subject to change is the current flowing through
the capacitor, as is seen from eq[6-5b]. Beginning from
a maximum current I at t = 0, the current decays away
N6-1
Note 06
exponentially. Here τ is interpreted as a constant of
decay—the time taken for the current to drop to the
factor 1/e of I. Since R and C are in series this current
is common to both elements.
resembles Figure 6-2b multiplied by a constant. As the
capacitor charges, the current decreases, and so does
vR (t). When the capacitor is fully charged at t = ∞, i(∞)
= 0 and vR (∞) = 0.
Discharging a Capacitor
A capacitor’s discharge curve is the horizontal inverse
of its charge curve. Suppose S is in position B long
enough for the capacitor to acquire a maximum
charge consistent with C and V . At time t = 0 S is
moved to position A to begin the discharge. Applying
KVR around the loop at a clocktime t when the
instantaneous charge on the capacitor is q(t) and an
instantaneous current i(t) is flowing,
vC (t) + v R (t) = 0 ,
(a)
or
€
q(t)
dq(t)
+R
= 0.
C
dt
…[6-7a]
Differentiating eq[6-7a] and rearranging we have
€
di(t) i(t)
+
= 0.
dt
RC
…[6-7b]
Thus the solutions of eqs[6-7] are
€
(b)
Figure 6-2. The charge curve (a) and current curve (b) of a
charging capacitor. The circuit constants here are R = 100
Ω , C = 1.0 F, V = 5 volts. The time constant is 100 s
The voltage across the capacitor is also a transient
signal. When the capacitor is charging, this voltage,
given by
vC (t) =
q(t) Q
= (1− e−t / RC ) ,
C
C
= V (1− e−t / RC )
…[6-6]
€
is also seen to change exponentially, in a manner
similar to eq[6-5a]. The graph of eq[6-6] resembles
Figure 6-2a
€ multiplied by a constant.
The voltage across the resistor also changes as the
current changes. This voltage is
v R (t) = i(t)R = IR e−t / RC = Ve−t / RC ,
which you can see resembles eq[6-5b]. Thus v R (t)
€ N6-2
and
Thus
€
i(t) = Ie−t / RC ,
…[6-8a]
qC (t) = Qe−t / RC .
…[6-8b]
vC (t) = Ve−t / RC .
…[6-8c]
€ see that the current, charge and voltage
You can
curves for a discharging capacitor all have the same
shape—the
€ shape of Figure 6-2b.
Thus you can see how an RC circuit responds to an
applied step voltage. (You can understand why this is
called step function analysis.) For a step-up voltage,
vC(t) requires a finite time to rise to the level of the
applied voltage V . The time lag can be seen as a
“shaping” or “distorting” of v C(t) in relation to the
voltage applied. Distortion may not be significant for
a single opening or closing of a switch (like S), but if S
is opened and closed at high speed (simulating a
series of applied voltage pulses) then the existing time
constants may degrade the signal. The information
“carried” in the pulsing signal may no longer be
recognizeable, and therefore lost for practical
purposes.
Time constants are particularly important in digital
electronics where the two states zero and one are
Note 06
represented by different voltage levels. Since every
electronic circuit, such as a memory chip or a
computer bus, has its own inherent capacitance, it is
impossible for the voltage to switch instantaneously
between these levels. To ensure that the voltage has
time to complete the transition before the new level is
read, a digital circuit is driven synchronously by a
clock pulse. The time constant of the circuit sets an
upper limit on the frequency of the clock pulse,
beyond which the voltage levels are not sufficiently
well defined to represent the two states unambiguously. This explains why it is not possible to increase
the speed of a computer indefinitely simply by
increasing the frequency of the clock. 1
We now come to circuits involving the inductor. The
terms charging and discharging do not apply to an inductor since charge does not accumulate on anything
like plates in a capacitor. But we shall see that the
energy delivered an inductor by a changing current
goes into constructing a magnetic field within the inductor. When the voltage is removed the current continues for a short time. This current comes from the
decay of the same magnetic field. Because of induction this build up and decay has its own inertia and is
not instantaneous. Energy is “stored” in the sense that
the formation and decay of the magnetic field takes a
finite time to take place.
RL Circuits
Example Problem 6-1
Charging a Capacitor
Energizing an Inductor
A 5.0 µF capacitor is connected to a source of emf of
50 V through a 1.0 MΩ resistor in order to be charged.
What is the voltage on the capacitor after 1 second?
Solution:
This is a straightforward application of eq[6-6]. The
time constant τ = RC = 106 (Ω)×5.0×1.0×10–6 (F) = 5.0 s.
Thus
vC (1) = 50 1− e−1/ 5 V = 9.06 V .
(
We have seen how the voltage across a real capacitor
connected to a voltage source takes a finite time to rise
to the voltage of the source. A similar effect occurs in
the case of an inductor. To see this consider Figure 6-3
where an inductor is connected to a voltage source via
switch S.
B
)
A
Solution:
The same time constant applies as in Example Problem 6-1, namely 5.0 s. Thus from eq[6-8c] we have
vC (t) = 1 = 50e−t / 5 ,
i
1
t = −5ln  = 19.6 s .
 50 
R
Figure 6-3. An inductor is connected to a voltage source.
When S is moved to position B current through the inductor rises to a maximum, but not instantaneously.
Initially, S is placed in position A to begin the
experiment with zero current flowing through the
inductor. S is moved to position B, say at a time t = 0,
to start current flow. Applying KVR around the loop
at a time t when the instantaneous current is i(t) we
have:
V −L
which, by solving for t yields
€
L
V
Example Problem 6-2
€Discharging a Capacitor
A 5.0 µF capacitor is initially charged to 50 V. A 1.0
MΩ resistor is connected across the capacitor to discharge it. How long does it take for the voltage on the
capacitor to fall to 1 V?
S
di(t)
− i(t)R = 0
dt
di(t) R
V
+ i(t) = .
dt
L
L
so that
€
…[6-9a]
…[6-9b]
The initial condition is
1
You
€ will be studying the charge and discharge of a capacitor in
Experiment 02, “Transients in RC Circuits”.
i(0) = 0 .
€
…[6-10]
N6-3
€
Note 06
Eq[6-9b] has the form of eq[6-2a] with the variable i
instead of q. Thus a solution is
i(t) = I (1− e−Rt / L ) ,
…[6-11]
where I = V/R. Eq[6-11] is sketched in Figure 6-4a.
Beginning at zero the current rises to within 1/e of the
maximum
€ current I in time τ = L/R seconds. Thus τ is
the time constant. Since L and R are in series this
current is common to both elements.
v L (t) = L
di(t)
,
dt
which, by substituting eq[6-11], becomes
v L (t) = Ve−Rt / L .
€
…[6-12]
Eq[6-12] is plotted in Figure 6-4b. At t = 0 when i(0) =
0, vL equals the applied voltage. But when current begins flowing
€ the voltage across the inductor decreases
as di/dt decreases. When di/dt = 0, vL = 0. At this stage
the applied voltage appears entirely across the
resistor. (Of course, we are assuming here that the inductor does not possess an internal resistance from its
wire windings.)
Deenergizing an Inductor
(a)
The curves which describe the situation when the
current through the inductor is decreasing are similar
to the above. We can imagine that at t = 0 switch S has
been in position B long enough to enable the current
to reach maximum. S is then moved to position A.
Applying KVR around the loop at some arbitrary
clocktime t when the instantaneous current is i(t):
L
or
€
The solution is
(b)
Figure 6-4. The current curve (a) and voltage curve (b) for a
series RL circuit. The exponential shape of the curve is due
to induction. Here the circuit constants are R = 100 Ω , L =
0.022 H, V = 5 volts.
The voltage across the resistor is given by
v R (t) = i(t)R = IR(1− e−Rt / L )
= V (1− e−Rt / L ) .
€
Thus vR (t) also resembles Figure 6-4a.
The voltage across the inductor is given by
€
N6-4
di(t)
+ i(t)R = 0 ,
dt
di(t) R
+ i(t) = 0 .
dt
L
i(t) = Ie−Rt / L ,
…[6-13]
which €
is of the form of eq[6-12]. The current continues
to flow in the same direction but dies away exponentially. The
€voltage across the resistor is proportional to
eq[6-13] and therefore also decreases in a similar
fashion.
The voltage across the inductor is
v L (t) = −Ve−Rt / L .
…[6-14]
Thus vL and V are of opposite polarity. At t = 0 when
the applied voltage vanishes and the current begins to
decrease,
€ the induced emf takes on the sign so as to
continue driving current in the same direction. This is
a consequence of Lenz’ law.
Note 06
Transients in RCL Circuits
We have seen in the previous sections that when a DC
source is applied to, or removed from, a capacitor or
inductor the resulting voltages and currents change
over the short term in an exponential way. These
voltage and current signals are called transients.
As the solutions of the differential equations that
describe these signals show, transients eventually give
way to steady state signals (as t tends to infinity).
(With reference to Figures 6-2 and 6-4, the voltages
and currents tend to maximum values or to zero
depending on whether the source is switched ON or
OFF.) The equations contain within themselves both
the transient and steady state solutions. This is a
special case that results from the fact that the external
voltage is a stepfunction; it is either ON or OFF and
doesn’t vary with time. Also, RC and RL circuits are
first order systems in that they contain only one
element capable of storing energy; the equations
describing them are first order differential equations.
We consider one more example of transients, this
time in a circuit containing all three elements: a
capacitor, inductor and resistor in series.
taneous current is i(t) and the charge on the capacitor
is q(t),
q(t)
di(t)
−L
− Ri(t) = 0 .
C
dt
Differentiating this expression and replacing dq(t)/dt =
–i(t) (since i(t) and q(t) are out of phase) we have
€
d 2i(t)
di(t) i(t)
L
+R
+
= 0.
2
dt
dt
C
This equation is second-order, linear, and homogeneous. It is second order since the highest derivative of
i(t)€is the second. It is linear since i(t) appears only to
first power and the coefficients (C , L , and R) do not
depend on time. It is homogeneous since the RHS is
zero. In mathematical parlance the equation’s solution
is called the complementary function.
Before solving the equation it is useful to put down
the initial conditions explicitly. Consistent with the
procedure of operating the switch,
Series RCL Circuit: Standard Method 2
The Standard Method of analyzing a series RCL
circuit (Figure 6-5) consists of the straightforward business of setting up the differential equation and then
of solving it.
A
and
€
q(0) = CV ,
…[6-16a]
i(0) = 0 ,
…[6-16b]
i'(0) =
€
V
.
L
…[6-16c]
We solve the equation in a general way by writing the
solution as
€
i(t) = Ae st ,
…[6-17]
B
S
V
…[6-15]
L
C
i
where A and s are real or complex constants to be
determined. Substituting eq[6-17] into eq[6-15] we get
€
R
s2 LAe st + sRAe st +
Figure 6-5. A series RCL circuit.
A st
e = 0,
C
which is satisfied when
This circuit resembles Figures 6-1 and 6-3 in that
provision exists for setting up the initial conditions.
To charge the capacitor to some initial, maximum,
value Q = CV switch S is placed in position A for a
sufficiently long time. When S is moved to position B,
say at time t = 0, charge Q is released by the capacitor
to flow through L and R. Applying KVR around the
loop at some arbitrary clocktime t when the instan-
€
Ls2 + Rs +
1
= 0.
C
…[6-18]
This equation, quadratic in s, is called the characteristic
equation. The roots of the equation are obtained from
the quadratic
formula:
€
s1,s2 = −
2
There is another method, called poles and zeroes or the method of
impedances, which we shall leave to a course in electronics.
 R 2 1
R
±   −
.
 2L  LC
2L
…[6-19]
N6-5
€
Note 06
Now according to the theory of differential equations
if either of the terms
i1 (t) = A1e s1 t
or
i2 (t) = A2e s2 t ,
where A1 and A 2 are constants, satisfies a linear homogeneous equation, then the sum of the two terms also
€satisfies the equation. The most general solution is
therefore their sum
i(t) = i1 (t) + i2 (t) = A1e s1 t + A2e s2 t . …[6-20]
A1 and A 2 are determined by the initial conditions and
s1 and s2 are set by the circuit constants: C, L and R.
€The values C, L and R are real and positive. But as
we shall see the values of s 1 and s 2 may be real or
complex. The solution is determined by the value of
the discriminant in eq[6-19]:
 R 2 1
D=  −
.
 2L  LC
Case 2: Underdamped Signal
If D < 0 then
 R 2
1
,
  <
 2L  LC
and s1, s 2 are complex conjugates. You should be able
to show that eq[6-20] reduces to
€
We can distinquish three cases depending on the sign
and value of D. We shall see that the corresponding
signals€
are termed overdamped, underdamped and critically damped and have the forms shown in Figure 6-7.
i
solution is called overdamped by studying Figure 6-7.
(It’s assumed for convenience that the initial condition
for all signals is zero current.) In the case of overdamping the current commences at zero, rises
quickly, passes through a maximum and then decays
slowly to zero again. The approach to zero is slower
than the critically damped signal (to be described
below) so the response in this case is dubbed
overdamped. In a physical situation, for example, a
d’Arsonval galvanometer, this kind of movement is
avoided.
(a)
i(t) = Ae−αt sin(ωt + φ ) ,
…[6-22]
where α is a constant to be determined below. This is
a sinusoidal signal of angular frequency ω with a
damped
amplitude. We shall expand on this
€
calculation following our discussion of Case 3. This is
arguably the most commonly-encountered motion of
a harmonic system. 3
Case 3: Critically damped Signal
(c)
If D = 0 then
t
 R 2
1
,
  =
 2L  LC
(b)
Figure 6-7. The three cases: (a) overdamped, (b) underdamped, and (c) critically damped signals. It is assumed
that the initial condition for all signals is zero current.
and s 1, s2 are real, negative and equal. In this case
eq[6-20] reduces to
Case 1: Overdamped Signal
Note the occurrence of t in the second term. The current begins at zero, passes through a maximum, and
then dies
€ away to zero, but faster than the overdamped signal. In fact, the signal decays to zero in the
shortest possible time, hence the term critically damped.
This is the most desireable motion of a d’Arsonval
type meter movement as it means that when a change
occurs the needle of the movement comes to a new
If D > 0 then
 R 2
1
,
  >
 2L  LC
and s1, s2 are real, negative and unequal. The general
solution, eq[6-20], becomes
€
i(t) = A1e−s1 t ± A2e−s2 t .
…[6-21]
You can see why the signal corresponding to this
€
N6-6
€
3
i(t) = A1e st + A2 te st .
…[6-23]
Eq[6-22] in the form of a time dependent current as here or a
time dependent voltage is also described as an impulse response
signal. You will study this case in Experiment 05, “Wave Filters”.
Note 06
equilibrium position in the shortest time, making
possible another measurement without delay.
More on Case 2
We now examine the underdamped signal in more
detail. It is convenient to define quantities α, ω, and ω0
in terms of the circuit constants as follows:
α=
so that
R
2L
ω 02 =
1
LC
ω2 =
1
R2
− 2,
LC 4L
Example Problem 6-3
Applying the Standard Method to a Real RCL Circuit
In the circuit shown in Figure 6-8 the capacitor is first
charged by moving switch S to position A and waiting
for a sufficiently long time. S is then moved to
position B to discharge the capacitor through L and R.
Show that the current that follows for a brief time
thereafter is a damped sinusoid. Find the phase factor,
amplitude and period.
ω 2 = ω 02 − α 2 .
A
€ Then from eq[6-19], the two s-values become
s€
1 = −α + jω
and
S
i(t) = A1e
€
1H
s2 = −α − jω ,
V
where j = √-1. Eq[6-20] then becomes
(−α + jω )t
B
+ A2e
(−α − jω )t
1/17 F
.
…[6-24]
i
2Ω
Figure 6-8. A series RCL circuit.
By applying the Euler relations
€
e ± jωt = cosωt + j sin ωt ,
eq[6-24] simplifies to
i(t)
€ = e−αt [( A1 + A2 ) cosωt + j ( A1 − A2 ) sinωt ] ,
€
Solution:
We can state that the current signal is a damped sinusoid if the roots of the characteristic equation are
complex conjugates. Let us find those roots. From
eqs[6-19],
 R 2 1
R
s1,s2 = −
±   −
 2L  LC
2L
where A1 and A2 are constants, possibly complex.
Since i(t) is a real signal the coefficients of sin ωt and
cosωt must also be reals. Let us call them B 1 and B 2.
Then we can write
i(t) = e−αt [ B1 cosωt + B2 sin ωt ] .
=−
€
= −1± −16 = −1± 4 j
= Ae−αt sin(ωt + φ ) .
€
You should be able to show how this follows (eq[622]). The factor ω is called the natural angular frequency
of oscillations
(with unit s–1). The natural linear freq€
uency is ω/2π (Hz). The factor α is called the damping
constant. Note that α-1 equals the time constant τ as we
defined previously.
This material, though developed for voltage and
current signals in an RCL circuit, also applies to a
mechanical harmonic oscillator. You will see how this
works yourself in Experiment 04, “The d’Arsonval
Galvanometer”.
 2 2
2
1
± 
 −
 2x1 1x 1
2x1
17
Thus €
α = +1 ; ω = 4 .
The roots
€ are, indeed, complex conjugates. From eq[624] the transient current is
€
i(t) = A1e(−1+4 j )t + A2e(−1−4 j )t
or, from the simplified form given by eq[6-22],
€
i(t) = Ae−t sin(4t + φ )
…[6-25]
Having found the solution the problem remains of
€
N6-7
Note 06
determining numerical values for A and φ. We can
find these from the initial conditions. First we find φ.
At t = 0 (just after the switch is closed), i(t) = 0. Thus
€
so that φ = 0. To calculate A we observe that at t = 0,
di(0)
= Ae−t 4 cos(4t + φ ) − Ae−t sin(4t + φ )
dt
i(t) =
= 4A
V −t
e sin(4t) ,
4
and the period of oscillations is given by
= Ae 0 4 cos(0) − Ae 0 sin(0)
€
€
…[6-26]
T=
1 2π 2π
=
=
= 1.57 s .
f
ω
4
Applying
KVR around the loop at t = 0 gives
€
vC (0) + v L (0) = 0
€
V −L
or
€
€
N6-8
di(0)
=0
dt
…[6-27]
and from eqs[6-26] and [6-27], A = V/4.
Thus eq[6-25] can be written more compactly as
i(0) = 0 = Ae 0 sin(0 + φ ) = Asin φ ,
€
di V V
= = =V
dt L 1
therefore
€
Practice Problems