ALTERNATING CURRENT

1. Alternating current (AC) is current that flows first in one direction along a wire, and then
in the reverse direction. The most common AC is sinuisoidal in which the current (and
voltage) follow a sine function, as in the graph below. The average value is zero because
the current flows for the same time in one direction as in the other.
However, the square of any AC wave is always positive. Thus its average is not zero. As
we shall see, upon averaging the square we get half the square of the peak value.
2
The calculation follows: if there is a sine wave of amplitude (height) equal to one and
frequency , ( 2 / , T=time period), then the squared amplitude is sin and its
average is:
ω ω = πT ωt
2 2
0 0 0
sin 1 sin 1 (1 cos2 ) 1 1 1
2 2 2
Taking the square root gives the root mean square value as 1/ 2 of the maximum value.
Of course, it does not matter whe
t Tdt t Tdt t T dt
T T T
ω
ω ω
−
< >≡ ∫ = ∫ = ∫ =
2 2
ther this is of the voltage or current:
0.707 , and 0.707 .
2 2 2 2
Exactly the same results are obtained for cosine waves. This is what one expects since
m m m m
rms m rms m
I I I I
ε ε
ε = = = ε = = =
the difference between sine and cosine is only that one starts earlier than the other.
m ε
t
ε
sin
0
m
average
ε ε ωt
ε ε
=
≡ < > =
rms ε
ε
t
0 ε
2
0 ε
2. AC is generated by a coil rotating in a magnetic field. We know from Faraday's Law,
, that a changing magnetic flux gives rise to an emf. Imagine a magnetic
field and a coil of area
B d
dt
A
ε Φ
= −
, rotating with frequency so that the flux through the coil at
any instant of time is cos . Then the induced emf is sin .
3. AC is particular useful because make it possib
B BA t BA t
transformers
ω
Φ = ω ε= ω ω
le to step up or step down
voltages. The basic transformer consists of two coils - the primary and secondary - both
wrapped around a core (typically iron) that enhances the magnetic field.
Suppose that the flux in the core is and that its rate of change is . Call the number
of turns in the primary and secondary and respectively. Then, from Faraday's law,
the prim
p s
t
N N
∂Φ
Φ
∂
p s
p
s p
s
ary emf is and the secondary emf is . The ratio is,
. So, the secondary emf is . If is less than one, then it is called
a step-down transformer beca
p s
p s s
s p p
N N
t t
N N N
N N N
ε ε
ε
ε ε
ε
∂Φ ∂Φ
= − = −
∂ ∂
= =
use the secondary voltage is less than the input voltage.
Else, it is a step-up transformer. Both types are used.
4. If this is a lossless transformer (and good transformers are 99% lossless), then
p s
the input
power must equal the output power, . From above, this shows that the ratio
of currents is .
5. Whatever the shape or size of a current carrying loop, the magnetic fl
p s
p
s p
s
I I
N
I I
N
ε =ε
=
ux that passes through
it is proportional to the current, Φ ∝I. The inductance L (called the self-inductance if
R
L
2
there is only one coil) is the constant of proportionality in the relation . The unit of
inductance is called Henry, 1 Henry= 1 Tesla metre /Ampere. Note that inductance, like
capacita
Φ = LI
nce, is purely geometrical and depends only upon the shape and sizes of wires.
It does not depend on the current.
6. Let us calculate the inductance of a long coil wound with n turns per unit lengt
( )( )
0
2
0
2
0
h. As
calculated earlier, the field is , and the flux passing through turns of the
coil of length and area is . From the definition, we find:
B
B
B B nI N
l A N nlBA nIAl
L N nlIA
I I
μ
μ
μ
μ
=
Φ = =
Φ
= = =
( )
2
0
2 2
7 2
(inductance of long solenoid).
Example: Find the inductance of a coil with 3500 turns, length 10 cm, and radius 5cm.
0.05
Solution: 4 10 3500 1.21 1.21 .
0.10
n lA
L T m m T m H
A m A
π
π − ⋅ ⋅
= ⋅ ⋅ ⋅ = =
7. When the current changes through an inductor, by Faraday's
Law it induces an emf equal to ( ) .
Let us use this to calculate the current through the circuit
shown he
B d dLI LdI
dt dt dt
Φ
− =− =−
2
re, where a resistor is present. Then, by using
Kirchoff's Law, or . In
words: the power expended by the battery equals energy
dissipated in the resistor + work
IR L dI I I R LI dI
dt dt
ε= + ε = +
done on inductor.
8. Before we go on to the AC case, suppose that we suddenly connect a battery to an
inductor so that the voltage suddenly increases from zero to across the circuit. Then,
LdI IR has
dt
ε
+ =ε ( ) ( / )
/
solution: 1 where . You can easily see that
this true using 1 . The solution shows that the current increases from zero to
a maximum of , i.e. that given by Ohm
t
t
I t e L
R R
dI e
dt R
R
τ
τ
ε
τ
ε
τ
ε
−
−
= − =
=
[ ] [ ]
[ ]
's Law. We need to pay a little attention to the
"time constant" ,which is the time after which the constant approaches 63% of its final
value. Units: henry volt.second / ampere volt
ohm ohm
L
R
τ
τ = = = = second second.
ampere.ohm
⎛ ⎞
⎜ ⎟ =
⎝ ⎠
L
C
Note that when t= , then I (1e1) (1 0.37) 0.63
R R R
ε ε ε
τ = − − = − =
9. When we pass current through an inductor a changing magnetic field is produced. This,
by Faraday's Law, induces an emf across the coil. So work has to be done to force the
current through.
0 0
How much work? The power, or rate of doing work, is emf current.
Let be the work done in passing current . Then, . Let us
integrate . Then,
B
B
B
U
B B
U I dU L dI I LI dI
dt dt dt
dU LI dI dU LIdI
×
=⎛⎜ ⎞⎟ =
⎝ ⎠
= ∫ = 2
2
1 . This is an important
2
result. It tells us that an inductor carrying current requires work 1 . By conservation
2
of energy, this is also the energy stored in the inductor.
I
B U LI
L I LI
∫ ⇒ =
2
Compare this result with the result
1 for a capacitor. Notice that and .
2
10. Let us use the result derived earlier for the inductance of a solenoid and the magnetic field
in it,
E U CV C L V I
L
= ↔ ↔
2 2 2 2
0 0 0 0
2
0
and . Putting this into 1 gives 1( )( / ) .
2 2
Divide the energy by the volume of the solenoid, . This directly gives the
volume 2
energy density (energy per
B B
B
n lA B nI U LI U n lA B n
U B
μ μ μ μ
μ
= = = =
=
unit volume) contained in a magnetic field.
11.
We know that energy can be stored in a capacitor as well
as in an inductor. What happens when we connect them up
together and put some
Electromagnetic oscillations in an LC circuit.
charge on the capacitor? As it
discharges, it creates a current that transfers energy to the
inductor. The total energy remains constant, of course.
2
2
2 2 2
2
2 2
This means that the sum 1 1 is constant. Differentiate this:
2 2
0 1 1 . Since , and , 1 0
2 2
To make this look nicer, put
B E
U U U LI q
C
dU d LI q I dq dI d q d q q
dt dt C dt dt dt dt LC
ω
= + = +
⎛ ⎞
= = ⎜ + ⎟ = = ∴ + =
⎝ ⎠
2
2 2
2
1 0. We have seen this equation
many times earlier. The solution is: cos . The important result here is that the m
d q q
LC dt
q q t
ω
ω
≡ ⇒ + =
=
charge, current, and voltage will oscillate with frequency 1 . This oscillation will
go on for forever if there is no resistance in the circuit.
LC
ω =