Example 4.3 A circuit has a 2 - resistor, a 2 mH inductor and a 0.15 ¹F
capacitor connected in series with a power supply that generates a square wave
emf " (t) with a 1 V amplitude and a period of 0.3 ms. What is the current in
the circuit?
Figure 4.7: A series LRC circuit with a periodic applied EMF
The circuit is shown in Figure 4.7. Kirchho¤zs rules applied to the circuit
give the di¤erential equation
L
where I =
dQ
.
dt
dI
Q
+ RI +
= " (t)
dt
C
Write the equation in standard form:
d2 Q
R dQ
Q
"
+
+
=
2
dt
L dt
LC
L
The coe¢cients are:
2® ´
and
!20 ´
R
2=
= 1 £ 10 3 s¡1
L
2 mH
1
1
1
=
= £ 1010 s¡2
LC
(2 mH) (0:15 ¹F)
3
Then:
10 5
!0 = p rad/s = 5:77 £ 104 rad/s
3
We have already found the series for the square wave emf " (equation ??)
0
1
µ
¶
+1
X
1
1
i
2n¼t
A
" (t) = " 0 @ +
exp i
2
¼
n
¿
n= ¡1; n o d d
1
where ¿ is the period of 0.3 ms and E0 = 1 V. Next write the solution as a
Fourier series:
µ
¶
1
X
2n¼t
Q (t) =
qn exp i
¿
n=¡1
(Note: it is much easier to use the series in exponential form here, because
the equation has both ¤rst and second derivatives. When we di¤erentiate
the series, each term in the equation contains simple multiples of the original
exponential terms. In contrast, the odd order derivatives mix the sines and
d
cosines : dx
sin x = cos x. Instead of one equation for each coe¢cient cn ; we
would have two equations to solve simultaneously for the coe¢cients of the sines
and cosines.)
Now substitute the series for E (t) and Q (t) into the di¤erential equation:
µ
¶2
1
1
1
X
X
X
2n¼
2n¼
¡
qn e i2n¼ t=¿ + 2®
i
qn ei2n¼t=¿ + !20
qn e i2n¼ t=¿
¿
¿
n=¡1
n=¡1
n=¡1
0
1
+1
X
E0 @ 1
1
i i2n¼t=¿ A
=
+
e
L 2
¼ n=¡ 1; n o dd n
Next we make use of the
of the exponentials by multiplying the
¡ orthogonality
¢
t
whole equation by exp ¡i 2m¼
and
integrating
over one period. Only the
¿
terms with n = m survive. Since this is true for any integer m; we can equate
the coe¢cients of each exponential separately.
The constant (n = 0) term gives
! 20 q0 =
The other terms are given by:q
à µ
!
¶2
2n¼
4®in¼
2
qn ¡
+
+ !0
¿
¿
qn
E0
E0C
) q0 =
2L
2
=
=
=
=
=
E0 i
¼L n
0
if
if
n is odd
n is even
E0 ¿ 2 i
1
³
´
¼L n ¡ (2n¼)2 + 4®¿ in¼ + ! 2 ¿ 2
0
³
´
2
2 2
E0 ¿ 2 i ¡ (2n¼) ¡ 4®¿ in¼ + ! 0 ¿
h
i2
¼L n 2 2
2
2
! 0 ¿ ¡ (2n¼)
+ (4®¿ n¼)
h
i
2
2 2
E0 ¿ 2 4®¿ n¼ + !0 ¿ ¡ (2n¼ ) i
h
i2
n¼L 2 2
! 0 ¿ ¡ (2n¼)2 + (4®¿ n¼) 2
h
i
2
2 2
E0 ¿ 2 4®¿ ¼ + ! 0 ¿ ¡ (2n¼) i=n
h
i2
¼L
2
2
! 20 ¿ 2 ¡ (2n¼)
+ (4®¿ n¼)
2
n odd
Notice that the real part of qn is even in n while the imaginary part is odd.
This is exactly what we expect if the resulting series is to be real. The real
parts combine to give cosines while the imaginary parts combine to give sines:
h
i
2
2 2
µ
¶
1
2
4®¿
¼
+
!
¿
¡
(2n¼)
i=n
X
0
E0 C
E0 ¿
2n¼t
Q (t) =
+
exp i
h
i2
2
¼L
¿
2
2 ¡ (2n¼ )2
n=¡ 1; 6=0; o d d ! 2
¿
+
(4®¿
n¼)
0
h
i
2
2n¼ t
1
2n¼t
2 2
1
E0 C
2E0 ¿ 2 X 4®¿ ¼ cos ¿ ¡ n ! 0 ¿ ¡ (2n¼) sin ¿
=
+
h
i2
2
¼L n=1
2
2
! 20 ¿ 2 ¡ (2n¼)
+ (4®¿ n¼)
Finally we can di¤erentiate to get I: Then:
4E0 ¿
I (t) = ¡
L
1
X
n=1; o d d
h
i
2
t
2 2
t
4n®¿ ¼ sin 2n¼
cos 2n¼
¿ + ! 0 ¿ ¡ (2n¼)
¿
h
i2
! 20 ¿ 2 ¡ (2n¼)2 + (4®¿ n¼)2
The constant in front of the sum is:
4E0 ¿
4 (1 V) (0:1 ms)
V ¢s
=
= 0:2
= 0:2 A
L
(2 mH)
H
Thus:
I (t) = ¡(0:2 A)
1
X
n=1, o d d
h
i
t
2 ¿ 2 ¡ (2n¼)2 cos 2n¼t
4n®¿ ¼ sin 2n¼
+
!
0
¿
¿
h
i2
2
2
!20 ¿ 2 ¡ (2n¼)
+ (4®¿ n¼)
Notice that if the natural frequency ! 0 of the circuit times the period ¿ of
the emf is very close to 2n¼ for some integer n; then the current will be very
large: there
at that frequency. In our example, ! 0 ¿ =¼ =
³
´ ¡ is a resonance
¢
5
10
¡3
p s¡1
0:3
£
10
s
=¼
=
5:513 4 rad, which is close to 6=2£3. Thw solu3
tion is shown in Figure 4.8. The current is dominated by the n = 3 term, which
has three times the frequency of the square wave emf.
3
0.004
I(A)
0.002
0
0.05
0.1
t 0.15
(ms)
0.2
0.25
0.3
-0.002
-0.004
The heavy solid line is the complete solution. The green dotted line is the
n = 1 term, red dashed n = 3; and blue-green dot-dash is the n ¡ 5 term. The
emf is shown as a thin line for reference. The vertical scale does not apply to
this term.
4