AC Generators
Alternating Current and
Transformers

Can be thought of as an electric motor in
reverse.

Converts mechanical energy to electrical
energy
Rotating a coil changes the flux and,
therefore, produces an emf.
emf.
Flux is maximum when coil is
perpendicular to B and varies as a cosine
curve



Peak Voltage and Current

ε0 = peak voltage (EMF)




Root Mean Square (RMS)

Occurs when flux = 0
Current will vary with voltage
I0 = ε0/R
Power will also vary since P = εI, but will
always be positive while I and ε vary from
positive to negative
Slip ring rather than split ring comutator




Averages of power, voltage and current
would be helpful, but would be zero for
voltage and current
So we square, average and take square
root to get a value
Irms = I0/√2
εrms = ε0/√2
Pavg = εrms Irms
1
BackBack-emf


A rotating motor armature is also
changing it’s flux and, therefore, inducing
an emf.
emf.
Lenz’s law says that the emf induced will
oppose the flux that created it.


Induced current will oppose current turning
the motor
So a rotating motor has more resistance
than one that isn’t rotating.

Why lights flicker when motors cut on and
why motors burn out when not turning.


A method to change
voltage for distribution
Two coils of wire
connected by an iron
core


The
Transformer
A laminated core reduces
power loss due to eddy
currents
AC in first (primary)
creates a changing
magnetic field, which
creates a changing flux in
2nd (secondary) coil, thus
inducing a current.
The Transformer (Cont.)
Vp/Vs = Np/Ns
 Ip/Is = Ns/Np

Practice Problems:
1.
2.
An AC circuit provides a peak voltage of 40 V
and a peak current of 2.5 A. Calculate the rms
voltage, rms current and the average power of
the circuit.
A transformer for home use of a portable radio
reduces 120120-V ac to 9.09.0-V ac. The secondary
coil contains 30 turns and the radio draws 400
mA of current. Calculate: (a) the number of
turns in the primary coil; (b) the current in the
primary coil; and (c) the power transformed.
2