Transformer is just wire coiled around metal
Magnetic field is generated
Secondary Voltage is
V2 = (N2/N1) V1
Secondary Current is
I2 = (N1/N2) I1
But Power in = Power out
– negligible power lost in transformer
Works only for AC, not DC
• My laptop computer requires about 20 Volts AC, which comes
from an adapter (transformer) that is plugged into the wall
• What is the approximate ratio of the number of turns on this
transformer, given 120 VAC from wall socket? Which coil has
more turns, the primary or the secondary?
Current Induction between Coils
Electrical Power Distribution
Household AC Power
• Resisters: Electrical Friction, energy released as heat, but not
• I = V/R
• Capacitors: Store Electrical Energy
• Electric companies bill us for Energy: E = P * t measured in
units of kilowatt-hours
• D - cell battery stores about 27 watt-hours
• Car Battery stores 120 watt-hours
Electricity is a Medium for Transporting Energy
Where does the energy come from?
Is energy “lost” in the transmission wires?
What’s the goal, in terms of energy transfer?
Putting Electricity to Work
• Recall the power consumed by an electrical device is given by
the product of the current through it times the voltage drop
P = VI
• Many ways to get the same useful work done, i.e., same power
output from electrical device
Segment of an Electrical Power Transmission Cable
• Recall Power dissipated in a resistor is P = I2R
• How can we minimize power dissipated in the cable?
– Minimize R
• Short cables with large cross sections
• Use high conductivity materials (silver is good!)
• Economic considerations limit the cross section and materials
• Distribution requirements establish needed lengths
– Minimize I
• In order to do this, while keeping power delivered to the household
appliances (P = VI) the same, must raise the voltage difference
between the 2 transmission lines
• Which as a bigger impact, halving I or R?
Power Dissipated in an Electricity Distribution System
Estimate resistance of power lines: say 0.001 Ohms per meter, times 200
km = 0.001 W/m 2105 m = 20 Ohms
With one light bulb on, we can figure out the current it draws using P = VI so
I = P/V = 120 Watts/12 Volts = 10 Amps
Power in transmission line is P = I2R = 102 20 = 2,000 Watts!!
“Efficiency” is e = 120 Watts/2120 Watts = 0.6%!!!
What could we change in order to do better?
Answer: Must reduce either resistance or the current
1. Reduce resistance in the power lines
– Already we’re using pretty hefty copper lines, not very cost-effective to
do anything else (superconductors?).
2. Raising the voltage in the system reduces current!
– Repeating the above calculation with 12,000 Volts delivered to the house
I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving
P = I2R = (0.01)220 = 2010-4 Watts, so
P = 0.002 Watts of power dissipated in transmission line
Efficiency in this case is e = 120 Watts/120.002 = 99.998%
Need a Way to Convert!
• We need a way to transform from a high voltage electrical
distribution system to a low voltage electricity within a
• So, use a transformer!
High voltage in
Low voltage out...
A way to provide high efficiency, safe low voltage:
step-up to 500,000 V
back to 5,000 V
step-down to 120 V
High Voltage Transmission Lines
Low Voltage to Consumers
Compared to the huge force that
attracts an iron tack to a strong
magnet, the force that the tack
exerts on the magnet is:
a) relatively small
b) equally huge
The pair of forces between the tack
and the magnet comprises a single
interaction and both are equal in
magnitude and opposite in directionNewton's Third Law.
When current flows in the wire that is
placed in the magnetic field shown,
the wire is forced upward. If the wire
is made to form a loop as shown
below, the loop will tend to:
a) rotate clockwise
b) rotate counter-clockwise
c) remain at rest
The left side is forced up while the right side is
forced down as shown. If you make the loop
rotate against a spring and attach a pointer to
it, you have a simple electric meter. At
maximum, it can only make a half turn.
But if you make the current change direction
(alternate) at every half turn, it will rotate
continuously as long as the alternating current
persists. Then you have a motor.
What happens to the
reading on the
Galvanometer when the
switch in circuit 1 is
a) first closed
b) kept closed
c) opened again?
When the switch is first closed, a current is
established in coil 1 and creates a magnetic field
which extends to coil 2. This build-up of field in
coil 2 induces current which is registered in the
Galvanometer. The current is brief, however,
because once the field is stabilized and no
further charge takes place, no current is induced
and the Galvanometer reads zero current. When
the switch is opened, the current ceases in coil 1
and the magnetic field in the coil and the part
that extends to coil 2 collapses. This change
induces a pulse of current in the opposite
direction which is registered on the