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Unformatted text preview: 13. Lecture 13
13.1 Electric generators and alternate current
We discussed before that a cable moving in a magnetic ﬁeld acts as a battery since
the magnetic ﬁeld creates a force that moves the charges along the cable. This is the
principle of the electric generator. It is more eﬃcient to move the cable in circles so
the motion is repetitive. In ﬁgs.73, 74, 75 we have a simple generator (see ﬁg.76 for
an actual example). A rectangular loop moves inside a magnetic ﬁeld an a voltage is
generated between terminals (a) and (b). We can ﬁnd the voltage from Faraday’s law
or equivalent from the Lorentz force. Figure 73: Schematic AC generator. A cable in the shape of a loop rotates inside a magnetic
ﬁeld. View in perspective. The magnetic ﬂux going through the loop is given by
Φ = AB  cos φ (13.1)
where A is the area of the loop, B  is the modulus of the magnetic ﬁeld (created
for example by a permanent magnet) and φ is the angle between the normal to the
loop and the magnetic ﬁeld. If the loop is horizontal the ﬂux is maximum and if it is
vertical the ﬂux is zero. If the loop rotates with angular velocity ω we have φ = ω t.
The ﬂux changes in time and, according to Faraday’s law a voltage is appears between
the terminals:
∆Φ
∆ cos ω t
Va − Vb =
= AB 
(13.2)
∆t
∆t
The only complication is that we need to compute the variation in time of cos ω t.
This can be done by using a mechanical analog. Indeed, if the position of an object
is given by
x = cos ω t
(13.3) – 87 – Figure 74: Schematic AC generator. A cable in the shape of a loop rotates inside a magnetic
ﬁeld. Front view. Figure 75: Schematic AC generator. The variation in ﬂux creates a potential between
terminals (a) and (b). The polarity alternates as seen in this ﬁgure. The symbols at the end
of the conductor indicate if the current goes into the page
or out of the page . then its velocity is given by
vx = ∆x
∆ cos ω t
=
∆t
∆t (13.4) We know how to do this, we just need to consider a particle moving in circles as in
ﬁg.77. If it rotates with angular velocity
omega, the position is given by
x = R cos ω t (13.5) y = R sin ω t (13.6) – 88 – Figure 76: Demo. Actual AC generator. Cranking the handle creates enough power to light
up a small light bulb. The modulus of the velocity is given by the fact that it goes around the circle in time
π
T = 2ω . The length of the circle is = 2π R so the velocity is
  =
v 2π R
= ωR
T (13.7) The direction of the velocity is as in the ﬁgure. Projecting over the x and y axis we
get for the velocity:
vx = −ω R sin ω t
vy = ω R cos ω t (13.8)
(13.9) We then derived the following very important formulas
∆ cos ω t
= −ω sin ω t
∆t
∆ sin ω t
= ω cos ω t
∆t – 89 – (13.10)
(13.11) Figure 77: For a particle rotating in circle sit is easy to compute the velocity. This gives us
same useful formulas that we can apply in other situations. The notation is not very precise but what it means is that is something moves as cos ω t
then its velocity is −ω sin ω t and similarly for the sine. This mechanical analog allows
us then to compute the voltage as
Va − Vb = −AB ω sin ω t (13.12) It is clearly seen form the nature of the motion that the relative polarity of (a) and
(b) alternates. For that reason this is called an AC generator. AC stands for alternate
current. If connected to a resistor the current circulates in one direction and then the
other, that is why it is is called alternate current. The household current alternates 60
times a second. This is called the frequency and is measured in Hertz:
1Hz = 1 1
s (13.13) The household current therefore has a frequency of 60Hz. It should be noticed that the
function
V = V0 sin ω t
(13.14) – 90 – π
repeats itself when we shift t → t + 2ω (since sine is a function that repeats itself when
the argument is shifted by 2π ). therefore the period and frequency of the current are
deﬁned as:
2π
1
ω
T=
, f= =
(13.15)
ω
T
2π 13.2 AC resistor circuit
An AC generator connected to a resistor dissipates power. Notice that power is dissipated no matter the direction of the current. According to Ohm’s law we have
∆V = IR (13.16) ∆V = V0 sin ω t (13.17) 1
V0 sin ω t
R (13.18) For the generator we have
therefore
I=
the power dissipated is 12 2
V sin ω t
(13.19)
R0
It changes in time but we can compute the average power dissipated. To average over
one cycle we notice that, if instead of a sine we had a cosine, the power dissipated
would be the same. But we have (denoting average with parenthesis):
P = I ∆V = sin2 ω t + cos2 ω t = 1
We get then
P =
where we deﬁned ⇒ V02
Vrms
=
2R
R 1
Vrms = √ V0
2 sin2 ω t = 1
2 (13.20) (13.21)
(13.22) which is known as root mean square voltage (and V0 is known as peak voltage). – 91 – ...
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This note was uploaded on 12/07/2011 for the course PHY 219 taught by Professor Na during the Fall '11 term at Purdue UniversityWest Lafayette.
 Fall '11
 NA
 Charge, Current, Force

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