Transcript 4-04 IOT - The 9 Core Technologies

DRILL
April 30, 2009
The power loss in a transmission cable is given by the
equation P= I2R.
A. If the original current is doubled, while the resistance
remains constant, what will happen to the power loss?
B. If the original current is cut in half, while the resistance
remains constant, what will happen to the power loss?
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The power loss in a transmission cable is given by the
equation P= I2R.
A. If the original current is doubled, while the resistance
remains constant, what will happen to the power loss?
When current is doubled, the new current is 2 I.
In the power equation, we must square current.
Therefore, we square 2 I, which becomes 2Ix2I or 4I2.
The new power is P=4I2R, which is FOUR times the
original power loss.
Let’s look at a simple example ……
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Let’s look at a simple example ……
Given: The resistance of a power line is 10 ohms, and the
current is 10 amperes.
Now, we double the current, while
keeping the same resistance.
P = I2R
P = (10 A)2(10 ohms)
P = 100 A2 (10 ohms)
P = 1000 Watts
P= I2R
P= (20 A)2(10 ohms)
P= 400 A2(10 ohms)
P= 4000 Watts
Conclusion: If the current is doubled, while the
resistance is held constant, the power loss will be
quadrupled.
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The power loss in a transmission cable is given by the
equation P= I2R.
B. If the original current is cut in half, while the resistance
remains constant, what will happen to the power loss?
When current is halved, the new current is 0.5 I.
In the power equation, we must square current.
Therefore, we square 0.5 I, which becomes 0.5Ix0.5I or
0.25I2.
The new power is P=.25I2R, which is ONE-FOURTH times the
original power loss.
Let’s look at a simple example ……
IOT
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POLY ENGINEERING
Let’s look at a simple example ……
Given: The resistance of a power line is 10 ohms, and the
current is 10 amperes.
Now, we halve the current, while
keeping the same resistance.
P = I2R
P = (10 A)2(10 ohms)
P = 100 A2 (10 ohms)
P = 1000 Watts
P= I2R
P= (5 A)2(10 ohms)
P= 25 A2(10 ohms)
P= 250 Watts
Conclusion: If the current is halved, while the
resistance is held constant, the power loss will be
one-fourth of the original value.
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Let’s review what we learned yesterday about the
sources of electricity:
There are six ways to generate (create) electricity:
1. Friction
2. Chemical
3. Light
4. Heat
5. Pressure
6. Magnetism
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Let’s review what we learned yesterday about the
resistance of a wire:
All metal wires have resistance.
The resistance of a wire is based on 4 factors:
1. Material
2. Length
3. Cross-sectional area
4. Temperature
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1. Material:
Some materials conduct electricity better than other materials.
For example, gold conducts electricity better than silver, which
conducts electricity better than copper.
However, gold and silver are extremely expensive compared to
copper, so they are only used in very special cases.
Material technology (one of the 9 Core technologies) is very
important in Electricity and Electronics.
Copper wire
Gold bars
Silver bars
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2. Length:
A long wire will have more resistance
than a shorter wire of the same material,
cross-sectional area, and temperature.
When two or more resistors are connected in series,
their combined resistance is greater than any of the
individual resistors.
RT= R1 + R2 + … + Rn
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Imagine that a long wire is simply two shorter wires
connected end-to-end in series. Therefore, a long wire has
a resistance which is the sum of the resistances of the
shorter pieces.
One of the reasons for the development of modern microcircuits is to reduce the resistance, thus causing the power
requirements to be minimized.
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3. Cross-sectional Area:
A thick wire will have less resistance
than a thin wire of the same material,
length, and temperature.
When two or more resistors are connected in parallel,
their combined resistance is less than any of the
individual resistors.
1 1 1
1
RT= R1 + R2 + … + Rn
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Imagine that a thick wire is simply two or more wires
bundled together side-by-side in parallel. Therefore,
a thick wire has a resistance which is less than the
resistances of the thinner pieces.
You have probably noticed that the wire in a light bulb
is very thin, thus giving it a high resistance which causes
it to get hot and glow. The same thing occurs in an
electric toaster, where the wire becomes red hot.
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4. Temperature:
In general, as the temperature of a wire increases, the
resistance increases. The opposite is true as well. As the
temperature of a wire decreases, the resistance decreases.
The changes of resistance with temperature can be
explained by looking at the Chemistry involved. When a
metal is heated, the molecules move farther apart and
move faster. Therefore, for electrons to jump from one
atom to another (electrical current) in a heated metal, they
must move farther, and they must hit a 'moving target' (the
next atom). This makes it more difficult for electrons to
move, thus increasing the resistance.
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Conversely, when a metal is cooled, the molecules move
closer together, and become less active. Therefore, for
electrons to jump from one atom to another (electrical
current) in a cooled metal, they do not have to move as far,
and they can easily hit a more stationary target (the next
atom). This makes it easier for electrons to move, thus
lowering the resistance.
Modern engineers and scientists are developing lowtemperature superconductors so that electrical power
requirements can be minimized.
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When two or more resistors are connected in series,
their combined resistance is greater than any of the
individual resistors.
RT= R1 + R2 + … + Rn
Solve the following:
A red-red-red resistor, a brown-black-red resistor,
and a yellow-violet-orange resistor are connected in series.
What is the resistance of this combination?
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RT= R1 + R2 + … + Rn
Solve the following:
A red-red-red resistor, a brown-black-red resistor,
and a yellow-violet-orange resistor are connected in series.
What is the resistance of this combination?
red-red-red
= 2200
brown-black-red
= 1000
yellow-violet-orange = 47000
RT= R1 + R2 + … + Rn
= 2200 + 1000 + 47000
RT = 50200 ohms
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When two or more resistors are connected in parallel,
their combined resistance is less than any of the
individual resistors.
1 1 1
1
RT= R1 + R2 + … + Rn
Solve the following:
An orange-black-red resistor, a red-black-red resistor,
and a blue-black-red resistor are connected in parallel.
What is the resistance of this combination?
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1 1 1
1
RT= R1 + R2 + … + Rn
Solve the following:
An orange-black-red resistor, a red-black-red resistor,
and a blue-black-red resistor are connected in parallel.
What is the resistance of this combination?
orange-black-red = 3000
red-black-red
= 2000
blue-black-red = 6000
1 1 1
1
RT= R1 + R2 + … + Rn
1
1
1
= 3000 + 2000 + 6000
2
3
1
= 6000 + 6000 + 6000
6
RT = 1000 ohms
1
= 6000 = 1000
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How would you find the equivalent resistance of the
following combination?
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HOMEWORK:
Study for a quiz on
Electricity/Electronics tomorrow.
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