SCHOOL OF MATHEMATICS
MATHEMATICS FOR PART I ENGINEERING
Self Study Course
MODULE 26
APPLICATIONS TO ELECTRICAL CIRCUITS
Module Topics
1. Complex numbers and alternating currents
2. Complex impedance
3. Differential equations for RLC circuits
4. Forced oscillations and resonance
5. Complex solutions of differential equations
6. Phasors
A:
Work Scheme based on JAMES (THIRD EDITION)
1. The first topic uses results from module 22 on complex numbers. Before starting on this topic you should
make sure that you understand the relevant topics. If you need to revise this material look at the module
again or study section 3.3 of James.
2. Turn to p.190 and study section 3.5 on alternating currents in electrical circuits. Work through Example
13.27.
The key results of this section are as follows:
When an alternating current with frequency ω/2π flows in a circuit it is given by
¡
¢
i = I sin ωt = Im Iejωt .
Similarly we have
¡
¢
I cos ωt = Re Iejωt
The corresponding voltage depends on the resistance, capacitance and inductance of the circuit. However
these devices not only determine the magnitude of the voltage but also change the phase of the voltage
relative to the current. We may therefore calculate the voltage by multiplying Iejωt by a complex number Z
and then looking at the imaginary part. Note using a complex number deals with the change of phase since
Im(ejφ ejωt ) = Im(ej(ωt+φ) ) = sin(ωt + φ),
and
Re(ejφ ejωt ) = Re(ej(ωt+φ) ) = cos(ωt + φ).
The complex number Z is called the complex impedance of the device. The formula for Z for various
devices is given as follows:
(
R
for a resistor
j
Z = − ωC
for a capacitor
jωL for an inductor
For a simple RLC circuit (with these devices in series)
R
L
Fig. 1.
–1–
C
we simply take the sum of these terms so that
Z = R + jωL −
j
ωC
Using this result the actual voltage is given by
¡
¢
v = Im IZejωt = I|Z| sin(ωt + φ)
where |Z| is the impedance of the circuit and the phase φ is given by the argument of Z.
Note: it is often convenient to introduce the reactance S given by
1
ωC
S = ωL −
So that
Z = R + jS
One example of the use of complex impedance is given in Example 3.27. Here is another example:
Example A: Calculate the complex impedance of the element of the circuit shown below with a resistor
of R = 20 Ω and a capacitor of C = 212 µF, when an alternating current of frequency 50 Hz flows. Use this
to find the impedance and the phase.
Fig. 2.
The complex impedance is the sum of the individual impedance’s. Thus
Z =R−
j
ωC
Here R = 20 Ω, ω = 2π × 50 rad s−1 and C = 212 µF, so that 1/Cω = 15
Z = 20 − 15j
Hence the impedance is |Z| =
p
(15)2 + (20)2 = 25Ω and the phase is φ = tan−1 (−15/20) = −0.64 rad.
***Do Exercise 3.5.1 0n p.192***
3. Turn to p.675 and study section 10.2.4 on Simple Electrical Circuits.
We again summarise the key results:
The relationship between voltage V , current i and resistance R for a pure resistor is given by
V = iR
Similarly for a pure capacitor of capacitance C with charge q we have
V =
q
C
Finally for a pure inductor of inductance L we have
V =L
di
dt
–2–
If we now consider the circuit shown below
A
B
Fig. 3.
then when the switch is in contact with B then by Kirchhoff ’s voltage law the total potential difference
around the circuit must be zero so we have
L
di
q
+ Ri +
=0
dt
C
(1)
The principle of conservation of charge tells us that the current flowing is equal to the rate of change of
charge so that
dq
i=
(2)
dt
Substituting for i (and its time derivative) in (1) using (2) gives
L
dq
1
d2 q
+R + q =0
2
dt
dt
C
(3)
This is a differential equation for the charge q. However it is more convenient to work with the current i and
differentiating (3) with respect to time gives
L
d2 i
di
1
+R + i=0
2
dt
dt C
(4)
Alternatively we can use V = q/C in equation (3) to obtain an equation for the voltage
LC
dV
d2 V
+ RC
+V =0
dt2
dt
(5)
We have therefore shown that a simple analysis of a RLC circuit gives rise to a second order (homogeneous)
linear differential equation with constant coefficients.
Example B: A series circuit (as in Fig. 3) contains an inductor for which L = 1 H, a resistor for which
R = 1 kΩ, and a capacitor for which C = 6.25 µF. The capacitor holds a charge of 1.5 × 10−3 C, at time
t = 0 a switch is moved from A to B and the capacitor discharges through the circuit. Find q and i as
functions of t.
The differential equation to be solved is
dq
1
d2 q
+ 103
+
q=0
2
dt
dt
6.25 × 10−6
This is a differential equation with constant coefficients so we look at the auxiliary equation
m2 + 1, 000m + 160, 000 = 0
–3–
which has roots m1 = −200 and m2 = −800. Hence
q = Ae−200t + Be−800t
(∗)
The initial charge is q0 = 1.5 × 10−3 so that
A + B = 1.5 × 10−3
(i)
To find the other initial condition we differentiate (*) to obtain an equation for the current
i = −200Ae−200t − 800Be−800t
The initial current is zero so that we have
−200A − 800B = 0
(ii)
Solving (i) and (ii) for A and B we get
A = 2 × 10−3 ,
Hence
B = −5 × 10−4
q = 2 × 10−3 e−200t − 5 × 10−4 e−800t
and
i = −0.4e−200t + 0.4e−800t
***Do Exercise A: A series circuit consists of an inductor for which L = 0.02 H and a capacitor for
which C = 8 × 10−6 F. The capacitor holds a charge of 16 × 10−4 C and at time t = 0 a switch is closed
allowing the capacitor to discharge through the circuit. Find the charge and the current in the circuit at
time t.
***Do Exercise B: For the problem given above also find the voltage drop across the inductor and across
the capacitor and show that the sum of these two voltages is zero in accordance with Kirchhoff’s law.
4. Linear differential equations with constant coefficients are covered in module 13 (Differential Equations
III) in order to be able to study oscillations in RLC circuits you need to make sure that you understand this
material. If you need to revise this topic look again at module 13 or else turn to p.723 and read section
10.9 of J. Note that you will also need the material from section 10.9.3 which deals with the inhomogeneous
equations with constant coefficients.
***Do Exercise 53 on p.727 and Exercise 61(c) and 61(f ) on p.734***
5. Read section 10.10.4 on p.745 which deals with oscillations in electrical circuits. The key result is the
following. Consider the RLC circuit shown below
R
C
Vi
L
Fig. 4.
–4–
Vo
Suppose that a voltage Vi (t) is applied across the input terminals then the voltage Vo (t) across the output
terminals is given by
d 2 Vo
dVo
LC 2 + RC
+ Vo = Vi (t)
(6)
dt
dt
which is second order inhomogeneous differential equation for constant coefficients. In many cases the input
signal is of the form Vi = V cos ωt (where V is a constant) so that the equation reduces to
LC
d 2 Vo
dVo
+ RC
+ Vo = V cos ωt
dt2
dt
(7)
To solve this we need to find a complementary function and a particular integral. The complementary
function is just a solution of the homogeneous equation (5) and in section 10.10.2 of J it is shown that in
this case (since R and C are positive) this is always given by a decaying function of time. For this reason
the complementary function is called the transient solution. In order to find the particular integral we
look for a solution of the form
P cos ωt + Q sin ωt
where P and Q are constants which depend upon ω. This can be written in the alternative form
AV
cos(ωt + δ)
C
where A(ω) is a constant that depends upon ω and δ gives the phase relative to the input. This is a sinusoidal
signal which oscillates with amplitude AV /C. Note that unlike the transient solution this solution does not
decay so that this solution gives the long term behaviour of the solution. For this reason it is called the
steady-state response.
A direct calculation shows that
A=
1
1
=
ω|Z|
ω(R2 + S 2 )1/2
where R is the resistance, S is the reactance and |Z| is the impedance, and the phase is given by
µ
δ = tan−1
R
S
¶
Therefore if a sinusoidal voltage is applied to the input terminals the voltage produced is also sinusoidal but
with an amplitude (and phase) which depends on the frequency of the input. Such a device is called a filter.
Note that the corresponding current is obtained by differentiating the voltage and multiplying by C so that
the modulus of the output current is the modulus of the input voltage divided by the impedance.
We now give an example of how equation (6) may be used to calculate the voltage in an alternating RLC
circuit.
Example C: An RLC-circuit as shown in Fig. 4 consists of a resistor with R = 11 Ω, an inductor with
L = 0.1 H, and a capacitor with C = 10−2 F. An input voltage Vi (t) = 1000 cos 400t is applied. Find the
resultant steady state voltage. Find the amplitude of this voltage and the phase lag compared to the the
voltage that produced it.
The differential equation for Vo (t) is given by
10−3
dVo
d2 Vo
+ 0.11
+ Vo = 1000 cos 400t
dt2
dt
To find the steady state solution we look for a particular solution of the form
Vo = P cos 400t + Q sin 400t
–5–
Differentiating this and substituting into the equation we get
(44Q − 159P ) cos 400t − (44P + 159Q) sin 400t = 1000 cos 400t
Hence
−159P + 44Q = 1000
44P + 159Q = 0
which has solution
P = −5.84,
Q = 1.62
so that Vo = −5.84 cos 400t + 1.62 sin 400t. It is more useful to write this in terms of modulus and phase as
Vo = 6.06 cos(400t + 0.27)
So that the amplitude is 6.06 and the voltage is 0.27 in advance of the voltage that produced it (since δ is
negative).
***Do Exercise C: An RLC-circuit as shown in Fig. 4 consists of a resistor with R = 100 Ω, an inductor
with L = 0.1 H, and a capacitor with C = 10−5 F. An input voltage Vi (t) = 100 cos 500t is applied. Find
the resultant steady state voltage and current and the amplitude of this current.
6. Complex Solutions The following sections deal with complex solutions and phasors which are not
described in J. As we have seen the equation for the output current is
L
d2 q
dq
1
+ R + q = Vi (t)
2
dt
dt
C
(8)
where Vi (t) is the input voltage. Two cases of interest are
Vi (t) = V cos(ωt)
(a)
Vi (t) = V sin(ωt)
(b)
and
If we solve equation (8) with a complex source given by
Vi (t) = V (cos(ωt) + j sin(ωt)) = V ejωt
(c)
then we obtain a complex solution q(t). However the real part of this is gives the solution corresponding
to source (a), while the imaginary part of this solution gives the solution corresponding to source (b). It
turns out that it is easier to solve (8) for the complex source (c) and then take the real part rather than
solve for the real source (a).
For the case of the complex source we want to solve
L
dq
1
d2 q
+ R + q = V ejωt
dt2
dt
C
(9)
The steady-state solution is given by the particular solution and we find this by looking for a solution of the
form
q(t) = KV ejωt
where K is a complex constant to be determined. Differentiating this gives
dq
= jωKV ejωt ,
dt
and
–6–
d2 q
= −ω 2 KV ejωt
dt2
Substituting into (9) gives
(−ω 2 L + jRω +
and hence using S = ωL −
1
)KV ejωt = V ejωt
C
1
, we must have
ωC
jω(R + jS)K = 1
So that
K=
1
jωZ
where Z is the complex impedance. Thus the corresponding complex current is given by
ic (t) =
dq
V
= ejωt
dt
Z
To find the actual (real) current we write the complex impedance in polar form as
Z = |Z|ejθ
where
|Z| =
p
µ
R2 + S 2 ,
Then the complex current produced is
ic (t) =
and
θ = tan−1
S
R
¶
V j(ωt−θ)
e
|Z|
So that the real current produced is
i(t) = Re(ic (t)) =
V
cos(ωt − θ)
|Z|
To summarise an input voltage of Re(V ejωt ) produces an output current of Re(
V jωt
e ).
Z
7. Phasors
If we are given an alternating current i(t) = Io cos(ωt + φ) in a circuit which is oscillating with angular
frequency ω, then all we need to know to find the current is the quantity Io ejφ and the angular frequency
ω. This information is provided by the current phasor
˜
I(jω)
= Io ejφ
since the actual current is given by
˜ jωt ) = Re(Io ej(ωt+φ) )
i(t) = Re(Ie
In the same way the voltage v(t) = V0 cos(ωt + θ) is determined by the voltage phasor
Ṽ (jω) = V0 ejθ
In the example of the RLC-circuit considered in section 6 we see that
Ṽ (jω)
˜
I(jω)
=
Z(jω)
–7–
or
˜
Ṽ (jω) = Z(jω)I(jω)
˜
where Ṽ (jω) is the phasor for the input voltage, I(jω)
is the phasor for the output current and Z(jω) is the
complex impedance of the circuit. Note by taking the modulus of this expression we immediately see that
the amplitude of the input voltage is the impedance times the amplitude of the output current. By looking
at the argument we also see that the phase of the output current is arg Z behind that of the input voltage.
Example D: Use the phasor method to calculate the amplitude and phase of the output current for the
circuit considered in Example C. Use this to compute the modulus of the output voltage.
In this example R = 11, ω = 400 and C = 10−2 . Hence
S = ωL −
1
= 159/4
ωC
and hence the complex impedance is Z = 14 (44 + 159j). Thus
|Z| =
p
R2 + S 2 = 41.243,
and
arg Z = tan−1 (159/44) = 1.3rad
Hence the amplitude of the current is 1000/41.243 = 24.25 and the phase is 1.3 rad. Hence the output
current is
i(t) = 24.25 cos(400t − 1.3)
To find the voltage we integrate and divide by C. Hence the modulus of the output voltage is given by
24.25
= 6.06
400 × 0.01
in agreement with the previous answer. Note the phase difference of the voltage is π/2 − 1.3 = 0.27 again
in agreement with the previous calculation (the extra π/2 coming from the fact that a cos becomes a sin on
integration).
–8–
Specimen Test 26
1.
An alternating current i = 2 sin 100t flows through a circuit consisting of a resistor with resistance
R = 50 Ω, and an inductor with inductance L = 0.5 H, as shown below
R
L
(i) Calculate the complex impedance Z of the circuit.
(ii) Calculate the amplitude V of the voltage.
(iii) Show that the voltage may be written in the form v(t) = V sin(100t + φ) where V is the amplitude and
φ is the phase which you should calculate.
2.
The complex impedance of two circuits in parallel with complex impedances Z1 and Z2 is given by
1
1
1
=
+
Z
Z1
Z2
If Z1 = 1 + 2j and Z2 = 2 − j calculate Z in the form Z = a + bj and use this to calculate the (real)
impedance of the circuit.
3.
A simple RLC-circuit shown below contains and inductor with L = 1 H, a resistor for which R = 103 Ω,
and a capacitor for which C = 4 × 10−6 F. With the switch in position A, the battery maintains a charge
of 5 × 10−4 C in the capacitor. At time t = 0 the switch is moved to B and the capacitor discharges
through the circuit.
A
B
(i) Find the differential equation for the charge q(t) and find the general solution.
(ii) Find the expression for the corresponding general solution for the current i(t).
(iii) Hence find the solutions q(t) and i(t) which satisfy the initial conditions for this circuit at t = 0.
–9–
4.
The RLC-circuit shown below
R
C
Vi
Vo
L
has R = 8 Ω, L = 0.5 H, C = 0.1 F and input voltage given by Vi = 100 sin 2t.
(i) Find the differential equation satisfied by the output current io (t).
(ii) Find a particular solution for the differential equation and hence find the steady state output current
and calculate its amplitude.
(iii) Confirm your answer for the amplitude of the output current by using the method of phasors, and
calculate the phase of the current relative to the input voltage.
– 10 –